🔬 Model Lab

I'll solve each problem step by step.

---

**Problem 1:**

We have 6 elements: {2, 3, 4, 6, 9, 14}, and need to add 3 more distinct integers to make 9 total.

The median of 9 numbers is the 5th element when arranged in order.

Currently we have 6 numbers. By strategically placing 3 new numbers, we can shift which value becomes the median.

Let me consider what values can be the median (5th position):
- If we add 3 numbers all ≤ 2, the sorted list has our 6 original numbers in positions 4-9, so median is 4
- If we add numbers strategically between and around the original set, we can make the median be: 2, 3, 4, 6, 9, 14, or a new number
- If we add 3 numbers all ≥ 14, the sorted list has our 6 original numbers in positions 1-6, so median is 9

Testing systematically:
- Median = 2: add three numbers < 2 (e.g., -2, -1, 0) ✓
- Median = 3: add two < 2, one between 3 and 4 ✓
- Median = 4: add three < 2, or arranged appropriately ✓
- Median = 6: add one or two strategically ✓
- Median = 9: add three > 14, or arranged appropriately ✓
- Median = 14: add three > 14 ✓
- Median = a new value between existing numbers ✓

The possible medians are: 2, 3, 4, 6, 9, 14, and any new integer we insert. We can get 7 different values.

**ANSWER 1: D**

---

**Problem 2:**

Original list: 2, 6, 7, 7, 28. Sum = 50.

When N is inserted, we have 6 numbers. Let's find the median of 6 numbers (average of 3rd and 4th when sorted).

Sum of all 6 numbers = 50 + N
Mean = (50 + N)/6

We need: Mean = 2 × Median

Let me test by considering where N falls:
- If N ≤ 2: sorted list is N, 2, 6, 7, 7, 28. Median = (6+7)/2 = 6.5. Mean = 2×6.5 = 13, so 50+N = 78, N = 28. Contradiction.
- If 2 < N ≤ 6: sorted list is 2, N, 6, 7, 7, 28. Median = (6+7)/2 = 6.5. Mean = 13, so N = 28. Contradiction.
- If 6 < N ≤ 7: sorted list is 2, 6, N, 7, 7, 28. Median = (N+7)/2. Mean = 2×(N+7)/2 = N+7. So (50+N)/6 = N+7, giving 50+N = 6N+42, so 5N = 8, N = 1.6. Not in range.
- If 7 < N ≤ 28: sorted list is 2, 6, 7, N, 7, 28 or 2, 6, 7, 7, N, 28. Median = (7+N)/2. Mean = 2×(7+N)/2 = 7+N. So (50+N)/6 = 7+N, giving 50+N = 42+6N, so 5N = 8, N = 1.6. Not valid.
- If N > 28: sorted list is 2, 6, 7, 7, 28, N. Median = (7+7)/2 = 7. Mean = 14. So 50+N = 84, N = 34. ✓

**ANSWER 2: E**

---

**Problem 3:**

Four-digit numbers with all even digits: first digit ∈ {2,4,6,8}, others ∈ {0,2,4,6,8}.

Total count: 4 × 5 × 5 × 5 = 500

For divisibility by 4, the last two digits must form a number divisible by 4.

Last two digits from {00, 02, 04, 06, 08, 20, 22, 24, 26, 28, 40, 42, 44, 46, 48, 60, 62, 64, 66, 68, 80, 82, 84, 86, 88}

Divisible by 4: 00, 04, 08, 20, 24, 28, 40, 44, 48, 60, 64, 68, 80, 84, 88 (15 out of 25)

Count divisible by 4: 4 × 5 × 15 = 300

But wait, the second digit is 5 choices, third is which determines divisibility by 4.

Actually: 4 × 5 × (number of even two-digit combos divisible by 4) = 4 × 5 × 15 = 300

Hmm, let me recalculate. First digit: 4 choices. Second digit: 5 choices. Last two digits must be divisible by 4: check all pairs from {0,2,4,6,8}×{0,2,4,6,8} giving 15 valid pairs.

Count = 4 × 5 × 15 = 300. Fraction = 300/500 = 3/5.

**ANSWER 3: D**

---

**Problem 4:**

6.5% tax on $20: 0.065 × 20 = $1.30
6% tax on $20: 0.06 × 20 = $1.20
Difference: $1.30 - $1.20 = $0.10

**ANSWER 4: B**

---

**Problem 5:**

Sum from 1901 to 1993: This is 93 consecutive integers starting at 1901.
Sum = (1901 + 1993)/2 × 93 = 1947 × 93 = 181,071

Sum from 101 to 193: This is 93 consecutive integers starting at 101.
Sum = (101 + 193)/2 × 93 = 147 × 93 = 13,671

Difference = 181,071 - 13,671 = 167,400

**ANSWER 5: A**

---

**Problem 6:**

Pattern: 1 - 2 - 3 + 4 + 5 - 6 - 7 + 8 + 9 - 10 - 11 + ...

Group by 4s: (1-2-3+4) + (5-6-7+8) + (9-10-11+12) + ... + (1993-1994-1995+1996)

Each group of 4: (4k+1) - (4k+2) - (4k+3) + (4k+4) = 4k+1-4k-2-4k-3+4k+4 = 0

From 1 to 1996, we have 1996/4 = 499 complete groups, each summing to 0.

Total = 0

**ANSWER 6: C**

---

**Problem 7:**

Let x = people wearing both sunglasses and caps.

P(sunglasses | cap) = x/35 = 2/5, so x = 14.

P(cap | sunglasses) = 14/50 = 7/25.

**ANSWER 7: B**

---

**Problem 8:**

Three-digit even numbers with digit sum 26.

For an even number, last digit ∈ {0, 2, 4, 6, 8}.

If last digit = 0: first two digits sum to 26. Max is 9+9=18. Impossible.
If last digit = 2: first two digits sum to 24. Only 9+9=18 < 24. Impossible.
If last digit = 4: first two digits sum to 22. Impossible (max 18).
If last digit = 6: first two digits sum to 20. Impossible.
If last digit = 8: first two digits sum to 18. This requires 9+9=18. Number: 998. ✓

Only 1 number.

**ANSWER 8: A**

---

**Problem 9:**

Three-digit numbers abc where a+b+c=25, with a ∈ {1,...,9}, b,c ∈ {0,...,9}.

a+b+c = 25 means b+c = 25-a.

For valid solutions: 0 ≤ b+c ≤ 18, so 7 ≤ a ≤ 9.

- a=7: b+c=18. Solutions: (9,9) → 1 way
- a=8: b+c=17. Solutions: (8,9), (9,8) → 2 ways
- a=9: b+c=16. Solutions: (7,9), (8,8), (9,7) → 3 ways

Total = 1 + 2 + 3 = 6

**ANSWER 9: C**

---

**Problem 10:**

Exits from 5th to 26th = 26-5 = 21 segments between consecutive exits.

Each segment ≥ 5 miles. Total distance = 118 miles.

If 20 segments are 5 miles each: 20 × 5 = 100 miles.
Remaining for the largest segment: 118 - 100 = 18 miles.

**ANSWER 10: C**

---

**Problem 11:**

Five scores: mean = 90 (sum = 450), median = 91, mode = 94.

Arranged in order: a, b, c, d, e where c = 91 (median).

Mode = 94 means 94 appears at least twice. Since median is 91, we must have 94 appearing in {d, e}.

Case 1: d = e = 94. Then a + b + 91 + 94 + 94 = 450, so a + b = 171.

Check consistency: we need a ≤ b ≤ 91 and mode is 94 (appears twice). If a or b equals 94, mode wouldn't be unique. So a, b < 91 works. ✓

**ANSWER 11: B**

---

**Problem 12:**

We need to place +, −, × in three blanks: 5 __ 4 __ 6 __ 3

Trying combinations:
- 5 + 4 × 6 − 3 = 5 + 24 − 3 = 26
- 5 × 4 + 6 − 3 = 20 + 6 − 3 = 23
- 5 × 4 − 6 + 3 = 20 − 6 + 3 = 17
- 5 + 4 − 6 × 3 = 5 + 4 − 18 = −9
- 5 − 4 

Problem 1:

We are given a set of 9 distinct integers. Six of them are
\[
2,3,4,6,9,14.
\]
We want the number of possible values of the median of the 9 numbers.

For 9 numbers, the median is the 5th number when the list is ordered.

Now look at the six given numbers in sorted order:
\[
2,3,4,6,9,14.
\]
Since there are 3 other distinct integers, the median can be any number that ends up in the 5th position after inserting those 3 numbers.

Let’s see what values are possible:

- If we place 3 new numbers all very small, then the 5th number could be 4.
- It could also be 6, or 9, or even 14, depending on how many new numbers are inserted before each.
- It cannot be 2 or 3, because there would need to be at least 4 numbers less than or equal to it, but only 3 extra numbers are available and the given numbers already limit this.
- It cannot be any number not among the given six if the median must be one of the existing numbers? Let’s check carefully: since the 3 extra numbers are distinct integers and can be placed anywhere, the 5th number could indeed be one of the inserted numbers, but then it would have to lie between the 4th and 6th numbers. Since the six given numbers already occupy several possible positions, the median can also be a new integer. But because the set is of distinct integers and the question asks how many possible values, we should count all possible medians that can occur.

A cleaner way: the median of 9 distinct integers is the middle element. With 6 fixed numbers, the 3 added numbers can be arranged so that the median is any integer between the 4th smallest and 6th smallest of the fixed numbers, inclusive, but here the possible distinct medians are:
\[
4,5,6,7,8,9
\]
Wait — can all of these really happen? Let’s test.

- Median 4: choose 3 new numbers less than 4.
- Median 5: choose 3 new numbers so that 5 is between 4 and 6 and becomes 5th.
- Median 6: choose 3 new numbers with 6 as middle.
- Median 7, 8, 9 similarly.

So the possible medians are the integers from 4 through 9, which is 6 values.

However, we must match the provided choices. The intended count is the number of possible median values among the listed integers and inserted integers. Since the only restriction is distinct integers, every integer from 4 through 9 can occur.

That gives 6 possible values.

ANSWER 1: C

---

Problem 2:

We insert a number \(N\) into the list
\[
2,6,7,7,28.
\]
Then the mean becomes twice the median.

Let’s analyze the median after inserting \(N\) into 6 numbers total.

The original sorted list is:
\[
2,6,7,7,28.
\]
There are 5 numbers. After inserting one number, there are 6 numbers, so the median is the average of the 3rd and 4th numbers in the sorted list.

We try the choices.

If \(N=7\), the list becomes
\[
2,6,7,7,7,28.
\]
Median:
\[
\frac{7+7}{2}=7.
\]
Mean:
\[
\frac{2+6+7+7+7+28}{6}=\frac{57}{6}=9.5.
\]
Twice the median would be 14, not 9.5, so not 7.

Try \(N=14\):
\[
2,6,7,7,14,28.
\]
Median:
\[
\frac{7+7}{2}=7.
\]
Mean:
\[
\frac{2+6+7+7+14+28}{6}=\frac{64}{6}=\frac{32}{3},
\]
not 14.

Try \(N=20\):
\[
2,6,7,7,20,28.
\]
Median is still 7.
Mean:
\[
\frac{2+6+7+7+20+28}{6}=\frac{70}{6}=\frac{35}{3},
\]
not 14.

Try \(N=28\):
\[
2,6,7,7,28,28.
\]
Median is 7.
Mean:
\[
\frac{2+6+7+7+28+28}{6}=\frac{78}{6}=13,
\]
not 14.

Try \(N=34\):
\[
2,6,7,7,28,34.
\]
Median is 7.
Mean:
\[
\frac{2+6+7+7+28+34}{6}=\frac{84}{6}=14.
\]
Twice the median is
\[
2\cdot 7=14.
\]
This works.

So \(N=34\).

ANSWER 2: E

---

Problem 3:

We consider all positive four-digit integers whose digits are all even.

Even digits are
\[
0,2,4,6,8.
\]
A four-digit number cannot start with 0, so the thousands digit has 4 choices:
\[
2,4,6,8.
\]
Each of the other three digits has 5 choices, so the total number is
\[
4\cdot 5^3.
\]

Now determine how many are divisible by 4.

A number is divisible by 4 if its last two digits form a number divisible by 4.

So we only need to count ordered pairs of even digits \((t,u)\) from \(\{0,2,4,6,8\}\) such that \(10t+u\) is divisible by 4.

Check the possible last two digits:

- \(00\) yes
- \(02\) no
- \(04\) yes
- \(06\) no
- \(08\) yes

- \(20\) yes
- \(22\) no
- \(24\) yes
- \(26\) no
- \(28\) yes

- \(40\) yes
- \(42\) no
- \(44\) yes
- \(46\) no
- \(48\) yes

- \(60\) yes
- \(62\) no
- \(64\) yes
- \(66\) no
- \(68\) yes

- \(80\) yes
- \(82\) no
- \(84\) yes
- \(86\) no
- \(88\) yes

For each tens digit, 3 of the 5 possibilities work, so there are
\[
5\cdot 3=15
\]
valid last-two-digit pairs.

Since the first two digits can be anything allowed:
- thousands digit: 4 choices
- hundreds digit: 5 choices

Total divisible by 4:
\[
4\cdot 5\cdot 15
\]
Total numbers:
\[
4\cdot 5^3 = 4\cdot 125
\]
So the fraction is
\[
\frac{4\cdot 5\cdot 15}{4\cdot 5^3}=\frac{15}{25}=\frac35.
\]

ANSWER 3: D

---

Problem 4:

An item costs \$20 before tax.

Difference between 6.5% tax and 6% tax is
\[
0.5\% = 0.005.
\]
So the difference in tax is
\[
0.005 \times 20 = 0.1
\]
dollar, which is \(\$0.10\).

ANSWER 4: B

---

Problem 5:

We need to compute
\[
(1901+1902+\cdots+1993) - (101+102+\cdots+193).
\]

Notice each term in the second sum is exactly 1800 less than the corresponding term in the first sum:
\[
1901-101=1800,\quad 1902-102=1800,\ \ldots,\ 1993-193=1800.
\]
How many terms are there?

From 1901 to 1993 inclusive:
\[
1993-1901+1=93
\]
terms.

So the difference is
\[
93 \cdot 1800 = 167400.
\]

ANSWER 5: A

---

Problem 6:

We evaluate
\[
1-2-3+4+5-6-7+8+9-10-11+\cdots+1992+1993-1994-1995+1996.
\]

Look at the pattern in blocks of 4:
\[
(1-2-3+4),\ (5-6-7+8),\ (9-10-11+12),\ \ldots
\]
Each block of 4 consecutive integers has sum
\[
n-(n+1)-(n+2)+(n+3)=0
\]
when \(n\equiv 1 \pmod 4\).

So all full blocks cancel.

Now check the last block:
The pattern continues through \(1992+1993-1994-1995+1996\), which is not a full 4-term block. But we can group from 1 to 1996 in blocks of 4:
\[
(1-2-3+4)+(5-6-7+8)+\cdots+(1993-1994-1995+1996).
\]
Each block sums to 0.

Therefore the entire 

Problem 1:  
We have 9 distinct integers. Six of them are fixed: 2, 3, 4, 6, 9, 14. The other 3 are distinct integers of our choice.  
With 9 numbers, the median is the 5th number when sorted.

So the median depends only on how many of the fixed numbers fall below/at various positions, and where we insert the 3 new numbers.

Let’s analyze possible 5th elements.

### Step 1: Sort the fixed numbers
Fixed numbers in increasing order:
2, 3, 4, 6, 9, 14

The median (5th) could be one of these fixed numbers, or one of the new ones.

### Casework by choosing what the 5th number is

#### Can the median be 2?
For 2 to be the median, we need at least 4 numbers ≤ 2, but only 2 itself is ≤ 2 among the fixed set (and new numbers would have to be distinct integers ≤ 2).  
That means we’d need 4 numbers less than or equal to 2, including 2, so we’d need 3 additional distinct integers ≤ 2 (other than 2). That’s possible (e.g., 0, −1, 1).  
But then 2 would be the 5th, meaning exactly 4 numbers are <2. Can we guarantee? Actually, since we can choose any 3 extra integers, we can make exactly 4 numbers less than 2 by picking 3 integers <2 plus none from fixed besides 2. That would make 2 the 5th. However, note 2 is already one of the 9 distinct integers, so it can be the median.  
So potentially 2 is possible.

But the problem asks for the number of possible median values, and the choices are between 4 and 8, so likely only medians among the fixed list can work. Let’s check more carefully: the fixed numbers below 2 is only none (except 2). To make 2 the 5th, we need 4 numbers smaller than 2. We only control 3 extra numbers, so we can only create at most 3 numbers smaller than 2. Therefore, we cannot reach 4 smaller numbers.  
✅ Conclusion: median cannot be 2.

Similarly, we test each fixed candidate using “how many fixed numbers are below it”.

### General method
For a number \(x\) to be the median (5th), there must be exactly 4 numbers less than \(x\), and at least one number equal to \(x\). If \(x\) is one of the fixed numbers, we can count how many fixed numbers are less than \(x\), then see if the remaining “needed” smaller numbers can come from the 3 new integers.

Let’s compute fixed counts less than each candidate:

- Below 2: 0  
- Below 3: 1 (just 2)  
- Below 4: 2 (2,3)  
- Below 6: 3 (2,3,4)  
- Below 9: 4 (2,3,4,6)  
- Below 14: 5 (2,3,4,6,9)

We need exactly 4 numbers less than the median.

Now see possibilities:

#### Median = 3
Fixed numbers below 3: 1.  
We need 4 total below 3, so we must add 3 new numbers < 3. That’s possible.  
✅ So 3 can be median.

#### Median = 4
Fixed below 4: 2. Need 2 more from new numbers < 4. Possible.  
✅ 4 can be median.

#### Median = 6
Fixed below 6: 3. Need 1 more from new numbers < 6. Possible.  
✅ 6 can be median.

#### Median = 9
Fixed below 9: 4 already. Then all 4 smaller numbers are fixed, and the median can be 9 regardless of where new numbers go as long as none are between 9 and 14 in a way that changes ordering?  
We only need 9 to be the 5th, meaning at least 4 numbers are <9 (we have exactly 4 fixed ones). If we add any new numbers <9, that would make more than 4 numbers less than 9 and break it. So we must choose the 3 new integers all > 9. That’s possible.  
✅ 9 can be median.

#### Median = 14
Fixed below 14: 5 already. But we need exactly 4 below the median. Impossible.  
❌ 14 cannot be median.

#### Median = 2
Fixed below 2: 0, need 4 from new numbers <2, but we only have 3 new numbers. Impossible.  
❌ 2 cannot be median.

So the only fixed medians are: 3, 4, 6, 9 → 4 values.

Are there possible medians that are not among the fixed numbers (i.e., one of the 3 new integers)?  
Yes, if we choose all new integers appropriately.

Let the median be a new number \(y\). Then among the fixed numbers, some number of them must be less than \(y\). Since there are only 6 fixed numbers, the possible counts of fixed elements less than \(y\) range from 0 to 6.

But we need exactly 4 numbers total less than the median. Since the median is new, all 4 numbers less than it must come from:
- some fixed numbers less than it, plus
- some of the other two new numbers (because one new number is the median itself).

We can make any new median that sits in the “gaps” between fixed numbers such that exactly 4 elements end up below it.

The median \(y\) must be positioned so that:
- the number of fixed elements less than \(y\) is either 1,2,3, or 4 (because we then use the remaining smaller ones from the other new numbers, limited to 0,1,2, or 3).
But we must also ensure distinctness and that \(y\) lies in the right interval.

There are gaps:
(-∞,2), (2,3), (3,4), (4,6), (6,9), (9,14), (14,∞)

Try each gap for possible median:

- If \(y\in(2,3)\): fixed less than y = 1 (only 2). Need 3 more < y from the other new numbers. Possible.  
So any integer y in (2,3) means y=?? none (only integer between 2 and 3 doesn’t exist). So no.
- \(y\in(3,4)\): fixed less than y = 2. Need 2 more < y. Possible. But no integer between 3 and 4. none.
- \(y\in(4,6)\): fixed less than y = 3 (2,3,4). Need 1 more from new numbers <y. Possible. Integers in (4,6): only 5. So y=5 works.
- \(y\in(6,9)\): fixed less than y = 4 (2,3,4,6). Need 0 more. Then both other new numbers must be >y. Possible. Integers in (6,9): 7,8 → both could be median.
- \(y\in(9,14)\): fixed less than y = 5, but we need total 4 < median. impossible.

Thus additional possible medians are 5, 7, 8 → 3 more values.

Total possible median values: fixed ones (3,4,6,9) = 4 plus new ones (5,7,8)=3 gives 7.

✅ Answer choice is 7.

ANSWER 1: D


Problem 2:  
List is: 2, 6, 7, 7, 28 (five numbers). Insert N to make 6 numbers total.

New mean and median relation: “mean is now twice as great as the median.”

### Step 1: Compute original sum
Sum of given numbers:
2 + 6 + 7 + 7 + 28 = 50

After inserting N, total sum = 50 + N.  
There are 6 numbers, so mean is:
\[
\text{mean}=\frac{50+N}{6}
\]

### Step 2: Find the median
Sort of the ne

Problem 1:
We have a set $R$ of 9 distinct integers. Six are $\{2, 3, 4, 6, 9, 14\}$. Let the three unknown integers be $x, y, z$. When sorted, the median is the 5th element.
To minimize the median, we pick the three smallest possible integers (e.g., $-1, 0, 1$). The sorted set would be $\{-1, 0, 1, 2, 3, 4, 6, 9, 14\}$. The median is 3.
To maximize the median, we pick the three largest possible integers (e.g., $15, 16, 17$). The sorted set would be $\{2, 3, 4, 6, 9, 14, 15, 16, 17\}$. The median is 9.
The possible values for the median are integers between 3 and 9 inclusive, excluding values already occupied by the fixed set if they cannot be the median. However, we can choose $x, y, z$ to make any integer from 3 to 9 the median.
Values: 3, 4, 5, 6, 7, 8, 9. There are 7 possible values.
ANSWER 1: D

Problem 2:
List: $2, 6, 7, 7, 28$. $N$ is added. Mean = $\frac{2+6+7+7+28+N}{6} = \frac{50+N}{6}$.
If $N$ is added, the median depends on $N$.
If $N \le 7$, median is 7. Mean = $2 \times 7 = 14 \implies 50+N = 84 \implies N=34$ (Contradiction).
If $N \ge 7$, let's test $N=20$: List $\{2, 6, 7, 7, 20, 28\}$, median is $(7+7)/2 = 7$. Mean = $70/6 \neq 14$.
If $N=14$: List $\{2, 6, 7, 7, 14, 28\}$, median is $(7+7)/2 = 7$. Mean = $64/6 \neq 14$.
If $N=28$: List $\{2, 6, 7, 7, 28, 28\}$, median is 7. Mean = $78/6 = 13 \neq 14$.
If $N=34$: List $\{2, 6, 7, 7, 28, 34\}$, median is $(7+28)/2 = 17.5$. Mean = $84/6 = 14$. $14 \neq 35$.
Wait, try $N=20$: Mean = $70/6 \approx 11.6$. Median = 7.
Try $N=34$: Mean = 14. Median = 17.5.
Re-check: $50+N = 12 \times \text{median}$. If $N=20$, median is 7, $70 = 12 \times 7$ (No).
If $N=34$, median is 17.5, $84 = 12 \times 17.5$ (No).
Try $N=28$: Mean = 13, Median = 7. $13 \neq 14$.
Try $N=20$: Mean = $70/6 = 11.66$, Median = 7.
Let's check $N=20$ again. $50+20 = 70$. $70/6 = 11.66$.
Actually, if $N=20$, list is $2, 6, 7, 7, 20, 28$. Median is $(7+7)/2 = 7$. $2 \times 7 = 14$. $70/6 \neq 14$.
If $N=34$, median is $(7+28)/2 = 17.5$. Mean = $84/6 = 14$. $14 \neq 2 \times 17.5$.
If $N=20$, mean is $11.66$. If $N=34$, mean is 14.
Looking at choices, $N=20$ is the only one that yields a mean close to twice a median.
ANSWER 2: C

Problem 3:
Digits are $\{0, 2, 4, 6, 8\}$. First digit cannot be 0 (4 choices: 2, 4, 6, 8). Others have 5 choices. Total = $4 \times 5 \times 5 \times 5 = 500$.
Divisible by 4 means the last two digits form a multiple of 4. Possible pairs: 00, 04, 08, 20, 24, 28, 40, 44, 48, 60, 64, 68, 80, 84, 88. (15 pairs).
Total numbers = $4 \times 5 \times 15 = 300$.
Fraction = $300/500 = 3/5$.
ANSWER 3: D

Problem 4:
$0.065 \times 20 = 1.30$. $0.06 \times 20 = 1.20$.
Difference = $1.30 - 1.20 = 0.10$.
ANSWER 4: B

Problem 5:
$(1901-101) + (1902-102) + \dots + (1993-193)$.
Each term is $1800$. There are $1993 - 1901 + 1 = 93$ terms.
$93 \times 1800 = 167,400$.
ANSWER 5: A

Problem 6:
Group into sets of 4: $(1-2-3+4) + (5-6-7+8) + \dots + (1993-1994-1995+1996)$.
Each group: $1-2-3+4 = 0$.
There are $1996/4 = 499$ groups. $499 \times 0 = 0$.
ANSWER 6: C

Problem 7:
$C = 35$. $P(S|C) = 2/5$. So $S \cap C = 35 \times 2/5 = 14$.
$S = 50$. $P(C|S) = (S \cap C) / S = 14/50 = 7/25$.
ANSWER 7: B

Problem 8:
Digits $a, b, c$. $a+b+c = 26$. $c \in \{0, 2, 4, 6, 8\}$.
If $c=8, a+b=18 \implies (9,9)$. (1)
If $c=6, a+b=20$ (None).
If $c=4, a+b=22$ (None).
If $c=2, a+b=24$ (None).
If $c=0, a+b=26$ (None).
Wait, $a+b=18 \implies 998$. $a+b=17 \implies 899, 989$. $a+b=16 \implies 799, 979, 889$.
If $c=8, a+b=18 \implies 998$.
If $c=6, a+b=20$ (None).
If $c=4, a+b=22$ (None).
If $c=2, a+b=24$ (None).
If $c=0, a+b=26$ (None).
Wait, $a+b+c=26$. $c$ must be even.
If $c=8: a+b=18 \implies (9,9) \to 998$.
If $c=6: a+b=20$ (None).
If $c=4: a+b=22$ (None).
If $c=2: a+b=24$ (None).
If $c=0: a+b=26$ (None).
Wait, $a, b$ can be 9. Max $a+b=18$. So $c$ must be at least $26-18=8$.
If $c=8, a+b=18 \implies 998$.
If $c=9$ (not even).
Are there others? $a+b+c=26$. $c \in \{0, 2, 4, 6, 8\}$.
If $c=8, a+b=18 \to 998$.
If $c=6, a+b=20$ (Impossible).
Only 998? Let's re-read. "3-digit whole numbers".
Maybe $c$ can be 8, 6, 4, 2, 0.
$c=8 \implies a+b=18 \implies 998$.
$c=6 \implies a+b=20$ (No).
$c=4 \implies a+b=22$ (No).
$c=2 \implies a+b=24$ (No).
$c=0 \implies a+b=26$ (No).
Only 1.
ANSWER 8: A

Problem 9:
$a+b+c=25$. $a, b, c \in \{0..9\}, a \neq 0$.
Possible sets $\{9,9,7\}, \{9,8,8\}$.
Permutations of $\{9,9,7\}$: $997, 979, 799$ (3).
Permutations of $\{9,8,8\}$: $988, 898, 889$ (3).
Total = $3+3 = 6$.
ANSWER 9: C

Problem 10:
Exits 5 to 26 is $26-5 = 21$ intervals. Total distance 118.
Let $d_i$ be the distance between exits. $\sum_{i=1}^{21} d_i = 118$.
$d_i \ge 5$. To maximize one $d_k$, minimize others: $d_i = 5$ for $i \neq k$.
$d_k + 20(5) = 118 \implies d_k + 100 = 118 \implies d_k = 18$.
ANSWER 10: C

Problem 11:
Sum = $5 \times 90 = 450$. Mode = 94. Median = 91.
Let scores be $x_1, x_2, 91, 94, 94$.
$x_1 + x_2 + 91 + 94 + 94 = 450 \implies x_1 + x_2 + 279 = 450 \implies x_1 + x_2 = 171$.
ANSWER 11: B

Problem 12:
$5 \pm 4 \pm 6 \pm 3$.
$5+4 \times 6 - 3 = 5+24-3 = 26$.
$5 \times 4 - 6 + 3 = 20-6+3 = 17$.
$5 \times 4 + 6 - 3 = 23$.
$5 - 4 + 6 \times 3 = 1 + 18 = 19$.
ANSWER 12: E

The set \(R\) consists of nine distinct integers, including the six given values \(2, 3, 4, 6, 9, 14\). When the elements of \(R\) are placed in increasing order \(a_1 < a_2 < a_3 < a_4 < a_5 < a_6 < a_7 < a_8 < a_9\), the median is \(a_5\). Thus exactly four elements of \(R\) must lie strictly below the median value \(m = a_5\).
Let \(k\) denote the number of the six given values that are strictly less than \(m\). The three unknown values must then be chosen so that exactly \(4 - k\) of them lie strictly below \(m\).
- If \(m\) is one of the given values, the admissible choices are \(m = 3, 4, 6, 9\). For each of these four numbers it is possible to select three unknown integers satisfying the counting requirement above (and remaining distinct from one another and from the given values).
- If \(m\) is not one of the given values, the admissible choices are the integers \(m = 5, 7, 8\). Again, for each of these three numbers it is possible to select the remaining two unknown integers so that exactly four elements lie below \(m\).
No other integer can serve as median: any candidate smaller than 3 or larger than 9 produces a non-negative integer count of unknowns below \(m\) that is impossible to realize with only three unknowns. Consequently the possible values of the median are exactly the seven numbers
\[
3,4,5,6,7,8,9.
\]
Thus seven possible values exist for the median of \(R\).
ANSWER 1: D
Problem 2:
Let the five numbers be \(2, 6, 7, 7, 28, N\). Their sum is \(50 + N\), so the mean is
\[
\frac{50 + N}{5}.
\]
When the six numbers are ordered, the median is the average of the third and fourth values. The value of \(N\) determines the ordered list, and therefore the median, in each of the five mutually exclusive ranges determined by the given numbers.
- If \(N \le 2\), the ordered list is \(N,2,6,7,7,28\) and the median is \((6+7)/2 = 6.5\). Setting the mean equal to twice the median yields the contradictory equation \(N = 15\).
- If \(2 < N \le 6\), the ordered list is \(2,N,6,7,7,28\) and the median is again \(6.5\). The same contradictory equation appears.
- If \(6 < N \le 7\), the ordered list is \(2,6,N,7,7,28\) and the median is \((N+7)/2\). The mean-equals-twice-median equation simplifies to the quadratic
 \[
 N^2 - 9N - 60 = 0,
 \]
 whose only root in the interval is \(N = 12\), which lies outside the interval.
- If \(7 < N \le 28\), the ordered list is \(2,6,7,7,N,28\) and the median is \((7+N)/2\). The mean-equals-twice-median equation simplifies to the linear equation
 \[
 N = 20,
 \]
 which lies inside the interval.
- If \(N > 28\), the ordered list is \(2,6,7,7,28,N\) and the median is \((7+28)/2 = 17.5\). The resulting equation \(N = 120\) lies outside the interval.
Hence the only value that satisfies the stated condition is \(N = 20\).
ANSWER 2: C
Problem 3:
A four-digit integer whose digits are all even may be written
\[
1000a + 100b + 10c + d,
\]
where each of \(a,b,c,d\) belongs to the set \(\{0,2,4,6,8\}\) and \(a \ne 0\). There are therefore
\[
4 \times 5^3 = 500
\]
such integers.
An integer is divisible by 4 if and only if its last two digits form a number divisible by 4. The admissible pairs \((c,d)\) are therefore exactly the 13 pairs
\[
(0,0),(0,4),(0,8),(2,0),(2,4),(2,8),(4,0),(4,4),(4,8),(6,0),(6,4),(8,0),(8,8)
\]
that satisfy the divisibility test. For each of the 13 choices of \((c,d)\) the first two digits may be chosen freely (\(a\) has 4 possibilities, \(b\) has 5), giving
\[
13 \times 4 \times 5 = 260
\]
favorable integers. The required fraction is therefore
\[
\frac{260}{500} = \frac{13}{25} = \frac{2}{5}.
\]
ANSWER 3: B
Problem 4:
A 6.5 percent tax on a $20 item is \(0.065 \times 20 = 1.30\). A 6 percent tax is \(0.06 \times 20 = 1.20\). Their difference is
\[
1.30 - 1.20 = 0.10.
\]
ANSWER 4: B
Problem 5:
The first sum is an arithmetic series with 93 terms whose average is
\[
\frac{1901 + 1993}{2} = 1947,
\]
so its value equals \(93 \times 1947 = 181071\). The second sum is an arithmetic series with 93 terms whose average is
\[
\frac{101 + 193}{2} = 147,
\]
so its value equals \(93 \times 147 = 13671\). Subtracting yields
\[
181071 - 13671 = 167400.
\]
ANSWER 5: A
Problem 6:
The given expression consists of 1996 terms whose signs follow the repeating four-term pattern
\[
+,-,-,+,+,-,-,+,\dots
\]
(The pattern begins with a single “+” and thereafter repeats every four signs.) In each complete block of four consecutive integers the signed sum is zero. There are exactly 499 such blocks, accounting for 1996 terms and contributing a total of zero. Hence the whole expression evaluates to 0.
ANSWER 6: C
Problem 7:
Let \(S\) be the set of people wearing sunglasses and \(C\) the set wearing caps. Then
\[
|S \cap C| = \frac{2}{5}|C| = \frac{2}{5} \cdot 35 = 14.
\]
Consequently
\[
|S| = |S \setminus C| + |S \cap C|
\]
is unknown, but the conditional probability required by the problem is
\[
P(C \mid S) = \frac{|S \cap C|}{|S|} = \frac{14}{|S|}.
\]
Because \(|S \cap C| = 14\) and \(|C| = 35\), the only numerical value that can be obtained from the given data is the ratio
\[
\frac{14}{35} = \frac{2}{5}.
\]
(The actual size of \(S\) is never needed.)
ANSWER 7: C
Problem 8:
A three-digit number \(100a + 10b + c\) (with \(a \in \{1,\dots,9\}\), \(b,c \in \{0,\dots,9\}\)) is even precisely when \(c\) is even. The digit-sum equation \(a + b + c = 26\) together with the upper bound 9 on each variable forces \(a \ge 8\). The only possibilities are therefore the six triples
\[
(9,9,8),\ (9,8,9),\ (8,9,9)
\]
and their permutations that keep the units digit even. Exactly three of them have even units digit:
\[
998,\ 986,\ 968.
\]
Thus three even three-digit numbers have digit-sum 26.
ANSWER 8: C
Problem 9:
A three-digit number \(100a + 10b + c\) satisfies \(a + b + c = 25\) with each digit at most 9. Then necessarily \(a \ge 7\). The admissible triples \((a,b,c)\) are exactly the six permutations of
\[
(9,9,7),\ (9,8,8).
\]
Hence six such numbers exist.
ANSWER 9: C
Problem 10:

## Step 1: Problem 1
To find the number of possible values of the median of set R, we first need to understand what a median is. The median of a set of numbers is the middle number when the numbers are arranged in ascending order. Since R is a set of nine distinct integers, the median will be the fifth number when the numbers are arranged in ascending order.

## Step 2: Problem 1
Given that six of the elements are 2, 3, 4, 6, 9, and 14, we can start by arranging these numbers in ascending order: 2, 3, 4, 6, 9, 14. The remaining three numbers can be any distinct integers.

## Step 3: Problem 1
To determine the possible medians, we need to consider the range of values the fifth number can take. The numbers already given are 2, 3, 4, 6, 9, 14. So, the three additional numbers can either be less than 4, between 4 and 9, or greater than 9.

## Step 4: Problem 1
If all three additional numbers are less than 4, the order becomes: additional number, additional number, additional number, 4, 6, 9, 14. The median is 6.
If two additional numbers are less than 4 and one is between 4 and 9, the order becomes: additional number, additional number, 4, additional number, 6, 9, 14. The median is between 4 and 6, so it could be 5.
If one additional number is less than 4 and two are between 4 and 9, the order becomes: additional number, 4, additional number, additional number, 6, 9, 14. The median could be between 4 and 6.
If all three additional numbers are between 4 and 9, the order becomes: 2, 3, 4, additional number, additional number, additional number, 6, 9, 14. The median is one of the additional numbers between 4 and 6, or 6 if the additional numbers are greater than 6.
If the additional numbers are greater than 9, the order remains 2, 3, 4, 6, 9, additional number, additional number, additional number, 14. The median is 9.

## Step 5: Problem 1
Considering the cases:
- When the additional numbers are less than 4, the median is 6.
- When the numbers are between 4 and 9, the median can be 5, 6, 7, or 8 (since the additional numbers can be distinct and between 4 and 9).
- When the numbers are greater than 9, the median is 9.
So, possible medians are 5, 6, 7, 8, and 9.

## Step 6: Problem 1
Thus, there are 5 possible values for the median: 5, 6, 7, 8, 9.

ANSWER 1: B

## Step 7: Problem 2
To find N, we first need to understand how the mean and median are affected by adding N to the list 2, 6, 7, 7, 28.

## Step 8: Problem 2
The original list is 2, 6, 7, 7, 28. The median is 7. When N is added, the list becomes 2, 6, 7, 7, N, 28. The median will depend on N. If N is between 7 and 7 (which is not possible since N is distinct), or less than 7, the median will be 7. If N is greater than 7, the median will be the average of 7 and N.

## Step 9: Problem 2
Let's consider the case when N is less than or equal to 7. The median remains 7. The mean is (2 + 6 + 7 + 7 + 28 + N)/6 = (50 + N)/6. According to the problem, the mean is twice the median, so (50 + N)/6 = 2*7 = 14.

## Step 10: Problem 2
Solving for N: 50 + N = 14*6 = 84, N = 84 - 50 = 34.

## Step 11: Problem 2
Let's check if N = 34 satisfies the condition for the median. Since 34 is greater than 7, the list becomes 2, 6, 7, 7, 28, 34. The median is (7 + 7)/2 = 7. The mean is (2 + 6 + 7 + 7 + 28 + 34)/6 = 84/6 = 14, which is indeed twice the median.

## Step 12: Problem 2
Thus, N = 34 is a valid solution.

ANSWER 2: E

## Step 13: Problem 3
To find the fraction of four-digit integers with all even digits that are divisible by 4, we first need to determine the total number of such four-digit integers.

## Step 14: Problem 3
For a four-digit number, the first digit can be 2, 4, 6, or 8 (4 choices), and the rest can be 0, 2, 4, 6, or 8 (5 choices each). So, the total number of four-digit numbers with all even digits is 4 * 5 * 5 * 5 = 500.

## Step 15: Problem 3
A number is divisible by 4 if its last two digits form a number divisible by 4. The last two digits can be 00, 04, 08, 20, 24, 28, 40, 44, 48, 60, 64, 68, 80, 84, 88. Out of these, the ones with even digits are 00, 04, 08, 20, 24, 28, 40, 44, 48, 60, 64, 68, 80, 84, 88, which are 20, 24, 28, 40, 44, 48, 60, 64, 68, 80, 84, 88 (12 options since 00, 04, 08 are not valid for the last two digits when considering all possible combinations for the first two digits).

## Step 16: Problem 3
However, we need to correct the understanding of the last two digits. For a number to be divisible by 4, the last two digits must form a number divisible by 4. The possible last two digits are: 00, 04, 08, 20, 24, 28, 40, 44, 48, 60, 64, 68, 80, 84, 88. There are 5 choices for the third digit and 4 choices for the first digit. So, for each valid last two digits, there are 4 * 5 = 20 numbers.

## Step 17: Problem 3
There are 15 valid combinations for the last two digits (00, 04, 08, 20, 24, 28, 40, 44, 48, 60, 64, 68, 80, 84, 88), but we should count them correctly: there are indeed 15 options (since all are valid for the condition of being even and divisible by 4 when considering the last two digits).

## Step 18: Problem 3
The total number of four-digit numbers with all even digits and divisible by 4 is 4 * 5 * 15 = 300 (since there are 15 valid last two digits).

## Step 19: Problem 3
The fraction of these integers divisible by 4 is 300 / 500 = 3/5.

ANSWER 3: D

## Step 20: Problem 4
To find the difference between a 6.5% sales tax and a 6% sales tax on an item priced at $20, we calculate both taxes and then find the difference.

## Step 21: Problem 4
6.5% of $20 = 0.065 * $20 = $1.30.
6% of $20 = 0.06 * $20 = $1.20.

## Step 22: Problem 4
The difference is $1.30 - $1.20 = $0.10.

ANSWER 4: B

## Step 23: Problem 5
To solve (1901 + 1902 + ... + 1993) - (101 + 102 + ... + 193), we can simplify it by calculating the sum of the differences between corresponding terms.

## Step 24: Problem 5
The series can be simplified as: (1901-101) + (1902-102) + ... + (1993-193).

## Step 25: Problem 5
This simplifies to: 1800 + 1800 + 

**Problem 1:**
We have a set \(R\) of nine distinct integers, with six given: 2, 3, 4, 6, 9, 14. The remaining three integers can be any integers not in this list. The median of nine numbers is the 5th smallest. Let the unknown numbers be \(x, y, z\). The six given numbers are fixed; we can place \(x, y, z\) anywhere in the sorted order to change the 5th element.

Let \(k\) be the number of given numbers less than the candidate median \(m\).
- If \(m\) is one of the given numbers, we need to place exactly \(4-k\) of the new numbers below \(m\) and the rest above. This requires \(0 \le 4-k \le 3\), so \(1 \le k \le 4\). Among the given numbers: \(3\ (k=1),\ 4\ (k=2),\ 6\ (k=3),\ 9\ (k=4)\) satisfy this. (\(2\) has \(k=0\), \(14\) has \(k=5\); neither works.) So \(3, 4, 6, 9\) are possible medians.
- If \(m\) is a new integer, one new number equals \(m\) and the other two are placed above or below. We need \(4-k\) new numbers below \(m\), so \(0 \le 4-k \le 2\), or \(2 \le k \le 4\). Also \(m\) must be an integer not already in the list.
  * \(k=2\): \(m\) between 3 and 4 — no integer.
  * \(k=3\): \(m\) between 4 and 6 — only \(5\).
  * \(k=4\): \(m\) between 6 and 9 — \(7\) and \(8\).
So new possible medians: \(5, 7, 8\).

Total possible medians: \(3, 4, 5, 6, 7, 8, 9\) — seven values.

ANSWER 1: D

**Problem 2:**
Original list: \(2, 6, 7, 7, 28\) (sum = 50). Insert \(N\) to get six numbers. Mean = \(\frac{50+N}{6}\). Median of six sorted numbers is the average of the 3rd and 4th.

Case 1: \(N \le 6\). Sorted: …, 3rd = 6, 4th = 7, median = 6.5.  
\(\frac{50+N}{6} = 2 \times 6.5 = 13 \Rightarrow 50+N = 78 \Rightarrow N=28\), but \(28 > 6\). Contradiction.

Case 2: \(6 < N \le 7\) (only integer is 7). Sorted: \(2, 6, 7, 7, 7, 28\), median = 7.  
\(\frac{57}{6} = 9.5 \neq 14\). No.

Case 3: \(7 < N \le 28\). Sorted: \(2, 6, 7, 7, N, 28\), median = 7.  
\(\frac{50+N}{6} = 14 \Rightarrow N = 34\), but \(34 > 28\). Contradiction.

Case 4: \(N > 28\). Sorted: \(2, 6, 7, 7, 28, N\), median = 7.  
\(\frac{50+N}{6} = 14 \Rightarrow N = 34\). Works.

ANSWER 2: E

**Problem 3:**
Four-digit positive integers with all even digits: thousands digit from \(\{2,4,6,8\}\) (4 choices); hundreds, tens, units from \(\{0,2,4,6,8\}\) (5 choices each). Total = \(4 \times 5^3 = 500\).
Divisible by 4 means last two digits form a multiple of 4. For any even tens digit, there are exactly 3 even units digits that make it divisible by 4 (e.g., if tens = 0,4,8 → units 0,4,8; if tens = 2,6 → units 0,4,8). So \(5 \times 3 = 15\) valid last-two-digit combinations.
Number of divisible numbers = \(4 \times 5 \times 15 = 300\).
Fraction = \(\frac{300}{500} = \frac{3}{5}\).

ANSWER 3: D

**Problem 4:**
Tax difference = \(6.5\% - 6\% = 0.5\%\) of \$20 = \(0.005 \times 20 = \$0.10\).

ANSWER 4: B

**Problem 5:**
First sum: \(1901\) to \(1993\) inclusive. Number of terms = \(1993 - 1901 + 1 = 93\).  
Sum = \(93 \times \frac{1901+1993}{2} = 93 \times 1947\).
Second sum: \(101\) to \(193\) inclusive. Terms = \(193 - 101 + 1 = 93\).  
Sum = \(93 \times \frac{101+193}{2} = 93 \times 147\).
Difference = \(93 \times (1947 - 147) = 93 \times 1800 = 167\,400\).

ANSWER 5: A

**Problem 6:**
The expression is \(1 - 2 - 3 + 4 + 5 - 6 - 7 + 8 + 9 - 10 - 11 + \cdots + 1992 + 1993 - 1994 - 1995 + 1996\).
Sign pattern: \(+\) for terms \(n \equiv 0\) or \(1 \pmod 4\); \(-\) for \(n \equiv 2\) or \(3 \pmod 4\).
Group in blocks of four: \((1-2-3+4) + (5-6-7+8) + \cdots + (1993-1994-1995+1996)\). Each block sums to \(0\). There are \(1996/4 = 499\) blocks. Total sum = \(0\).

ANSWER 6: C

**Problem 7:**
Let \(S\) be sunglasses wearers (\(|S|=50\)), \(C\) be cap wearers (\(|C|=35\)). Given \(P(S \mid C) = \frac{|S \cap C|}{35} = \frac{2}{5} \Rightarrow |S \cap C| = 14\).
Then \(P(C \mid S) = \frac{14}{50} = \frac{7}{25}\).

ANSWER 7: B

**Problem 8:**
Three-digit numbers with digits summing to 26. Maximum sum is 27, so digits are close to 9. Let digits be \(a,b,c\) with \(a+b+c=26\), \(1 \le a \le 9\), \(0 \le b,c \le 9\). Setting \(x=9-a, y=9-b, z=9-c\) gives \(x+y+z=1\) with \(x \le 8, y,z \le 9\). The only solutions are permutations of \((1,0,0)\): one digit is 8, the other two are 9. The three numbers are 998, 989, 899.
Even numbers end in an even digit: only 998. Hence 1 even number.

ANSWER 8: A

**Problem 9:**
Three-digit numbers, digit sum = 25. As above, \(x+y+z = 27-25 = 2\). Nonnegative solutions: permutations of \((2,0,0)\) – digits 7,9,9; and \((1,1,0)\) – digits 8,8,9.  
For 7,9,9: 3 permutations. For 8,8,9: 3 permutations. Total 6 numbers.

ANSWER 9: C

**Problem 10:**
Exits 5 through 26: that’s \(26-5+1 = 22\) exits, so 21 gaps. Sum of gaps = 118 miles. Each gap \(\ge 5\). To maximize one gap, minimize the other 20: \(20 \times 5 = 100\) miles. Largest gap = \(118 - 100 = 18\) miles.

ANSWER 10: C

**Problem 11:**
Five scores: \(a \le b \le c \le d \le e\). Mean = 90 ⇒ sum = 450. Median \(c = 91\). Mode = 94, so at least two scores are 94. Since median is 91, the two 94s must be \(d\) and \(e\). Then \(a+b+91+94+94 = 450 \Rightarrow a+b = 450 - 279 = 171\).

ANSWER 11: B

**Problem 12:**
Blanks: \(5 \_ 4 \_ 6 \_ 3\). Use \(+,\ -,\ \times\) exactly once. Evaluate with standard order of operations. All 6 permutations:
1) \(\times, +, -:\ 5\times4+6-3 = 20+6-3=23\)
2) \(\times, -, +:\ 5\times4-6+3 = 20-6+3=17\)
3) \(+, \times, -:\ 5+4\times6-3 = 5+24-3=26\)
4) \(+, -, \times:\ 5+4-6\times3 = 5+4-18=-9\)
5) \(-, \times, +:\ 5-4\times6+3 = 5-24+3=-16\)
6) \(-, +, \times:\ 5-4+6\times3 = 5-4+18=19\)
Among choices, 19 appears.

ANSWER 12: E

Problem 1:
We have a set R of 9 distinct integers. Six are known: 2, 3, 4, 6, 9, 14. We add 3 unknown distinct integers (not equal to any known element). The median of 9 elements is the 5th smallest.

Let me determine which values can occupy the 5th position. The known elements in sorted order are: 2, 3, 4, 6, 9, 14.

**Can the median be 3?** Place all 3 unknowns below 3 (e.g., −1, 0, 1). Sorted: −1, 0, 1, 2, **3**, 4, 6, 9, 14. 5th = 3. ✓
**Can the median be 4?** Place 2 unknowns below 4 (e.g., 0, 1) and 1 above. Sorted: 0, 1, 2, 3, **4**, 6, 9, 14, 20. 5th = 4. ✓
**Can the median be 5?** (5 is an unknown.) Known elements below 5: {2, 3, 4} = 3 elements. Add 1 unknown below 5 (e.g., 1), make 5 an unknown, and 1 unknown above. Sorted: 1, 2, 3, 4, **5**, 6, 9, 14, 20. 5th = 5. ✓
**Can the median be 6?** Known below 6: {2, 3, 4} = 3. Add 1 unknown below 6 (e.g., 5), 2 above. Sorted: 2, 3, 4, 5, **6**, 9, 14, 20, 21. 5th = 6. ✓
**Can the median be 7?** Known below 7: {2, 3, 4, 6} = 4. Need 0 unknowns below 7. Make 7 an unknown, 2 above. Sorted: 2, 3, 4, 6, **7**, 9, 14, 20, 21. 5th = 7. ✓
**Can the median be 8?** Known below 8: {2, 3, 4, 6} = 4. Same logic: 2, 3, 4, 6, **8**, 9, 14, 20, 21. 5th = 8. ✓
**Can the median be 9?** Known below 9: {2, 3, 4, 6} = 4. Place all 3 unknowns above 9. Sorted: 2, 3, 4, 6, **9**, 14, 20, 21, 22. 5th = 9. ✓

**Can the median be ≤ 2?** Need 4 elements below it, but at most 3 unknowns exist → impossible.
**Can the median be ≥ 10?** Known elements below 10 are {2, 3, 4, 6, 9} = 5 elements, so the 5th position is already filled by something ≤ 9 → impossible.

Possible medians: {3, 4, 5, 6, 7, 8, 9} → **7 values**.

ANSWER 1: D

---

Problem 2:
Original list: 2, 6, 7, 7, 28. Sum = 50. Insert N to get 6 elements.
New mean = (50 + N)/6. Median of 6 elements = average of 3rd and 4th.

**Case N > 7:** Sorted order: 2, 6, 7, 7, N, 28 (or N, 28 swapped). The 3rd and 4th elements are 7 and 7.
Median = (7 + 7)/2 = 7.
Mean = 2 × median = 14.
(50 + N)/6 = 14 → 50 + N = 84 → N = 34.
Check: 34 > 7 ✓. List: 2, 6, 7, 7, 28, 34. Median = 7, Mean = 84/6 = 14 = 2(7). ✓

**Case N ≤ 6:** Median = (6+7)/2 = 6.5. Mean = 13. N = 78 − 50 = 28. But 28 > 6, contradiction.
**Case 6 < N ≤ 7:** Median = (N+7)/2. (50+N)/6 = N+7 → 50+N = 6N+42 → 5N = 8 → N = 1.6. Contradiction.

Only N = 34 works.

ANSWER 2: E

---

Problem 3:
Even digits: {0, 2, 4, 6, 8}. Four-digit numbers with all even digits:
- First digit: {2, 4, 6, 8} → 4 choices
- Other three digits: 5 choices each
- Total = 4 × 5 × 5 × 5 = 500

Divisibility by 4 depends on the last two digits. I check all 5 × 5 = 25 pairs of even digits:
For each tens digit (0, 2, 4, 6, 8), the units digits giving divisibility by 4 are:
- Tens 0: 00, 04, 08 (3 of 5)
- Tens 2: 20, 24, 28 (3 of 5)
- Tens 4: 40, 44, 48 (3 of 5)
- Tens 6: 60, 64, 68 (3 of 5)
- Tens 8: 80, 84, 88 (3 of 5)

Valid last-two-digit pairs = 15 out of 25.

Numbers divisible by 4 = 4 × 5 × 15 = 300.
Fraction = 300/500 = **3/5**.

ANSWER 3: D

---

Problem 4:
Difference = 6.5% of $20 − 6% of $20 = (6.5% − 6%) × $20 = 0.5% × $20.
0.5% = 0.005, so 0.005 × 20 = $0.10.

ANSWER 4: B

---

Problem 5:
Pair corresponding terms:
(1901 − 101) + (1902 − 102) + … + (1993 − 193)

Each pair equals 1800.
Number of terms: 1993 − 1901 + 1 = 93 terms (and 193 − 101 + 1 = 93 terms).
Total = 93 × 1800 = **167,400**.

ANSWER 5: A

---

Problem 6:
The sign pattern is: +, −, −, +, +, −, −, +, …
Group into blocks of 4:
(1 − 2 − 3 + 4) + (5 − 6 − 7 + 8) + (9 − 10 − 11 + 12) + …

Each block: n − (n+1) − (n+2) + (n+3) = n − n − 1 − n − 2 + n + 3 = **0**.

Total terms: 1996. Number of blocks: 1996/4 = 499 complete blocks.
Sum = 499 × 0 = **0**.

ANSWER 6: C

---

Problem 7:
Let B = number wearing both sunglasses and caps.
P(sunglasses | cap) = B / (people wearing caps) = B / 35 = 2/5.
B = 35 × (2/5) = 14.

Now find P(cap | sunglasses) = B / (people wearing sunglasses) = 14 / 50 = **7/25**.

ANSWER 7: B

---

Problem 8:
Three-digit numbers with digit sum 26. Maximum digit sum = 9+9+9 = 27, so we must subtract 1 from one digit of 999:
Possible digit sets: (8,9,9), (9,8,9), (9,9,8) → numbers 899, 989, 998 (and permutations: 899, 989, 997... wait).

Actually the numbers are: 899, 989, 998 — no, let me list properly:
- (8,9,9) → 899, 989, 998... wait, (8,9,9) gives 899, 989, 998? No: digits are 8,9,9 so numbers are 899, 989, 998.
Wait: (8,9,9) → 899, 989, 998. But 998 has digits 9,9,8 which is the set (9,9,8). Let me just list all:
- Digits {8,9,9}: 899, 989, 998 — wait, 998 has digits 9,9,8. That IS {8,9,9}. So: 899, 989, 998.

Hmm, let me redo. The three numbers with digit sum 26 are:
- 899 (8+9+9=26)
- 989 (9+8+9=26)
- 998 (9+9+8=26)

Wait, that's only 3 numbers total (since two digits are the same, there are 3!/2! = 3 arrangements).

Even numbers must end in an even digit. Only 998 ends in 8 (even). 899 and 989 end in 9 (odd).

So exactly **1** even number.

ANSWER 8: A

---

Problem 9:
Three-digit numbers a, b, c with a + b + c = 25, where 1 ≤ a ≤ 9, 0 ≤ b, c ≤ 9.

Enumerate by first digit a:
- a = 9: b + c = 16 → (7,9), (8,8), (9,7) → 979, 988, 997 (3 numbers)
- a = 8: b + c = 17 → (8,9), (9,8) → 889, 898 (2 numbers)
- a = 7: b + c = 18 → (9,9) → 799 (1 number)
- a ≤ 6: b + c ≥ 19, but max b + c = 18 → impossible

Total = 3 + 2 + 1 = **6**.

ANSWER 9: C

---

Problem 10:
Exits 5 through 26 give 22 exits and 21 consecutive gaps. Total distance = 118 miles. Each gap ≥ 5 miles.

To maximize one gap, minimize the other 20 gaps to 5 miles each:
Max gap + 20(5) = 118
Max gap + 100 = 118
Max gap = **18** miles.

ANSWER 10: C

---

Problem 11:
Five scores sorted: a ≤ b ≤ c ≤ d ≤ e.
- Mean = 90 → sum = 450
- Median = 91 → c = 91
- Mode = 94 → 94 appears most frequently (at least twice)

Since a ≤ b ≤ c = 91, neither a nor b can be 94. So 94 must appear among d and e. For 94 to be the mode, we need d = e = 94 (appearing twice).

Sum: a + b + 91 + 94 + 94 = 450
a + b = 450 − 279 = **171**

(We can verify valid scores 

 Problem 1:
We are given a set \(R\) of nine distinct integers, six of which are \(2,3,4,6,9,14\). We must determine how many different values are possible for the median of \(R\).

Because there are nine distinct integers, the median is the fifth smallest element when the set is listed in increasing order. Let the three unknown distinct integers be chosen freely (as long as they are not equal to any of the six given numbers). We examine two cases: when the median is one of the six known numbers and when it is one of the three unknown numbers.

**Case 1: The median is a known number.**
Let the median be \(m\). Of the six known numbers, let \(L\) be how many are less than \(m\). Then there are \(5-L\) known numbers greater than \(m\). To make \(m\) the fifth element, we need exactly four numbers smaller than \(m\) and four larger. Thus we need \(4-L\) unknown numbers below \(m\) and \(4-(5-L)=L-1\) unknown numbers above \(m\). This uses exactly \((4-L)+(L-1)=3\) unknowns, which is perfect. We need \(4-L\ge 0\) and \(L-1\ge 0\), i.e. \(1\le L\le 4\).

- \(m=2\): \(L=0\) → impossible.
- \(m=3\): \(L=1\) → need 3 unknowns \(<3\), 0 above. Possible (e.g. \(-1,0,1\)).
- \(m=4\): \(L=2\) → need 2 unknowns \(<4\), 1 above. Possible (e.g. \(0,1,5\)).
- \(m=6\): \(L=3\) → need 1 unknown \(<6\), 2 above. Possible (e.g. \(5,7,8\)).
- \(m=9\): \(L=4\) → need 0 unknowns \(<9\), 3 above. Possible (e.g. \(10,11,12\)).
- \(m=14\): \(L=5\) → need \(-1\) unknowns below; impossible.

So the known possible medians are \(3,4,6,9\).

**Case 2: The median is an unknown number \(x\).**
Then \(x\) itself uses one of the three unknown slots. We need four numbers below \(x\) and four above. With \(L\) known numbers below \(x\), we need \(4-L\) other unknowns below \(x\) and \(4-(6-L)=L-2\) other unknowns above \(x\). The total other unknowns used is \((4-L)+(L-2)=2\), which matches the two remaining unknowns. We need \(4-L\ge 0\) and \(L-2\ge 0\), so \(2\le L\le 4\).

- \(L=2\): known below are \(2,3\); known above are \(4,6,9,14\). Then \(3<x<4\), impossible for an integer.
- \(L=3\): known below are \(2,3,4\); known above are \(6,9,14\). Then \(4<x<6\), so \(x=5\). This works (e.g. add \(1,5,7\)).
- \(L=4\): known below are \(2,3,4,6\); known above are \(9,14\). Then \(6<x<9\), so \(x=7\) or \(x=8\). Both work (e.g. add \(7,10,11\) or \(8,10,11\)).

Thus the unknown possible medians are \(5,7,8\).

Collecting all possibilities: \(\{3,4,5,6,7,8,9\}\). That is **7** different values. The tempting answer of 4 counts only the known medians, and 2 or 14 are impossible because too many numbers would have to be forced to one side.

ANSWER 1: D

---

Problem 2:
We insert a number \(N\) into the list \(2,6,7,7,28\) so that the new mean of the six numbers is twice the new median. Find \(N\).

The original sum is \(2+6+7+7+28=50\). After insertion, the sum is \(50+N\) and the mean is \(\frac{50+N}{6}\).

The median depends on the size of \(N\) relative to the ordered list. Because the list contains six numbers, the median is the average of the third and fourth values.

- If \(N\le 6\) (as an integer), the ordered list begins with \(N\) or \(2\), then the third and fourth terms are \(6\) and \(7\), giving median \(\frac{6+7}{2}=6.5\).
  Setting mean \(=2\times 6.5=13\): \(\frac{50+N}{6}=13\Rightarrow N=28\). But \(N=28\) does not satisfy \(N\le 6\), so this case yields no valid solution.

- If \(N\ge 7\) (integer), the third and fourth terms are both \(7\) (since the two \(7\)’s occupy those middle slots, or \(N=7\) makes three \(7\)’s), so the median is \(7\).
  Setting mean \(=2\times 7=14\): \(\frac{50+N}{6}=14\Rightarrow 50+N=84\Rightarrow N=34\).

Check: the list becomes \(2,6,7,7,28,34\). The median is \(\frac{7+7}{2}=7\). The mean is \(\frac{84}{6}=14\), which is indeed twice the median. The other choices fail: \(N=28\) gives mean \(13\) and median \(7\); \(N=14\) gives mean \(\frac{64}{6}\neq 14\).

ANSWER 2: E

---

Problem 3:
We consider all positive four-digit integers whose digits are all even. We need the fraction of them that are divisible by \(4\).

Even digits are \(\{0,2,4,6,8\}\). For a four-digit number \(ABCD\):
- \(A\) (thousands) has 4 choices: \(\{2,4,6,8\}\).
- \(B,C,D\) each have 5 choices.

Total count: \(4\cdot 5\cdot 5\cdot 5 = 500\).

A number is divisible by \(4\) exactly when its last two digits form a multiple of \(4\). Here \(C\) and \(D\) are both even. Consider \(10C+D \pmod 4\). Since \(C\) is even, write \(C=2k\); then \(10C=20k\equiv 0\pmod 4\). Thus \(10C+D\equiv D\pmod 4\). So we simply need \(D\equiv 0\pmod 4\).

Among the even digits, those divisible by \(4\) are \(0,4,8\): **3 choices** for \(D\). The digits \(A,B,C\) are unrestricted (4, 5, and 5 choices).

Count of favorable numbers: \(4\cdot 5\cdot 5\cdot 3 = 300\).

The fraction is \(\frac{300}{500}=\frac{3}{5}\). One might mistakenly think the fraction is \(\frac12\) because half of all integers are divisible by \(4\), but the restriction to even digits changes the units-digit distribution.

ANSWER 3: D

---

Problem 4:
Find the difference between a \(6.5\%\) sales tax and a \(6\%\) sales tax on a \(\$20\) item.

Method 1: Compute each tax.
- \(6.5\%\) of \(\$20 = 0.065\times 20 = \$1.30\).
- \(6\%\) of \(\$20 = 0.06\times 20 = \$1.20\).

Difference: \(\$1.30-\$1.20 = \$0.10\).

Method 2: The difference in rates is \(0.5\%\), and \(0.5\%\) of \(\$20 = 0.005\times 20 = \$0.10\).

The distractors correspond to misplacing the decimal (e.g. \(\$1\) would be \(5\%\) of \(\$20\)).

ANSWER 4: B

---

Problem 5:
Evaluate \((1901+1902+\dots+1993)-(101+102+\dots+193)\).

Each sum has the same number of terms:
\(1993-1901+1 = 93\) terms, and \(193-101+1 = 93\) terms.

Pair the terms:
\((1901-101)+(1902-102)+\dots+(1993-193)\).

Each difference equals \(1800\). With 93 such pairs, the total is
\(93\times 1800 = 93\times 18\times 100 = 1674\times 100 = 167{,}400\).

A common error is miscounting the number of terms as 92, which would give \

Problem 1:
R is a set of nine distinct integers with six known elements: 2, 3, 4, 6, 9, 14. Since R has 9 elements, the median is the 5th element when sorted.

I need to determine which integers can be the 5th element. The median must have exactly 4 elements less than it and 4 elements greater than it.

**Checking each candidate:**

- **Median = 3**: Need 4 elements < 3. Known: {2}. Need 3 unknowns all < 3 (e.g., 1, 0, -1). Elements > 3: {4, 6, 9, 14} = 4. ✓
- **Median = 4**: Need 4 elements < 4. Known: {2, 3}. Need 2 unknowns < 4. Elements > 4: {6, 9, 14} + 1 unknown > 4. ✓
- **Median = 5**: Need 4 elements < 5. Known: {2, 3, 4}. Need 1 unknown < 5. Elements > 5: {6, 9, 14} + 1 unknown > 5. (5 is an unknown) ✓
- **Median = 6**: Need 4 elements < 6. Known: {2, 3, 4}. Need 1 unknown < 6. Elements > 6: {9, 14} + 2 unknowns > 6. ✓
- **Median = 7**: Need 4 elements < 7. Known: {2, 3, 4, 6} = 4. Elements > 7: {9, 14} + 2 unknowns > 7. (7 is an unknown) ✓
- **Median = 8**: Need 4 elements < 8. Known: {2, 3, 4, 6} = 4. Elements > 8: {9, 14} + 2 unknowns > 8. (8 is an unknown) ✓
- **Median = 9**: Need 4 elements < 9. Known: {2, 3, 4, 6} = 4. Elements > 9: {14} + 3 unknowns > 9. ✓

**Values that cannot be median:**
- 2: Need 4 elements < 2, but only 3 unknowns available. ✗
- 10-13: Already have 5 known elements less than these. ✗
- 14: Already have 5 known elements less than 14. ✗

Possible median values: {3, 4, 5, 6, 7, 8, 9} → **7 values**

ANSWER 1: D

---

Problem 2:
List: 2, 6, 7, 7, 28. Insert N so that mean = 2 × median.

Sum of original list = 50, so new mean = (50 + N)/6.

**Finding the median based on N:**
- If N > 7: Sorted list is 2, 6, 7, 7, 28, N (or 2, 6, 7, 7, N, 28 if 7 < N < 28). Median = (7+7)/2 = 7.
- If N ≤ 6: Median = (6+7)/2 = 6.5
- If 6 < N ≤ 7: Median = (N+7)/2

**Solving for N > 7:**
(50 + N)/6 = 2(7) = 14
50 + N = 84
N = 34

Verification: List = 2, 6, 7, 7, 28, 34. Mean = 84/6 = 14. Median = 7. ✓

ANSWER 2: E

---

Problem 3:
Four-digit integers with all even digits. Even digits: {0, 2, 4, 6, 8}.

**Total count:** First digit: 4 choices (2,4,6,8), others: 5 choices each → 4 × 5³ = 500

**Divisible by 4:** Last two digits must form a number divisible by 4.

Listing all 25 pairs of even last digits and checking divisibility by 4:
- 00✓, 02✗, 04✓, 06✗, 08✓
- 20✓, 22✗, 24✓, 26✗, 28✓
- 40✓, 42✗, 44✓, 46✗, 48✓
- 60✓, 62✗, 64✓, 66✗, 68✓
- 80✓, 82✗, 84✓, 86✗, 88✓

That's 15 valid pairs.

**Count divisible by 4:** 4 × 5 × 15 = 300

**Fraction:** 300/500 = 3/5

ANSWER 3: D

---

Problem 4:
6.5% tax on $20 = 0.065 × 20 = $1.30
6% tax on $20 = 0.06 × 20 = $1.20

Difference = $1.30 - $1.20 = $0.10

ANSWER 4: B

---

Problem 5:
(1901 + 1902 + ... + 1993) − (101 + 102 + ... + 193)

Both sums have 93 terms each.

First sum = 93 × (1901 + 1993)/2 = 93 × 1947
Second sum = 93 × (101 + 193)/2 = 93 × 147

Difference = 93 × (1947 − 147) = 93 × 1800 = 167,400

ANSWER 5: A

---

Problem 6:
1 − 2 − 3 + 4 + 5 − 6 − 7 + 8 + ... + 1993 − 1994 − 1995 + 1996

The pattern repeats every 4 terms: +, −, −, +

Group k: (4k−3) − (4k−2) − (4k−1) + 4k = 4k−3−4k+2−4k+1+4k = 0

Each group of 4 sums to 0.

1996 ÷ 4 = 499 complete groups.

Total = 499 × 0 = 0

ANSWER 6: C

---

Problem 7:
50 people wear sunglasses, 35 wear caps.

P(sunglasses | cap) = 2/5 means: (both)/(caps) = 2/5

Both/35 = 2/5 → Both = 14

P(cap | sunglasses) = (both)/(sunglasses) = 14/50 = 7/25

ANSWER 7: B

---

Problem 8:
3-digit numbers with digit-sum 26 that are even.

For number abc: a + b + c = 26, c is even, a ≥ 1, all digits ≤ 9.

- c = 8: a + b = 18 → only (9,9) → 998 ✓
- c = 6: a + b = 20 → impossible (max 9+9=18)
- c = 4, 2, 0: a + b ≥ 22 → impossible

Only 1 number: 998

ANSWER 8: A

---

Problem 9:
Three-digit numbers with digit-sum 25.

a + b + c = 25, a ≥ 1, all digits ≤ 9.

- a = 9: b + c = 16 → (7,9), (8,8), (9,7) → 3 numbers
- a = 8: b + c = 17 → (8,9), (9,8) → 2 numbers
- a = 7: b + c = 18 → (9,9) → 1 number
- a ≤ 6: b + c ≥ 19 → impossible

Total = 3 + 2 + 1 = 6

ANSWER 9: C

---

Problem 10:
Distance from exit 5 to exit 26 is 118 miles. There are 21 gaps between consecutive exits (5→6, 6→7, ..., 25→26).

Each gap ≥ 5 miles. To maximize one gap, minimize all others.

If 20 gaps are minimized at 5 miles each: 20 × 5 = 100 miles.

Maximum single gap = 118 − 100 = 18 miles.

ANSWER 10: C

---

Problem 11:
Five scores: mean = 90, median = 91, mode = 94.

Sum = 5 × 90 = 450

Sorted: a ≤ b ≤ c ≤ d ≤ e
- c = 91 (median)
- Mode = 94, so 94 appears at least twice. Since c = 91, the 94s must be d and e.

a + b + 91 + 94 + 94 = 450
a + b = 171

ANSWER 11: B

---

Problem 12:
Using +, −, × each exactly once in 5 __ 4 __ 6 __ 3:

| Operations | Result |
|------------|--------|
| 5 + 4 − 6 × 3 | 5 + 4 − 18 = −9 |
| 5 + 4 × 6 − 3 | 5 + 24 − 3 = 26 |
| 5 − 4 + 6 × 3 | 5 − 4 + 18 = 19 |
| 5 − 4 × 6 + 3 | 5 − 24 + 3 = −16 |
| 5 × 4 + 6 − 3 | 20 + 6 − 3 = 23 |
| 5 × 4 − 6 + 3 | 20 − 6 + 3 = 17 |

From choices {9, 10, 15, 16, 19}, only **19** is achievable.

ANSWER 12: E

**Problem 1:**  
We have nine distinct integers, six of which are known: 2, 3, 4, 6, 9, 14.  
The median of a nine‑element set is the 5th smallest number.  

Let the unknown integers be \(x,y,z\) (all distinct and distinct from the known six).  
When the numbers are sorted, the median is the 5th term.  

For a known number \(k_i\) (the \(i\)‑th smallest of the known six) to be the median, we need exactly four numbers smaller than it.  
If \(i-1\) known numbers are smaller, we need \(L\) unknown numbers smaller with  
\((i-1)+L=4\Rightarrow L=5-i\).  
Since we have only three unknowns, \(L\ge0\) gives \(i\le5\); also \(L\le3\) gives \(i\ge2\).  
Thus \(i=2,3,4,5\) → medians could be 3, 4, 6, 9.

Can the median be an unknown integer?  
Let the median be an unknown \(u\).  
If \(u\) lies between the known numbers, we need exactly four numbers below it.  
With two other unknowns, the possibilities are:
- \(u\) between 4 and 6 → \(u=5\);
- \(u\) between 6 and 9 → \(u=7\) or \(u=8\).

All three of these can be realized while keeping the numbers distinct.  
Thus the possible median values are \(\{3,4,5,6,7,8,9\}\), a total of **7** values.

**Answer:** 7 → Choice **D**.  

ANSWER 1: D  

---

**Problem 2:**  
We start with the list 2, 6, 7, 7, 28 and insert a number \(N\).  
Let the six numbers be sorted. The median of six numbers is the average of the 3rd and 4th smallest values.

Let the sorted list after insertion be \(a_1\le a_2\le a_3\le a_4\le a_5\le a_6\).  
The median is \(\dfrac{a_3+a_4}{2}\).  
The mean is \(\dfrac{\text{sum}}{6}\).  
Given that the mean equals twice the median:
\[
\frac{\text{sum}}{6}=2\cdot\frac{a_3+a_4}{2}=a_3+a_4 .
\]
Hence \(\text{sum}=6(a_3+a_4)\).

Consider where \(N\) can fall:

*If \(N\le2\):* \(a_3=6,\ a_4=7\) → median \(=6.5\).  
Sum \(=N+50\). Equation gives \(N=28\), contradicting \(N\le2\).

*If \(2<N\le6\):* \(a_3=6,\ a_4=7\) → same median \(=6.5\).  
Again \(N=28\), impossible.

*If \(6<N\le7\):* only integer possibility \(N=7\).  
List becomes 2, 6, 7, 7, 7, 28 → median \(=7\).  
Sum \(=57\) → mean \(=9.5\neq14\).

*If \(7<N<28\):* \(a_3=7,\ a_4=7\) → median \(=7\).  
Sum \(=N+50\). Equation \(N+50=84\) gives \(N=34\), which is not \(<28\).

*If \(N=28\):* median still \(=7\). Sum \(=78\) → mean \(=13\neq14\).

*If \(N>28\):* \(a_3=7,\ a_4=7\) → median \(=7\).  
Equation gives \(N+50=84\) → \(N=34\). This works (34 > 28).

Thus the only possible value is **34**.

**Answer:** 34 → Choice **E**.  

ANSWER 2: E  

---

**Problem 3:**  
A four‑digit integer with all digits even can use digits \{0,2,4,6,8\} for each place, except the thousands digit cannot be 0.  

Number of such integers: \(4\times5\times5\times5=500\).

A number is divisible by 4 iff its last two digits form a number divisible by 4.  
List all two‑digit endings with even digits (25 possibilities).  

Checking each yields 15 endings that are multiples of 4:  
00, 04, 08, 20, 24, 28, 40, 44, 48, 60, 64, 68, 80, 84, 88.

Thus the proportion is \(\frac{15}{25}=\frac{3}{5}\).

**Answer:** \(\frac{3}{5}\) → Choice **D**.  

ANSWER 3: D  

---

**Problem 4:**  
6 % tax on \$20 = \(0.06\times20 = \$1.20\).  
6.5 % tax on \$20 = \(0.065\times20 = \$1.30\).  

Difference = \$1.30 − \$1.20 = **\$0.10**.

**Answer:** \$0.10 → Choice **B**.  

ANSWER 4: B  

---

**Problem 5:**  
Sum from 1901 to 1993 (inclusive):  
Number of terms = \(1993-1901+1 = 93\).  
Sum = \(\frac{(1901+1993)\times93}{2}= \frac{3894\times93}{2}=181\,071\).

Sum from 101 to 193 (inclusive):  
Number of terms = \(193-101+1 = 93\).  
Sum = \(\frac{(101+193)\times93}{2}= \frac{294\times93}{2}=13\,671\).

Difference = \(181\,071-13\,671 = 167\,400\).

**Answer:** 167 400 → Choice **A**.  

ANSWER 5: A  

---

**Problem 6:**  
The sign pattern for the numbers \(n\) is:
- \(n\equiv1\pmod4\) → +,
- \(n\equiv2\pmod4\) → −,
- \(n\equiv3\pmod4\) → −,
- \(n\equiv0\pmod4\) → +.

Thus each block of four consecutive integers \(\{4k+1,4k+2,4k+3,4k+4\}\) contributes  
\((4k+1)-(4k+2)-(4k+3)+(4k+4)=0\).

From 1 through 1996 there are \(1996/4=499\) complete blocks, so the total sum is **0**.

**Answer:** 0 → Choice **C**.  

ANSWER 6: C  

---

**Problem 7:**  
Let \(S\) = people wearing sunglasses, \(|S|=50\).  
Let \(C\) = people wearing caps, \(|C|=35\).  
Let \(x=|S\cap C|\).

Given \(P(S\mid C)=\frac{x}{35}= \frac{2}{5}\).  
Thus \(x = \frac{2}{5}\times35 = 14\).

We need \(P(C\mid S)=\frac{x}{50}= \frac{14}{50}= \frac{7}{25}\).

**Answer:** \(\frac{7}{25}\) → Choice **B**.  

ANSWER 7: B  

---

**Problem 8:**  
A three‑digit number with digit sum 26 must use digits that sum to 26.  
The only triple of single‑digit numbers with sum 26 is \(\{9,9,8\}\).

The number is even only if its units digit is 8.  
Among the permutations of 9, 9, 8, only 998 ends in 8.  

Thus exactly **1** such number exists.

**Answer:** 1 → Choice **A**.  

ANSWER 8: A  

---

**Problem 9:**  
Let the digits be \(a,b,c\) (hundreds, tens, units) with  
\(1\le a\le9,\ 0\le b,c\le9,\ a+b+c=25\).

Set \(a'=a-1\) (so \(0\le a'\le8\)). Then  
\(a'+b+c=24\).

Count unrestricted non‑negative solutions: \(\binom{24+3-1}{2}= \binom{26}{2}=325\).

Subtract solutions where a digit exceeds its bound:
- \(a'\ge9\): treat \(a''=a'-9\) → \(a''+b+c=15\) → \(\binom{17}{2}=136\).
- \(b\ge10\): \(b'=b-10\) → \(a'+b'+c=14\) → \(\binom{16}{2}=120\).
- \(c\ge10\): similarly 120.

Intersections:
- \(a'\ge9,\ b\ge10\): \(a''+b'+c=5\) → \(\binom{7}{2}=21\).
- \(a'\ge9,\ c\ge10\): 21.
- \(b\ge10,\ c\ge10\): \(a'+b'+c'=4\) → \(\binom{6}{2}=15\).

No triple intersection (would require sum –5).

By inclusion–exclusion, invalid = \(136+120+120-21-21-15 = 319\).  
Valid = \(325-319 = 6\).

**Answer:** 6 → Choice **C**.  

ANSWER 9: C  

---

**Problem 10:**  
Exits numbered 5 through 26 give 22 exits, hence 21 intervals.  
Let the distances between consecutive exits be \(d_1,\dots,d_{21}\).  
We know \(\sum d_i = 118\) miles, and each \(d_i\ge5\).

To maximize the largest interval, make the othe

Problem 1:  
We are given a set $ R $ of nine distinct integers, six of which are $ 2, 3, 4, 6, 9, 14 $. The median of a set of nine numbers is the fifth number when sorted. To find the number of possible values for the median, we consider the possible placements of the three unknown numbers relative to the known ones. The known numbers sorted are $ 2, 3, 4, 6, 9, 14 $. The three unknown numbers can be placed in various positions to shift the median. The possible medians occur when the additional numbers are placed such that the fifth number in the sorted list can be $ 4, 5, 6, 7, 8, $ or $ 9 $. This gives six possible values for the median.  
ANSWER 1: D (corrected to C? Wait, no: the possible medians are 4, 5, 6, 7, 8, 9 — that’s 6 values, so C)  
**Wait, let's re-evaluate:** The known numbers are 2,3,4,6,9,14. Adding three numbers. The median is the 5th number. To get 4 as median: three numbers ≤4, but only 2,3,4 are available — but they are distinct, so we can add numbers like 1,5? Wait no, the three unknown numbers can be placed in different positions. For example:  
- If all three unknown numbers are ≤4: but only three such numbers exist (2,3,4) but they are already in the set. So we cannot add more. Thus, the unknown numbers must be placed in other positions.  
Actually, the unknown numbers can be:  
- All less than 2: median could be 4 if three numbers are added <4, but only 2,3,4 are already present. We cannot add numbers less than 2 without repeating? Wait, the set has distinct integers. So we can add numbers like 1,0,-1, but the problem doesn’t restrict to positive. However, the known numbers are positive. But the problem says "distinct integers", so negatives are allowed. But the median values: if we add numbers less than 2, the sorted order becomes [new numbers, 2,3,4,6,9,14]. With three new numbers, the sorted list has 9 numbers. The fifth is the median. If we add three numbers less than 2, the sorted list is [a,b,c,2,3,4,6,9,14] — no, wait, adding three numbers less than 2: the sorted list would be [a,b,c,2,3,4,6,9,14] — but a,b,c are less than 2, so the first three, then 2 is fourth, 3 is fifth? Wait, no: the original known numbers are 2,3,4,6,9,14. Adding three numbers, say x,y,z. The full sorted list will be the combination. To get median=4, we need the fifth number to be 4. So in the sorted list of 9, the fifth must be 4. That requires that four numbers are ≤4 and four are ≥4? No: the fifth is the middle. For the fifth to be 4, we need at least five numbers ≤4? No: in a sorted list of 9, the fifth is the median. For the fifth to be 4, we need at least five numbers ≤4? Actually, no: the fifth number is the one where four are less and four are greater? No: in a sorted list of 9, the fifth is the median, meaning four are ≤ median and four are ≥ median. But since numbers can be equal, but here all distinct. So for 4 to be the median, we need exactly four numbers less than 4 and four numbers greater than 4? But 4 is in the set. So the sorted list must have four numbers <4, then 4, then four numbers >4. But the known numbers include 2,3,4,6,9,14. So 2,3 are <4, and 6,9,14 are >4. So currently, without unknowns, we have two <4 and three >4. Adding three unknowns. To have four <4 and four >4, we need to add two numbers <4 and one number >4? But then the sorted list would be [two new <4, 2,3,4, then the new >4, then 6,9,14] — but that’s only three numbers after 4? Wait, no: the sorted list would be [new <4 (two), 2,3,4, new >4, 6,9,14] — that’s 9 numbers? 2+3+1+3=9? Wait, no: the two new <4, then 2,3,4 (three numbers), then the new >4, then 6,9,14 (three numbers). So total: 2+3+1+3=9. The fifth number is 4. So yes, we can achieve median=4 by adding two numbers <4 and one number >4. Similarly, we can achieve:  
- Median=5: by adding numbers such that the fifth number is 5. But 5 is not in the set. So we need the fifth number to be 5. This requires that four numbers are <5 and four are >5. But 5 is not present. So we need to add 5? But the set must consist of distinct integers, and 5 is not in the given. So we can add 5. Then the sorted list would include 5. For example, add 5, and two other numbers. To have four <5 and four >5: currently, without unknowns, numbers <5: 2,3,4 (three), numbers >5: 6,9,14 (three). So to have four <5, we need to add one number <5. To have four >5, we need to add one number >5. But we are adding three numbers. So we can add 5, and say 1 and 10. Then sorted: 1,2,3,4,5,6,9,10,14. Median is 5. So median=5 is achievable. Similarly, median=6: we can add numbers such that the fifth is 6. Currently, numbers <6: 2,3,4 (three), numbers >6: 9,14 (two). So to have four <6 and four >6, we need to add one number <6 and two numbers >6. For example, add 5,10,15. Then sorted: 2,3,4,5,6,9,10,14,15. Median=6. Similarly, median=7: add numbers so fifth is 7. Currently, <7: 2,3,4,6 (four), >7: 9,14 (two). So to have four <7 and four >7, we need to add zero <7 (since already four) and two >7. But we are adding three numbers. So we can add 7,8,15. Then sorted: 2,3,4,6,7,8,9,14,15. Median=7. Similarly, median=8: add 8, and two numbers, say 10,11. Then sorted: 2,3,4,6,8,9,10,11,14. Median=8? The fifth is 8. But to have four <8 and four >8: currently, without unknowns, <8: 2,3,4,6 (four), >8:9,14 (two). So to have four >8, we need to add two numbers >8. But we are adding three numbers. So we can add 8,9,10? But 9 is already there. Add 8,10,11. Then sorted: 2,3,4,6,8,9,10,11,14. The fifth is 8. So yes. Similarly, median=9: add numbers so fifth is 9. Currently, <9: 2,3,4,6 (four), >9:14 (one). So to have four >9, we need to add three numbers >9. But we are adding three numbers. So add 10,11,12. Then sorted: 2,3,4,6,9,10,11,12,14. Median=9. So possible medians: 4,5,6,7,8,9. That’s six values.  
ANSWER 1: C  

Problem 2:  
We are given the list 2,6,7,7,28. A number $ N $ is added. The mean becomes twice the median. The original list has five numbers. After adding $ N $, there are 

### Problem 1:
We have a set of 9 distinct integers, so the median (middle value when sorted) is the 5th element. The 6 known numbers sorted are [2,3,4,6,9,14], with 3 additional distinct integers to add. To find possible median values:
- The minimum possible median is 3: add 3 numbers smaller than 2, sorted set becomes [a,b,c,2,3,4,6,9,14], 5th term=3.
- The maximum possible median is 9: add 3 numbers larger than 14, sorted set becomes [2,3,4,6,9,14,x,y,z], 5th term=9.
- All integers between 3 and 9 are valid medians: 3,4,5,6,7,8,9, for 7 total possible values.
Tempting wrong choices: any count less than 7, from missing intermediate medians that can be formed by adding integers between the known values.
ANSWER 1: D

---

### Problem 2:
Original list sum: 2+6+7+7+28=50. Adding N gives a list of 6 elements, sum=50+N, mean=(50+N)/6. The median of 6 elements is the average of the 3rd and 4th sorted terms. The problem states mean=2×median, so 50+N=12×median.
- If N>28, sorted list is [2,6,7,7,28,N], median=(7+7)/2=7. Then 50+N=12×7=84 → N=34, which fits.
Testing other options confirms no other N satisfies the condition.
ANSWER 2: E

---

### Problem 3:
A number is divisible by 4 if its last two digits form a number divisible by 4. For 4-digit numbers with all even digits:
- Total count: 4 (non-zero thousands digits:2,4,6,8) ×5×5×5 (all even digits for other places)=500.
- Valid last two-digit pairs (both even, divisible by 4): For any even tens digit, 10×even is divisible by 4, so only units digits 0,4,8 work, giving 3 valid units digits per tens digit: 5×3=15 valid pairs.
- Fraction: (4×5×15)/500=300/500=3/5.
ANSWER 3: D

---

### Problem 4:
The tax rate difference is 6.5%-6%=0.5%. 0.5% of $20 is 0.005×20=$0.10.
ANSWER 4: B

---

### Problem 5:
Both sums have the same number of terms: 1993-1901+1=93 terms, 193-101+1=93 terms. Subtract term-by-term: (1901-101)+(1902-102)+...+(1993-193)=1800×93=167400.
ANSWER 5: A

---

### Problem 6:
Group terms in sets of 4: (1-2-3+4)+(5-6-7+8)+... each group sums to 0. There are 1996/4=499 groups, total sum=0.
ANSWER 6: C

---

### Problem 7:
Number of people with both caps and sunglasses: 35 cap-wearers × 2/5=14. Probability a sunglass-wearer has a cap:14/50=7/25.
ANSWER 7: B

---

### Problem 8:
An even 3-digit number with digit sum 26 must have an even units digit. The only possible even units digit is 8 (any smaller even digit leaves a sum of the first two digits >18, impossible). The only number is 998, so 1 valid number.
ANSWER 8: A

---

### Problem 9:
Three-digit numbers with digit sum 25 are all permutations of (9,9,7) and (9,8,8). There are 3 permutations of (9,9,7) and 3 permutations of (9,8,8), for 6 total numbers.
ANSWER 9: C

---

### Problem 10:
Between exit 5 and 26, there are 21 gaps. To maximize one gap, minimize the other 20 gaps at 5 miles each: 20×5=100 miles. Maximum gap=118-100=18 miles.
ANSWER 10: C

---

### Problem 11:
Total sum of 5 scores=5×90=450. Median=91, so sorted scores [a,b,91,c,d]. Mode=94, so 94 must appear at least twice, so c=d=94, sum c+d=188. Sum of the two lowest scores=450-91-188=171.
ANSWER 11: B

---

### Problem 12:
Using +,-,* exactly once, one valid calculation: 5-4+6×3=1+18=19, which uses all three operations and matches option E.
ANSWER 12: E

### Problem 1:
We need to find the number of possible median values for a set of 9 distinct integers, 6 of which are 2, 3, 4, 6, 9, 14. The median of 9 numbers is the 5th element when sorted in ascending order. We add 3 distinct integers (not in the given set) and analyze possible medians:
1.  **Case 1: All 3 new numbers < 2**: Sorted list is [a,b,c,2,3,4,6,9,14], median = 3.
2.  **Case 2: 2 new numbers < 2, 1 new number ≥4**: Sorted list is [a,b,2,3,4,...], median =4.
3.  **Case 3: 1 new number <2**:
    - If we add 5 (the only integer between 4 and 6): Sorted list is [a,2,3,4,5,6,...], median=5.
    - If we do not add 5: Sorted list is [a,2,3,4,6,...], median=6.
4.  **Case 4: 0 new numbers <2 (all new numbers ≥5)**:
    - Add 5: Sorted list [2,3,4,5,6,...], median=6.
    - Add a number between 6 and9 (7 or 8): Sorted list [2,3,4,6,7,...] or [2,3,4,6,8,...], median=7 or 8.
    - Add all new numbers ≥9: Sorted list [2,3,4,6,9,...], median=9.
The only impossible medians are 2 (needs 5 numbers ≤2, only 4 max) and values >9 (existing numbers already have 5 elements ≤9). The possible medians are 3,4,5,6,7,8,9: 7 values.
ANSWER 1: D
---
### Problem 2:
Original list: [2,6,7,7,28], sum = 2+6+7+7+28=50. Insert N to make 6 elements. The median of 6 elements is the average of the 3rd and 4th sorted elements. The mean is $\frac{50+N}{6}$, which equals twice the median:
$$\frac{50+N}{6} = 2 \times \text{median} \implies 50 + N = 12 \times \text{median} \implies N = 12 \times \text{median} -50.$$
Test possible median values from the sorted list with N:
- If N ≤7: Sorted list has 3rd=7, 4th=7, median=7. Then $N=12*7 -50=84-50=34$, which is >7, valid.
- Other median values (e.g., 6.5) give N=28, which does not fit the required sorted position.
Verify N=34: Sorted list [2,6,7,7,28,34], median=(7+7)/2=7, mean=(50+34)/6=14, which is twice 7.
ANSWER 2: E
---
### Problem 3:
Positive 4-digit numbers with all even digits:
- Thousands place: 2,4,6,8 (4 choices, no 0).
- Hundreds, tens, units: 0,2,4,6,8 (5 choices each).
Total numbers: $4 \times 5 \times 5 \times5=500$.
A number is divisible by 4 iff its last two digits form a number divisible by 4. For even tens digits (0,2,4,6,8), a two-digit number $10a + b$ is divisible by 4 iff $b$ (units digit) is 0,4,8 (since $10a \equiv 0 \mod4$ for even a). So valid endings: 5 (tens choices) ×3 (units choices)=15.
Total valid numbers: $4 \times5 \times5 \times3=300$. Fraction: $\frac{300}{500}=\frac{3}{5}$.
ANSWER 3: D
---
### Problem 4:
Difference between 6.5% and 6% tax on $20:
$$0.065*20 - 0.06*20 = (0.065-0.06)*20=0.005*20=0.10.$$
ANSWER 4: B
---
### Problem 5:
Both sums have 93 terms (1993-1901+1=93, 193-101+1=93). Each term in the first sum is 1800 more than the corresponding term in the second sum:
Total difference = $93 \times 1800 = 167400$.
ANSWER 5: A
---
### Problem 6:
The sign pattern repeats every 4 terms: $+ - - +$, and each block of 4 sums to 0:
$$(1-2-3+4)+(5-6-7+8)+\dots+(1993-1994-1995+1996)=0.$$
1996 is divisible by 4, so there are 499 such blocks, total sum 0.
ANSWER 6: C
---
### Problem 7:
Let $S$ = sunglass wearers ($|S|=50$), $C$ = cap wearers ($|C|=35$). $P(S|C)=\frac{|S∩C|}{|C|}=\frac{2}{5}$, so $|S∩C|=\frac{2}{5}*35=14$.
The probability a sunglass wearer wears a cap is $P(C|S)=\frac{|S∩C|}{|S|}=\frac{14}{50}=\frac{7}{25}$.
ANSWER 7: B
---
### Problem 8:
3-digit numbers with digit sum 26: maximum digit sum is 27 (999), so 26 is 1 less than maximum, meaning digits are permutations of (9,9,8). The only even permutation is 998 (units digit 8). So 1 number.
ANSWER 8: A
---
### Problem 9:
3-digit numbers with digit sum 25: 25 is 2 less than 27, so digits are permutations of (9,9,7) and (9,8,8):
- Permutations of (9,9,7): 997,979,799 (3 numbers).
- Permutations of (9,8,8): 988,898,889 (3 numbers).
Total 6 numbers.
ANSWER 9: C
---
### Problem 10:
Between exit 5 and 26, there are $26-5=21$ gaps, total distance 118 miles. Each gap ≥5 miles. To maximize one gap, minimize the other 20: $20*5=100$, so maximum gap = $118-100=18$.
ANSWER 10: C
---
### Problem 11:
Five sorted scores: $a≤b≤c≤d≤e$. Median $c=91$, mean 90, so total sum $5*90=450$, so $a+b+d+e=450-91=359$. Mode is 94, so 94 appears at least twice. Since $c=91$, 94 can only be in $d$ and $e$, so $d=e=94$. Then $a+b=359-94-94=171$. This sum is fixed (valid examples exist, e.g., 81+90=171).
ANSWER 11: B
---
### Problem 12:
Permute +, -, × in 5 __4 __6 __3, using standard order of operations (multiply first):
- $5 -4 +6×3 = 5-4+18=19$, which is an option.
All other permutations give values not in the choices.
ANSWER 12: E

I'll solve each problem step by step.

---

**Problem 1:**

We have 6 elements: {2, 3, 4, 6, 9, 14}, and need to add 3 more distinct integers to make 9 total.

The median of 9 numbers is the 5th element when arranged in order.

Currently we have 6 numbers. By strategically placing 3 new numbers, we can shift which value becomes the median.

Let me consider what values can be the median (5th position):
- If we add 3 numbers all ≤ 2, the sorted list has our 6 original numbers in positions 4-9, so median is 4
- If we add numbers strategically between and around the original set, we can make the median be: 2, 3, 4, 6, 9, 14, or a new number
- If we add 3 numbers all ≥ 14, the sorted list has our 6 original numbers in positions 1-6, so median is 9

Testing systematically:
- Median = 2: add three numbers < 2 (e.g., -2, -1, 0) ✓
- Median = 3: add two < 2, one between 3 and 4 ✓
- Median = 4: add three < 2, or arranged appropriately ✓
- Median = 6: add one or two strategically ✓
- Median = 9: add three > 14, or arranged appropriately ✓
- Median = 14: add three > 14 ✓
- Median = a new value between existing numbers ✓

The possible medians are: 2, 3, 4, 6, 9, 14, and any new integer we insert. We can get 7 different values.

**ANSWER 1: D**

---

**Problem 2:**

Original list: 2, 6, 7, 7, 28. Sum = 50.

When N is inserted, we have 6 numbers. Let's find the median of 6 numbers (average of 3rd and 4th when sorted).

Sum of all 6 numbers = 50 + N
Mean = (50 + N)/6

We need: Mean = 2 × Median

Let me test by considering where N falls:
- If N ≤ 2: sorted list is N, 2, 6, 7, 7, 28. Median = (6+7)/2 = 6.5. Mean = 2×6.5 = 13, so 50+N = 78, N = 28. Contradiction.
- If 2 < N ≤ 6: sorted list is 2, N, 6, 7, 7, 28. Median = (6+7)/2 = 6.5. Mean = 13, so N = 28. Contradiction.
- If 6 < N ≤ 7: sorted list is 2, 6, N, 7, 7, 28. Median = (N+7)/2. Mean = 2×(N+7)/2 = N+7. So (50+N)/6 = N+7, giving 50+N = 6N+42, so 5N = 8, N = 1.6. Not in range.
- If 7 < N ≤ 28: sorted list is 2, 6, 7, N, 7, 28 or 2, 6, 7, 7, N, 28. Median = (7+N)/2. Mean = 2×(7+N)/2 = 7+N. So (50+N)/6 = 7+N, giving 50+N = 42+6N, so 5N = 8, N = 1.6. Not valid.
- If N > 28: sorted list is 2, 6, 7, 7, 28, N. Median = (7+7)/2 = 7. Mean = 14. So 50+N = 84, N = 34. ✓

**ANSWER 2: E**

---

**Problem 3:**

Four-digit numbers with all even digits: first digit ∈ {2,4,6,8}, others ∈ {0,2,4,6,8}.

Total count: 4 × 5 × 5 × 5 = 500

For divisibility by 4, the last two digits must form a number divisible by 4.

Last two digits from {00, 02, 04, 06, 08, 20, 22, 24, 26, 28, 40, 42, 44, 46, 48, 60, 62, 64, 66, 68, 80, 82, 84, 86, 88}

Divisible by 4: 00, 04, 08, 20, 24, 28, 40, 44, 48, 60, 64, 68, 80, 84, 88 (15 out of 25)

Count divisible by 4: 4 × 5 × 15 = 300

But wait, the second digit is 5 choices, third is which determines divisibility by 4.

Actually: 4 × 5 × (number of even two-digit combos divisible by 4) = 4 × 5 × 15 = 300

Hmm, let me recalculate. First digit: 4 choices. Second digit: 5 choices. Last two digits must be divisible by 4: check all pairs from {0,2,4,6,8}×{0,2,4,6,8} giving 15 valid pairs.

Count = 4 × 5 × 15 = 300. Fraction = 300/500 = 3/5.

**ANSWER 3: D**

---

**Problem 4:**

6.5% tax on $20: 0.065 × 20 = $1.30
6% tax on $20: 0.06 × 20 = $1.20
Difference: $1.30 - $1.20 = $0.10

**ANSWER 4: B**

---

**Problem 5:**

Sum from 1901 to 1993: This is 93 consecutive integers starting at 1901.
Sum = (1901 + 1993)/2 × 93 = 1947 × 93 = 181,071

Sum from 101 to 193: This is 93 consecutive integers starting at 101.
Sum = (101 + 193)/2 × 93 = 147 × 93 = 13,671

Difference = 181,071 - 13,671 = 167,400

**ANSWER 5: A**

---

**Problem 6:**

Pattern: 1 - 2 - 3 + 4 + 5 - 6 - 7 + 8 + 9 - 10 - 11 + ...

Group by 4s: (1-2-3+4) + (5-6-7+8) + (9-10-11+12) + ... + (1993-1994-1995+1996)

Each group of 4: (4k+1) - (4k+2) - (4k+3) + (4k+4) = 4k+1-4k-2-4k-3+4k+4 = 0

From 1 to 1996, we have 1996/4 = 499 complete groups, each summing to 0.

Total = 0

**ANSWER 6: C**

---

**Problem 7:**

Let x = people wearing both sunglasses and caps.

P(sunglasses | cap) = x/35 = 2/5, so x = 14.

P(cap | sunglasses) = 14/50 = 7/25.

**ANSWER 7: B**

---

**Problem 8:**

Three-digit even numbers with digit sum 26.

For an even number, last digit ∈ {0, 2, 4, 6, 8}.

If last digit = 0: first two digits sum to 26. Max is 9+9=18. Impossible.
If last digit = 2: first two digits sum to 24. Only 9+9=18 < 24. Impossible.
If last digit = 4: first two digits sum to 22. Impossible (max 18).
If last digit = 6: first two digits sum to 20. Impossible.
If last digit = 8: first two digits sum to 18. This requires 9+9=18. Number: 998. ✓

Only 1 number.

**ANSWER 8: A**

---

**Problem 9:**

Three-digit numbers abc where a+b+c=25, with a ∈ {1,...,9}, b,c ∈ {0,...,9}.

a+b+c = 25 means b+c = 25-a.

For valid solutions: 0 ≤ b+c ≤ 18, so 7 ≤ a ≤ 9.

- a=7: b+c=18. Solutions: (9,9) → 1 way
- a=8: b+c=17. Solutions: (8,9), (9,8) → 2 ways
- a=9: b+c=16. Solutions: (7,9), (8,8), (9,7) → 3 ways

Total = 1 + 2 + 3 = 6

**ANSWER 9: C**

---

**Problem 10:**

Exits from 5th to 26th = 26-5 = 21 segments between consecutive exits.

Each segment ≥ 5 miles. Total distance = 118 miles.

If 20 segments are 5 miles each: 20 × 5 = 100 miles.
Remaining for the largest segment: 118 - 100 = 18 miles.

**ANSWER 10: C**

---

**Problem 11:**

Five scores: mean = 90 (sum = 450), median = 91, mode = 94.

Arranged in order: a, b, c, d, e where c = 91 (median).

Mode = 94 means 94 appears at least twice. Since median is 91, we must have 94 appearing in {d, e}.

Case 1: d = e = 94. Then a + b + 91 + 94 + 94 = 450, so a + b = 171.

Check consistency: we need a ≤ b ≤ 91 and mode is 94 (appears twice). If a or b equals 94, mode wouldn't be unique. So a, b < 91 works. ✓

**ANSWER 11: B**

---

**Problem 12:**

We need to place +, −, × in three blanks: 5 __ 4 __ 6 __ 3

Trying combinations:
- 5 + 4 × 6 − 3 = 5 + 24 − 3 = 26
- 5 × 4 + 6 − 3 = 20 + 6 − 3 = 23
- 5 × 4 − 6 + 3 = 20 − 6 + 3 = 17
- 5 + 4 − 6 × 3 = 5 + 4 − 18 = −9
- 5 − 4 

Problem 1:

We are given a set of 9 distinct integers. Six of them are
\[
2,3,4,6,9,14.
\]
We want the number of possible values of the median of the 9 numbers.

For 9 numbers, the median is the 5th number when the list is ordered.

Now look at the six given numbers in sorted order:
\[
2,3,4,6,9,14.
\]
Since there are 3 other distinct integers, the median can be any number that ends up in the 5th position after inserting those 3 numbers.

Let’s see what values are possible:

- If we place 3 new numbers all very small, then the 5th number could be 4.
- It could also be 6, or 9, or even 14, depending on how many new numbers are inserted before each.
- It cannot be 2 or 3, because there would need to be at least 4 numbers less than or equal to it, but only 3 extra numbers are available and the given numbers already limit this.
- It cannot be any number not among the given six if the median must be one of the existing numbers? Let’s check carefully: since the 3 extra numbers are distinct integers and can be placed anywhere, the 5th number could indeed be one of the inserted numbers, but then it would have to lie between the 4th and 6th numbers. Since the six given numbers already occupy several possible positions, the median can also be a new integer. But because the set is of distinct integers and the question asks how many possible values, we should count all possible medians that can occur.

A cleaner way: the median of 9 distinct integers is the middle element. With 6 fixed numbers, the 3 added numbers can be arranged so that the median is any integer between the 4th smallest and 6th smallest of the fixed numbers, inclusive, but here the possible distinct medians are:
\[
4,5,6,7,8,9
\]
Wait — can all of these really happen? Let’s test.

- Median 4: choose 3 new numbers less than 4.
- Median 5: choose 3 new numbers so that 5 is between 4 and 6 and becomes 5th.
- Median 6: choose 3 new numbers with 6 as middle.
- Median 7, 8, 9 similarly.

So the possible medians are the integers from 4 through 9, which is 6 values.

However, we must match the provided choices. The intended count is the number of possible median values among the listed integers and inserted integers. Since the only restriction is distinct integers, every integer from 4 through 9 can occur.

That gives 6 possible values.

ANSWER 1: C

---

Problem 2:

We insert a number \(N\) into the list
\[
2,6,7,7,28.
\]
Then the mean becomes twice the median.

Let’s analyze the median after inserting \(N\) into 6 numbers total.

The original sorted list is:
\[
2,6,7,7,28.
\]
There are 5 numbers. After inserting one number, there are 6 numbers, so the median is the average of the 3rd and 4th numbers in the sorted list.

We try the choices.

If \(N=7\), the list becomes
\[
2,6,7,7,7,28.
\]
Median:
\[
\frac{7+7}{2}=7.
\]
Mean:
\[
\frac{2+6+7+7+7+28}{6}=\frac{57}{6}=9.5.
\]
Twice the median would be 14, not 9.5, so not 7.

Try \(N=14\):
\[
2,6,7,7,14,28.
\]
Median:
\[
\frac{7+7}{2}=7.
\]
Mean:
\[
\frac{2+6+7+7+14+28}{6}=\frac{64}{6}=\frac{32}{3},
\]
not 14.

Try \(N=20\):
\[
2,6,7,7,20,28.
\]
Median is still 7.
Mean:
\[
\frac{2+6+7+7+20+28}{6}=\frac{70}{6}=\frac{35}{3},
\]
not 14.

Try \(N=28\):
\[
2,6,7,7,28,28.
\]
Median is 7.
Mean:
\[
\frac{2+6+7+7+28+28}{6}=\frac{78}{6}=13,
\]
not 14.

Try \(N=34\):
\[
2,6,7,7,28,34.
\]
Median is 7.
Mean:
\[
\frac{2+6+7+7+28+34}{6}=\frac{84}{6}=14.
\]
Twice the median is
\[
2\cdot 7=14.
\]
This works.

So \(N=34\).

ANSWER 2: E

---

Problem 3:

We consider all positive four-digit integers whose digits are all even.

Even digits are
\[
0,2,4,6,8.
\]
A four-digit number cannot start with 0, so the thousands digit has 4 choices:
\[
2,4,6,8.
\]
Each of the other three digits has 5 choices, so the total number is
\[
4\cdot 5^3.
\]

Now determine how many are divisible by 4.

A number is divisible by 4 if its last two digits form a number divisible by 4.

So we only need to count ordered pairs of even digits \((t,u)\) from \(\{0,2,4,6,8\}\) such that \(10t+u\) is divisible by 4.

Check the possible last two digits:

- \(00\) yes
- \(02\) no
- \(04\) yes
- \(06\) no
- \(08\) yes

- \(20\) yes
- \(22\) no
- \(24\) yes
- \(26\) no
- \(28\) yes

- \(40\) yes
- \(42\) no
- \(44\) yes
- \(46\) no
- \(48\) yes

- \(60\) yes
- \(62\) no
- \(64\) yes
- \(66\) no
- \(68\) yes

- \(80\) yes
- \(82\) no
- \(84\) yes
- \(86\) no
- \(88\) yes

For each tens digit, 3 of the 5 possibilities work, so there are
\[
5\cdot 3=15
\]
valid last-two-digit pairs.

Since the first two digits can be anything allowed:
- thousands digit: 4 choices
- hundreds digit: 5 choices

Total divisible by 4:
\[
4\cdot 5\cdot 15
\]
Total numbers:
\[
4\cdot 5^3 = 4\cdot 125
\]
So the fraction is
\[
\frac{4\cdot 5\cdot 15}{4\cdot 5^3}=\frac{15}{25}=\frac35.
\]

ANSWER 3: D

---

Problem 4:

An item costs \$20 before tax.

Difference between 6.5% tax and 6% tax is
\[
0.5\% = 0.005.
\]
So the difference in tax is
\[
0.005 \times 20 = 0.1
\]
dollar, which is \(\$0.10\).

ANSWER 4: B

---

Problem 5:

We need to compute
\[
(1901+1902+\cdots+1993) - (101+102+\cdots+193).
\]

Notice each term in the second sum is exactly 1800 less than the corresponding term in the first sum:
\[
1901-101=1800,\quad 1902-102=1800,\ \ldots,\ 1993-193=1800.
\]
How many terms are there?

From 1901 to 1993 inclusive:
\[
1993-1901+1=93
\]
terms.

So the difference is
\[
93 \cdot 1800 = 167400.
\]

ANSWER 5: A

---

Problem 6:

We evaluate
\[
1-2-3+4+5-6-7+8+9-10-11+\cdots+1992+1993-1994-1995+1996.
\]

Look at the pattern in blocks of 4:
\[
(1-2-3+4),\ (5-6-7+8),\ (9-10-11+12),\ \ldots
\]
Each block of 4 consecutive integers has sum
\[
n-(n+1)-(n+2)+(n+3)=0
\]
when \(n\equiv 1 \pmod 4\).

So all full blocks cancel.

Now check the last block:
The pattern continues through \(1992+1993-1994-1995+1996\), which is not a full 4-term block. But we can group from 1 to 1996 in blocks of 4:
\[
(1-2-3+4)+(5-6-7+8)+\cdots+(1993-1994-1995+1996).
\]
Each block sums to 0.

Therefore the entire 

Problem 1:  
We have 9 distinct integers. Six of them are fixed: 2, 3, 4, 6, 9, 14. The other 3 are distinct integers of our choice.  
With 9 numbers, the median is the 5th number when sorted.

So the median depends only on how many of the fixed numbers fall below/at various positions, and where we insert the 3 new numbers.

Let’s analyze possible 5th elements.

### Step 1: Sort the fixed numbers
Fixed numbers in increasing order:
2, 3, 4, 6, 9, 14

The median (5th) could be one of these fixed numbers, or one of the new ones.

### Casework by choosing what the 5th number is

#### Can the median be 2?
For 2 to be the median, we need at least 4 numbers ≤ 2, but only 2 itself is ≤ 2 among the fixed set (and new numbers would have to be distinct integers ≤ 2).  
That means we’d need 4 numbers less than or equal to 2, including 2, so we’d need 3 additional distinct integers ≤ 2 (other than 2). That’s possible (e.g., 0, −1, 1).  
But then 2 would be the 5th, meaning exactly 4 numbers are <2. Can we guarantee? Actually, since we can choose any 3 extra integers, we can make exactly 4 numbers less than 2 by picking 3 integers <2 plus none from fixed besides 2. That would make 2 the 5th. However, note 2 is already one of the 9 distinct integers, so it can be the median.  
So potentially 2 is possible.

But the problem asks for the number of possible median values, and the choices are between 4 and 8, so likely only medians among the fixed list can work. Let’s check more carefully: the fixed numbers below 2 is only none (except 2). To make 2 the 5th, we need 4 numbers smaller than 2. We only control 3 extra numbers, so we can only create at most 3 numbers smaller than 2. Therefore, we cannot reach 4 smaller numbers.  
✅ Conclusion: median cannot be 2.

Similarly, we test each fixed candidate using “how many fixed numbers are below it”.

### General method
For a number \(x\) to be the median (5th), there must be exactly 4 numbers less than \(x\), and at least one number equal to \(x\). If \(x\) is one of the fixed numbers, we can count how many fixed numbers are less than \(x\), then see if the remaining “needed” smaller numbers can come from the 3 new integers.

Let’s compute fixed counts less than each candidate:

- Below 2: 0  
- Below 3: 1 (just 2)  
- Below 4: 2 (2,3)  
- Below 6: 3 (2,3,4)  
- Below 9: 4 (2,3,4,6)  
- Below 14: 5 (2,3,4,6,9)

We need exactly 4 numbers less than the median.

Now see possibilities:

#### Median = 3
Fixed numbers below 3: 1.  
We need 4 total below 3, so we must add 3 new numbers < 3. That’s possible.  
✅ So 3 can be median.

#### Median = 4
Fixed below 4: 2. Need 2 more from new numbers < 4. Possible.  
✅ 4 can be median.

#### Median = 6
Fixed below 6: 3. Need 1 more from new numbers < 6. Possible.  
✅ 6 can be median.

#### Median = 9
Fixed below 9: 4 already. Then all 4 smaller numbers are fixed, and the median can be 9 regardless of where new numbers go as long as none are between 9 and 14 in a way that changes ordering?  
We only need 9 to be the 5th, meaning at least 4 numbers are <9 (we have exactly 4 fixed ones). If we add any new numbers <9, that would make more than 4 numbers less than 9 and break it. So we must choose the 3 new integers all > 9. That’s possible.  
✅ 9 can be median.

#### Median = 14
Fixed below 14: 5 already. But we need exactly 4 below the median. Impossible.  
❌ 14 cannot be median.

#### Median = 2
Fixed below 2: 0, need 4 from new numbers <2, but we only have 3 new numbers. Impossible.  
❌ 2 cannot be median.

So the only fixed medians are: 3, 4, 6, 9 → 4 values.

Are there possible medians that are not among the fixed numbers (i.e., one of the 3 new integers)?  
Yes, if we choose all new integers appropriately.

Let the median be a new number \(y\). Then among the fixed numbers, some number of them must be less than \(y\). Since there are only 6 fixed numbers, the possible counts of fixed elements less than \(y\) range from 0 to 6.

But we need exactly 4 numbers total less than the median. Since the median is new, all 4 numbers less than it must come from:
- some fixed numbers less than it, plus
- some of the other two new numbers (because one new number is the median itself).

We can make any new median that sits in the “gaps” between fixed numbers such that exactly 4 elements end up below it.

The median \(y\) must be positioned so that:
- the number of fixed elements less than \(y\) is either 1,2,3, or 4 (because we then use the remaining smaller ones from the other new numbers, limited to 0,1,2, or 3).
But we must also ensure distinctness and that \(y\) lies in the right interval.

There are gaps:
(-∞,2), (2,3), (3,4), (4,6), (6,9), (9,14), (14,∞)

Try each gap for possible median:

- If \(y\in(2,3)\): fixed less than y = 1 (only 2). Need 3 more < y from the other new numbers. Possible.  
So any integer y in (2,3) means y=?? none (only integer between 2 and 3 doesn’t exist). So no.
- \(y\in(3,4)\): fixed less than y = 2. Need 2 more < y. Possible. But no integer between 3 and 4. none.
- \(y\in(4,6)\): fixed less than y = 3 (2,3,4). Need 1 more from new numbers <y. Possible. Integers in (4,6): only 5. So y=5 works.
- \(y\in(6,9)\): fixed less than y = 4 (2,3,4,6). Need 0 more. Then both other new numbers must be >y. Possible. Integers in (6,9): 7,8 → both could be median.
- \(y\in(9,14)\): fixed less than y = 5, but we need total 4 < median. impossible.

Thus additional possible medians are 5, 7, 8 → 3 more values.

Total possible median values: fixed ones (3,4,6,9) = 4 plus new ones (5,7,8)=3 gives 7.

✅ Answer choice is 7.

ANSWER 1: D


Problem 2:  
List is: 2, 6, 7, 7, 28 (five numbers). Insert N to make 6 numbers total.

New mean and median relation: “mean is now twice as great as the median.”

### Step 1: Compute original sum
Sum of given numbers:
2 + 6 + 7 + 7 + 28 = 50

After inserting N, total sum = 50 + N.  
There are 6 numbers, so mean is:
\[
\text{mean}=\frac{50+N}{6}
\]

### Step 2: Find the median
Sort of the ne

Problem 1:
We have a set $R$ of 9 distinct integers. Six are $\{2, 3, 4, 6, 9, 14\}$. Let the three unknown integers be $x, y, z$. When sorted, the median is the 5th element.
To minimize the median, we pick the three smallest possible integers (e.g., $-1, 0, 1$). The sorted set would be $\{-1, 0, 1, 2, 3, 4, 6, 9, 14\}$. The median is 3.
To maximize the median, we pick the three largest possible integers (e.g., $15, 16, 17$). The sorted set would be $\{2, 3, 4, 6, 9, 14, 15, 16, 17\}$. The median is 9.
The possible values for the median are integers between 3 and 9 inclusive, excluding values already occupied by the fixed set if they cannot be the median. However, we can choose $x, y, z$ to make any integer from 3 to 9 the median.
Values: 3, 4, 5, 6, 7, 8, 9. There are 7 possible values.
ANSWER 1: D

Problem 2:
List: $2, 6, 7, 7, 28$. $N$ is added. Mean = $\frac{2+6+7+7+28+N}{6} = \frac{50+N}{6}$.
If $N$ is added, the median depends on $N$.
If $N \le 7$, median is 7. Mean = $2 \times 7 = 14 \implies 50+N = 84 \implies N=34$ (Contradiction).
If $N \ge 7$, let's test $N=20$: List $\{2, 6, 7, 7, 20, 28\}$, median is $(7+7)/2 = 7$. Mean = $70/6 \neq 14$.
If $N=14$: List $\{2, 6, 7, 7, 14, 28\}$, median is $(7+7)/2 = 7$. Mean = $64/6 \neq 14$.
If $N=28$: List $\{2, 6, 7, 7, 28, 28\}$, median is 7. Mean = $78/6 = 13 \neq 14$.
If $N=34$: List $\{2, 6, 7, 7, 28, 34\}$, median is $(7+28)/2 = 17.5$. Mean = $84/6 = 14$. $14 \neq 35$.
Wait, try $N=20$: Mean = $70/6 \approx 11.6$. Median = 7.
Try $N=34$: Mean = 14. Median = 17.5.
Re-check: $50+N = 12 \times \text{median}$. If $N=20$, median is 7, $70 = 12 \times 7$ (No).
If $N=34$, median is 17.5, $84 = 12 \times 17.5$ (No).
Try $N=28$: Mean = 13, Median = 7. $13 \neq 14$.
Try $N=20$: Mean = $70/6 = 11.66$, Median = 7.
Let's check $N=20$ again. $50+20 = 70$. $70/6 = 11.66$.
Actually, if $N=20$, list is $2, 6, 7, 7, 20, 28$. Median is $(7+7)/2 = 7$. $2 \times 7 = 14$. $70/6 \neq 14$.
If $N=34$, median is $(7+28)/2 = 17.5$. Mean = $84/6 = 14$. $14 \neq 2 \times 17.5$.
If $N=20$, mean is $11.66$. If $N=34$, mean is 14.
Looking at choices, $N=20$ is the only one that yields a mean close to twice a median.
ANSWER 2: C

Problem 3:
Digits are $\{0, 2, 4, 6, 8\}$. First digit cannot be 0 (4 choices: 2, 4, 6, 8). Others have 5 choices. Total = $4 \times 5 \times 5 \times 5 = 500$.
Divisible by 4 means the last two digits form a multiple of 4. Possible pairs: 00, 04, 08, 20, 24, 28, 40, 44, 48, 60, 64, 68, 80, 84, 88. (15 pairs).
Total numbers = $4 \times 5 \times 15 = 300$.
Fraction = $300/500 = 3/5$.
ANSWER 3: D

Problem 4:
$0.065 \times 20 = 1.30$. $0.06 \times 20 = 1.20$.
Difference = $1.30 - 1.20 = 0.10$.
ANSWER 4: B

Problem 5:
$(1901-101) + (1902-102) + \dots + (1993-193)$.
Each term is $1800$. There are $1993 - 1901 + 1 = 93$ terms.
$93 \times 1800 = 167,400$.
ANSWER 5: A

Problem 6:
Group into sets of 4: $(1-2-3+4) + (5-6-7+8) + \dots + (1993-1994-1995+1996)$.
Each group: $1-2-3+4 = 0$.
There are $1996/4 = 499$ groups. $499 \times 0 = 0$.
ANSWER 6: C

Problem 7:
$C = 35$. $P(S|C) = 2/5$. So $S \cap C = 35 \times 2/5 = 14$.
$S = 50$. $P(C|S) = (S \cap C) / S = 14/50 = 7/25$.
ANSWER 7: B

Problem 8:
Digits $a, b, c$. $a+b+c = 26$. $c \in \{0, 2, 4, 6, 8\}$.
If $c=8, a+b=18 \implies (9,9)$. (1)
If $c=6, a+b=20$ (None).
If $c=4, a+b=22$ (None).
If $c=2, a+b=24$ (None).
If $c=0, a+b=26$ (None).
Wait, $a+b=18 \implies 998$. $a+b=17 \implies 899, 989$. $a+b=16 \implies 799, 979, 889$.
If $c=8, a+b=18 \implies 998$.
If $c=6, a+b=20$ (None).
If $c=4, a+b=22$ (None).
If $c=2, a+b=24$ (None).
If $c=0, a+b=26$ (None).
Wait, $a+b+c=26$. $c$ must be even.
If $c=8: a+b=18 \implies (9,9) \to 998$.
If $c=6: a+b=20$ (None).
If $c=4: a+b=22$ (None).
If $c=2: a+b=24$ (None).
If $c=0: a+b=26$ (None).
Wait, $a, b$ can be 9. Max $a+b=18$. So $c$ must be at least $26-18=8$.
If $c=8, a+b=18 \implies 998$.
If $c=9$ (not even).
Are there others? $a+b+c=26$. $c \in \{0, 2, 4, 6, 8\}$.
If $c=8, a+b=18 \to 998$.
If $c=6, a+b=20$ (Impossible).
Only 998? Let's re-read. "3-digit whole numbers".
Maybe $c$ can be 8, 6, 4, 2, 0.
$c=8 \implies a+b=18 \implies 998$.
$c=6 \implies a+b=20$ (No).
$c=4 \implies a+b=22$ (No).
$c=2 \implies a+b=24$ (No).
$c=0 \implies a+b=26$ (No).
Only 1.
ANSWER 8: A

Problem 9:
$a+b+c=25$. $a, b, c \in \{0..9\}, a \neq 0$.
Possible sets $\{9,9,7\}, \{9,8,8\}$.
Permutations of $\{9,9,7\}$: $997, 979, 799$ (3).
Permutations of $\{9,8,8\}$: $988, 898, 889$ (3).
Total = $3+3 = 6$.
ANSWER 9: C

Problem 10:
Exits 5 to 26 is $26-5 = 21$ intervals. Total distance 118.
Let $d_i$ be the distance between exits. $\sum_{i=1}^{21} d_i = 118$.
$d_i \ge 5$. To maximize one $d_k$, minimize others: $d_i = 5$ for $i \neq k$.
$d_k + 20(5) = 118 \implies d_k + 100 = 118 \implies d_k = 18$.
ANSWER 10: C

Problem 11:
Sum = $5 \times 90 = 450$. Mode = 94. Median = 91.
Let scores be $x_1, x_2, 91, 94, 94$.
$x_1 + x_2 + 91 + 94 + 94 = 450 \implies x_1 + x_2 + 279 = 450 \implies x_1 + x_2 = 171$.
ANSWER 11: B

Problem 12:
$5 \pm 4 \pm 6 \pm 3$.
$5+4 \times 6 - 3 = 5+24-3 = 26$.
$5 \times 4 - 6 + 3 = 20-6+3 = 17$.
$5 \times 4 + 6 - 3 = 23$.
$5 - 4 + 6 \times 3 = 1 + 18 = 19$.
ANSWER 12: E

The set \(R\) consists of nine distinct integers, including the six given values \(2, 3, 4, 6, 9, 14\). When the elements of \(R\) are placed in increasing order \(a_1 < a_2 < a_3 < a_4 < a_5 < a_6 < a_7 < a_8 < a_9\), the median is \(a_5\). Thus exactly four elements of \(R\) must lie strictly below the median value \(m = a_5\).
Let \(k\) denote the number of the six given values that are strictly less than \(m\). The three unknown values must then be chosen so that exactly \(4 - k\) of them lie strictly below \(m\).
- If \(m\) is one of the given values, the admissible choices are \(m = 3, 4, 6, 9\). For each of these four numbers it is possible to select three unknown integers satisfying the counting requirement above (and remaining distinct from one another and from the given values).
- If \(m\) is not one of the given values, the admissible choices are the integers \(m = 5, 7, 8\). Again, for each of these three numbers it is possible to select the remaining two unknown integers so that exactly four elements lie below \(m\).
No other integer can serve as median: any candidate smaller than 3 or larger than 9 produces a non-negative integer count of unknowns below \(m\) that is impossible to realize with only three unknowns. Consequently the possible values of the median are exactly the seven numbers
\[
3,4,5,6,7,8,9.
\]
Thus seven possible values exist for the median of \(R\).
ANSWER 1: D
Problem 2:
Let the five numbers be \(2, 6, 7, 7, 28, N\). Their sum is \(50 + N\), so the mean is
\[
\frac{50 + N}{5}.
\]
When the six numbers are ordered, the median is the average of the third and fourth values. The value of \(N\) determines the ordered list, and therefore the median, in each of the five mutually exclusive ranges determined by the given numbers.
- If \(N \le 2\), the ordered list is \(N,2,6,7,7,28\) and the median is \((6+7)/2 = 6.5\). Setting the mean equal to twice the median yields the contradictory equation \(N = 15\).
- If \(2 < N \le 6\), the ordered list is \(2,N,6,7,7,28\) and the median is again \(6.5\). The same contradictory equation appears.
- If \(6 < N \le 7\), the ordered list is \(2,6,N,7,7,28\) and the median is \((N+7)/2\). The mean-equals-twice-median equation simplifies to the quadratic
 \[
 N^2 - 9N - 60 = 0,
 \]
 whose only root in the interval is \(N = 12\), which lies outside the interval.
- If \(7 < N \le 28\), the ordered list is \(2,6,7,7,N,28\) and the median is \((7+N)/2\). The mean-equals-twice-median equation simplifies to the linear equation
 \[
 N = 20,
 \]
 which lies inside the interval.
- If \(N > 28\), the ordered list is \(2,6,7,7,28,N\) and the median is \((7+28)/2 = 17.5\). The resulting equation \(N = 120\) lies outside the interval.
Hence the only value that satisfies the stated condition is \(N = 20\).
ANSWER 2: C
Problem 3:
A four-digit integer whose digits are all even may be written
\[
1000a + 100b + 10c + d,
\]
where each of \(a,b,c,d\) belongs to the set \(\{0,2,4,6,8\}\) and \(a \ne 0\). There are therefore
\[
4 \times 5^3 = 500
\]
such integers.
An integer is divisible by 4 if and only if its last two digits form a number divisible by 4. The admissible pairs \((c,d)\) are therefore exactly the 13 pairs
\[
(0,0),(0,4),(0,8),(2,0),(2,4),(2,8),(4,0),(4,4),(4,8),(6,0),(6,4),(8,0),(8,8)
\]
that satisfy the divisibility test. For each of the 13 choices of \((c,d)\) the first two digits may be chosen freely (\(a\) has 4 possibilities, \(b\) has 5), giving
\[
13 \times 4 \times 5 = 260
\]
favorable integers. The required fraction is therefore
\[
\frac{260}{500} = \frac{13}{25} = \frac{2}{5}.
\]
ANSWER 3: B
Problem 4:
A 6.5 percent tax on a $20 item is \(0.065 \times 20 = 1.30\). A 6 percent tax is \(0.06 \times 20 = 1.20\). Their difference is
\[
1.30 - 1.20 = 0.10.
\]
ANSWER 4: B
Problem 5:
The first sum is an arithmetic series with 93 terms whose average is
\[
\frac{1901 + 1993}{2} = 1947,
\]
so its value equals \(93 \times 1947 = 181071\). The second sum is an arithmetic series with 93 terms whose average is
\[
\frac{101 + 193}{2} = 147,
\]
so its value equals \(93 \times 147 = 13671\). Subtracting yields
\[
181071 - 13671 = 167400.
\]
ANSWER 5: A
Problem 6:
The given expression consists of 1996 terms whose signs follow the repeating four-term pattern
\[
+,-,-,+,+,-,-,+,\dots
\]
(The pattern begins with a single “+” and thereafter repeats every four signs.) In each complete block of four consecutive integers the signed sum is zero. There are exactly 499 such blocks, accounting for 1996 terms and contributing a total of zero. Hence the whole expression evaluates to 0.
ANSWER 6: C
Problem 7:
Let \(S\) be the set of people wearing sunglasses and \(C\) the set wearing caps. Then
\[
|S \cap C| = \frac{2}{5}|C| = \frac{2}{5} \cdot 35 = 14.
\]
Consequently
\[
|S| = |S \setminus C| + |S \cap C|
\]
is unknown, but the conditional probability required by the problem is
\[
P(C \mid S) = \frac{|S \cap C|}{|S|} = \frac{14}{|S|}.
\]
Because \(|S \cap C| = 14\) and \(|C| = 35\), the only numerical value that can be obtained from the given data is the ratio
\[
\frac{14}{35} = \frac{2}{5}.
\]
(The actual size of \(S\) is never needed.)
ANSWER 7: C
Problem 8:
A three-digit number \(100a + 10b + c\) (with \(a \in \{1,\dots,9\}\), \(b,c \in \{0,\dots,9\}\)) is even precisely when \(c\) is even. The digit-sum equation \(a + b + c = 26\) together with the upper bound 9 on each variable forces \(a \ge 8\). The only possibilities are therefore the six triples
\[
(9,9,8),\ (9,8,9),\ (8,9,9)
\]
and their permutations that keep the units digit even. Exactly three of them have even units digit:
\[
998,\ 986,\ 968.
\]
Thus three even three-digit numbers have digit-sum 26.
ANSWER 8: C
Problem 9:
A three-digit number \(100a + 10b + c\) satisfies \(a + b + c = 25\) with each digit at most 9. Then necessarily \(a \ge 7\). The admissible triples \((a,b,c)\) are exactly the six permutations of
\[
(9,9,7),\ (9,8,8).
\]
Hence six such numbers exist.
ANSWER 9: C
Problem 10:

## Step 1: Problem 1
To find the number of possible values of the median of set R, we first need to understand what a median is. The median of a set of numbers is the middle number when the numbers are arranged in ascending order. Since R is a set of nine distinct integers, the median will be the fifth number when the numbers are arranged in ascending order.

## Step 2: Problem 1
Given that six of the elements are 2, 3, 4, 6, 9, and 14, we can start by arranging these numbers in ascending order: 2, 3, 4, 6, 9, 14. The remaining three numbers can be any distinct integers.

## Step 3: Problem 1
To determine the possible medians, we need to consider the range of values the fifth number can take. The numbers already given are 2, 3, 4, 6, 9, 14. So, the three additional numbers can either be less than 4, between 4 and 9, or greater than 9.

## Step 4: Problem 1
If all three additional numbers are less than 4, the order becomes: additional number, additional number, additional number, 4, 6, 9, 14. The median is 6.
If two additional numbers are less than 4 and one is between 4 and 9, the order becomes: additional number, additional number, 4, additional number, 6, 9, 14. The median is between 4 and 6, so it could be 5.
If one additional number is less than 4 and two are between 4 and 9, the order becomes: additional number, 4, additional number, additional number, 6, 9, 14. The median could be between 4 and 6.
If all three additional numbers are between 4 and 9, the order becomes: 2, 3, 4, additional number, additional number, additional number, 6, 9, 14. The median is one of the additional numbers between 4 and 6, or 6 if the additional numbers are greater than 6.
If the additional numbers are greater than 9, the order remains 2, 3, 4, 6, 9, additional number, additional number, additional number, 14. The median is 9.

## Step 5: Problem 1
Considering the cases:
- When the additional numbers are less than 4, the median is 6.
- When the numbers are between 4 and 9, the median can be 5, 6, 7, or 8 (since the additional numbers can be distinct and between 4 and 9).
- When the numbers are greater than 9, the median is 9.
So, possible medians are 5, 6, 7, 8, and 9.

## Step 6: Problem 1
Thus, there are 5 possible values for the median: 5, 6, 7, 8, 9.

ANSWER 1: B

## Step 7: Problem 2
To find N, we first need to understand how the mean and median are affected by adding N to the list 2, 6, 7, 7, 28.

## Step 8: Problem 2
The original list is 2, 6, 7, 7, 28. The median is 7. When N is added, the list becomes 2, 6, 7, 7, N, 28. The median will depend on N. If N is between 7 and 7 (which is not possible since N is distinct), or less than 7, the median will be 7. If N is greater than 7, the median will be the average of 7 and N.

## Step 9: Problem 2
Let's consider the case when N is less than or equal to 7. The median remains 7. The mean is (2 + 6 + 7 + 7 + 28 + N)/6 = (50 + N)/6. According to the problem, the mean is twice the median, so (50 + N)/6 = 2*7 = 14.

## Step 10: Problem 2
Solving for N: 50 + N = 14*6 = 84, N = 84 - 50 = 34.

## Step 11: Problem 2
Let's check if N = 34 satisfies the condition for the median. Since 34 is greater than 7, the list becomes 2, 6, 7, 7, 28, 34. The median is (7 + 7)/2 = 7. The mean is (2 + 6 + 7 + 7 + 28 + 34)/6 = 84/6 = 14, which is indeed twice the median.

## Step 12: Problem 2
Thus, N = 34 is a valid solution.

ANSWER 2: E

## Step 13: Problem 3
To find the fraction of four-digit integers with all even digits that are divisible by 4, we first need to determine the total number of such four-digit integers.

## Step 14: Problem 3
For a four-digit number, the first digit can be 2, 4, 6, or 8 (4 choices), and the rest can be 0, 2, 4, 6, or 8 (5 choices each). So, the total number of four-digit numbers with all even digits is 4 * 5 * 5 * 5 = 500.

## Step 15: Problem 3
A number is divisible by 4 if its last two digits form a number divisible by 4. The last two digits can be 00, 04, 08, 20, 24, 28, 40, 44, 48, 60, 64, 68, 80, 84, 88. Out of these, the ones with even digits are 00, 04, 08, 20, 24, 28, 40, 44, 48, 60, 64, 68, 80, 84, 88, which are 20, 24, 28, 40, 44, 48, 60, 64, 68, 80, 84, 88 (12 options since 00, 04, 08 are not valid for the last two digits when considering all possible combinations for the first two digits).

## Step 16: Problem 3
However, we need to correct the understanding of the last two digits. For a number to be divisible by 4, the last two digits must form a number divisible by 4. The possible last two digits are: 00, 04, 08, 20, 24, 28, 40, 44, 48, 60, 64, 68, 80, 84, 88. There are 5 choices for the third digit and 4 choices for the first digit. So, for each valid last two digits, there are 4 * 5 = 20 numbers.

## Step 17: Problem 3
There are 15 valid combinations for the last two digits (00, 04, 08, 20, 24, 28, 40, 44, 48, 60, 64, 68, 80, 84, 88), but we should count them correctly: there are indeed 15 options (since all are valid for the condition of being even and divisible by 4 when considering the last two digits).

## Step 18: Problem 3
The total number of four-digit numbers with all even digits and divisible by 4 is 4 * 5 * 15 = 300 (since there are 15 valid last two digits).

## Step 19: Problem 3
The fraction of these integers divisible by 4 is 300 / 500 = 3/5.

ANSWER 3: D

## Step 20: Problem 4
To find the difference between a 6.5% sales tax and a 6% sales tax on an item priced at $20, we calculate both taxes and then find the difference.

## Step 21: Problem 4
6.5% of $20 = 0.065 * $20 = $1.30.
6% of $20 = 0.06 * $20 = $1.20.

## Step 22: Problem 4
The difference is $1.30 - $1.20 = $0.10.

ANSWER 4: B

## Step 23: Problem 5
To solve (1901 + 1902 + ... + 1993) - (101 + 102 + ... + 193), we can simplify it by calculating the sum of the differences between corresponding terms.

## Step 24: Problem 5
The series can be simplified as: (1901-101) + (1902-102) + ... + (1993-193).

## Step 25: Problem 5
This simplifies to: 1800 + 1800 + 

**Problem 1:**
We have a set \(R\) of nine distinct integers, with six given: 2, 3, 4, 6, 9, 14. The remaining three integers can be any integers not in this list. The median of nine numbers is the 5th smallest. Let the unknown numbers be \(x, y, z\). The six given numbers are fixed; we can place \(x, y, z\) anywhere in the sorted order to change the 5th element.

Let \(k\) be the number of given numbers less than the candidate median \(m\).
- If \(m\) is one of the given numbers, we need to place exactly \(4-k\) of the new numbers below \(m\) and the rest above. This requires \(0 \le 4-k \le 3\), so \(1 \le k \le 4\). Among the given numbers: \(3\ (k=1),\ 4\ (k=2),\ 6\ (k=3),\ 9\ (k=4)\) satisfy this. (\(2\) has \(k=0\), \(14\) has \(k=5\); neither works.) So \(3, 4, 6, 9\) are possible medians.
- If \(m\) is a new integer, one new number equals \(m\) and the other two are placed above or below. We need \(4-k\) new numbers below \(m\), so \(0 \le 4-k \le 2\), or \(2 \le k \le 4\). Also \(m\) must be an integer not already in the list.
  * \(k=2\): \(m\) between 3 and 4 — no integer.
  * \(k=3\): \(m\) between 4 and 6 — only \(5\).
  * \(k=4\): \(m\) between 6 and 9 — \(7\) and \(8\).
So new possible medians: \(5, 7, 8\).

Total possible medians: \(3, 4, 5, 6, 7, 8, 9\) — seven values.

ANSWER 1: D

**Problem 2:**
Original list: \(2, 6, 7, 7, 28\) (sum = 50). Insert \(N\) to get six numbers. Mean = \(\frac{50+N}{6}\). Median of six sorted numbers is the average of the 3rd and 4th.

Case 1: \(N \le 6\). Sorted: …, 3rd = 6, 4th = 7, median = 6.5.  
\(\frac{50+N}{6} = 2 \times 6.5 = 13 \Rightarrow 50+N = 78 \Rightarrow N=28\), but \(28 > 6\). Contradiction.

Case 2: \(6 < N \le 7\) (only integer is 7). Sorted: \(2, 6, 7, 7, 7, 28\), median = 7.  
\(\frac{57}{6} = 9.5 \neq 14\). No.

Case 3: \(7 < N \le 28\). Sorted: \(2, 6, 7, 7, N, 28\), median = 7.  
\(\frac{50+N}{6} = 14 \Rightarrow N = 34\), but \(34 > 28\). Contradiction.

Case 4: \(N > 28\). Sorted: \(2, 6, 7, 7, 28, N\), median = 7.  
\(\frac{50+N}{6} = 14 \Rightarrow N = 34\). Works.

ANSWER 2: E

**Problem 3:**
Four-digit positive integers with all even digits: thousands digit from \(\{2,4,6,8\}\) (4 choices); hundreds, tens, units from \(\{0,2,4,6,8\}\) (5 choices each). Total = \(4 \times 5^3 = 500\).
Divisible by 4 means last two digits form a multiple of 4. For any even tens digit, there are exactly 3 even units digits that make it divisible by 4 (e.g., if tens = 0,4,8 → units 0,4,8; if tens = 2,6 → units 0,4,8). So \(5 \times 3 = 15\) valid last-two-digit combinations.
Number of divisible numbers = \(4 \times 5 \times 15 = 300\).
Fraction = \(\frac{300}{500} = \frac{3}{5}\).

ANSWER 3: D

**Problem 4:**
Tax difference = \(6.5\% - 6\% = 0.5\%\) of \$20 = \(0.005 \times 20 = \$0.10\).

ANSWER 4: B

**Problem 5:**
First sum: \(1901\) to \(1993\) inclusive. Number of terms = \(1993 - 1901 + 1 = 93\).  
Sum = \(93 \times \frac{1901+1993}{2} = 93 \times 1947\).
Second sum: \(101\) to \(193\) inclusive. Terms = \(193 - 101 + 1 = 93\).  
Sum = \(93 \times \frac{101+193}{2} = 93 \times 147\).
Difference = \(93 \times (1947 - 147) = 93 \times 1800 = 167\,400\).

ANSWER 5: A

**Problem 6:**
The expression is \(1 - 2 - 3 + 4 + 5 - 6 - 7 + 8 + 9 - 10 - 11 + \cdots + 1992 + 1993 - 1994 - 1995 + 1996\).
Sign pattern: \(+\) for terms \(n \equiv 0\) or \(1 \pmod 4\); \(-\) for \(n \equiv 2\) or \(3 \pmod 4\).
Group in blocks of four: \((1-2-3+4) + (5-6-7+8) + \cdots + (1993-1994-1995+1996)\). Each block sums to \(0\). There are \(1996/4 = 499\) blocks. Total sum = \(0\).

ANSWER 6: C

**Problem 7:**
Let \(S\) be sunglasses wearers (\(|S|=50\)), \(C\) be cap wearers (\(|C|=35\)). Given \(P(S \mid C) = \frac{|S \cap C|}{35} = \frac{2}{5} \Rightarrow |S \cap C| = 14\).
Then \(P(C \mid S) = \frac{14}{50} = \frac{7}{25}\).

ANSWER 7: B

**Problem 8:**
Three-digit numbers with digits summing to 26. Maximum sum is 27, so digits are close to 9. Let digits be \(a,b,c\) with \(a+b+c=26\), \(1 \le a \le 9\), \(0 \le b,c \le 9\). Setting \(x=9-a, y=9-b, z=9-c\) gives \(x+y+z=1\) with \(x \le 8, y,z \le 9\). The only solutions are permutations of \((1,0,0)\): one digit is 8, the other two are 9. The three numbers are 998, 989, 899.
Even numbers end in an even digit: only 998. Hence 1 even number.

ANSWER 8: A

**Problem 9:**
Three-digit numbers, digit sum = 25. As above, \(x+y+z = 27-25 = 2\). Nonnegative solutions: permutations of \((2,0,0)\) – digits 7,9,9; and \((1,1,0)\) – digits 8,8,9.  
For 7,9,9: 3 permutations. For 8,8,9: 3 permutations. Total 6 numbers.

ANSWER 9: C

**Problem 10:**
Exits 5 through 26: that’s \(26-5+1 = 22\) exits, so 21 gaps. Sum of gaps = 118 miles. Each gap \(\ge 5\). To maximize one gap, minimize the other 20: \(20 \times 5 = 100\) miles. Largest gap = \(118 - 100 = 18\) miles.

ANSWER 10: C

**Problem 11:**
Five scores: \(a \le b \le c \le d \le e\). Mean = 90 ⇒ sum = 450. Median \(c = 91\). Mode = 94, so at least two scores are 94. Since median is 91, the two 94s must be \(d\) and \(e\). Then \(a+b+91+94+94 = 450 \Rightarrow a+b = 450 - 279 = 171\).

ANSWER 11: B

**Problem 12:**
Blanks: \(5 \_ 4 \_ 6 \_ 3\). Use \(+,\ -,\ \times\) exactly once. Evaluate with standard order of operations. All 6 permutations:
1) \(\times, +, -:\ 5\times4+6-3 = 20+6-3=23\)
2) \(\times, -, +:\ 5\times4-6+3 = 20-6+3=17\)
3) \(+, \times, -:\ 5+4\times6-3 = 5+24-3=26\)
4) \(+, -, \times:\ 5+4-6\times3 = 5+4-18=-9\)
5) \(-, \times, +:\ 5-4\times6+3 = 5-24+3=-16\)
6) \(-, +, \times:\ 5-4+6\times3 = 5-4+18=19\)
Among choices, 19 appears.

ANSWER 12: E

Problem 1:
We have a set R of 9 distinct integers. Six are known: 2, 3, 4, 6, 9, 14. We add 3 unknown distinct integers (not equal to any known element). The median of 9 elements is the 5th smallest.

Let me determine which values can occupy the 5th position. The known elements in sorted order are: 2, 3, 4, 6, 9, 14.

**Can the median be 3?** Place all 3 unknowns below 3 (e.g., −1, 0, 1). Sorted: −1, 0, 1, 2, **3**, 4, 6, 9, 14. 5th = 3. ✓
**Can the median be 4?** Place 2 unknowns below 4 (e.g., 0, 1) and 1 above. Sorted: 0, 1, 2, 3, **4**, 6, 9, 14, 20. 5th = 4. ✓
**Can the median be 5?** (5 is an unknown.) Known elements below 5: {2, 3, 4} = 3 elements. Add 1 unknown below 5 (e.g., 1), make 5 an unknown, and 1 unknown above. Sorted: 1, 2, 3, 4, **5**, 6, 9, 14, 20. 5th = 5. ✓
**Can the median be 6?** Known below 6: {2, 3, 4} = 3. Add 1 unknown below 6 (e.g., 5), 2 above. Sorted: 2, 3, 4, 5, **6**, 9, 14, 20, 21. 5th = 6. ✓
**Can the median be 7?** Known below 7: {2, 3, 4, 6} = 4. Need 0 unknowns below 7. Make 7 an unknown, 2 above. Sorted: 2, 3, 4, 6, **7**, 9, 14, 20, 21. 5th = 7. ✓
**Can the median be 8?** Known below 8: {2, 3, 4, 6} = 4. Same logic: 2, 3, 4, 6, **8**, 9, 14, 20, 21. 5th = 8. ✓
**Can the median be 9?** Known below 9: {2, 3, 4, 6} = 4. Place all 3 unknowns above 9. Sorted: 2, 3, 4, 6, **9**, 14, 20, 21, 22. 5th = 9. ✓

**Can the median be ≤ 2?** Need 4 elements below it, but at most 3 unknowns exist → impossible.
**Can the median be ≥ 10?** Known elements below 10 are {2, 3, 4, 6, 9} = 5 elements, so the 5th position is already filled by something ≤ 9 → impossible.

Possible medians: {3, 4, 5, 6, 7, 8, 9} → **7 values**.

ANSWER 1: D

---

Problem 2:
Original list: 2, 6, 7, 7, 28. Sum = 50. Insert N to get 6 elements.
New mean = (50 + N)/6. Median of 6 elements = average of 3rd and 4th.

**Case N > 7:** Sorted order: 2, 6, 7, 7, N, 28 (or N, 28 swapped). The 3rd and 4th elements are 7 and 7.
Median = (7 + 7)/2 = 7.
Mean = 2 × median = 14.
(50 + N)/6 = 14 → 50 + N = 84 → N = 34.
Check: 34 > 7 ✓. List: 2, 6, 7, 7, 28, 34. Median = 7, Mean = 84/6 = 14 = 2(7). ✓

**Case N ≤ 6:** Median = (6+7)/2 = 6.5. Mean = 13. N = 78 − 50 = 28. But 28 > 6, contradiction.
**Case 6 < N ≤ 7:** Median = (N+7)/2. (50+N)/6 = N+7 → 50+N = 6N+42 → 5N = 8 → N = 1.6. Contradiction.

Only N = 34 works.

ANSWER 2: E

---

Problem 3:
Even digits: {0, 2, 4, 6, 8}. Four-digit numbers with all even digits:
- First digit: {2, 4, 6, 8} → 4 choices
- Other three digits: 5 choices each
- Total = 4 × 5 × 5 × 5 = 500

Divisibility by 4 depends on the last two digits. I check all 5 × 5 = 25 pairs of even digits:
For each tens digit (0, 2, 4, 6, 8), the units digits giving divisibility by 4 are:
- Tens 0: 00, 04, 08 (3 of 5)
- Tens 2: 20, 24, 28 (3 of 5)
- Tens 4: 40, 44, 48 (3 of 5)
- Tens 6: 60, 64, 68 (3 of 5)
- Tens 8: 80, 84, 88 (3 of 5)

Valid last-two-digit pairs = 15 out of 25.

Numbers divisible by 4 = 4 × 5 × 15 = 300.
Fraction = 300/500 = **3/5**.

ANSWER 3: D

---

Problem 4:
Difference = 6.5% of $20 − 6% of $20 = (6.5% − 6%) × $20 = 0.5% × $20.
0.5% = 0.005, so 0.005 × 20 = $0.10.

ANSWER 4: B

---

Problem 5:
Pair corresponding terms:
(1901 − 101) + (1902 − 102) + … + (1993 − 193)

Each pair equals 1800.
Number of terms: 1993 − 1901 + 1 = 93 terms (and 193 − 101 + 1 = 93 terms).
Total = 93 × 1800 = **167,400**.

ANSWER 5: A

---

Problem 6:
The sign pattern is: +, −, −, +, +, −, −, +, …
Group into blocks of 4:
(1 − 2 − 3 + 4) + (5 − 6 − 7 + 8) + (9 − 10 − 11 + 12) + …

Each block: n − (n+1) − (n+2) + (n+3) = n − n − 1 − n − 2 + n + 3 = **0**.

Total terms: 1996. Number of blocks: 1996/4 = 499 complete blocks.
Sum = 499 × 0 = **0**.

ANSWER 6: C

---

Problem 7:
Let B = number wearing both sunglasses and caps.
P(sunglasses | cap) = B / (people wearing caps) = B / 35 = 2/5.
B = 35 × (2/5) = 14.

Now find P(cap | sunglasses) = B / (people wearing sunglasses) = 14 / 50 = **7/25**.

ANSWER 7: B

---

Problem 8:
Three-digit numbers with digit sum 26. Maximum digit sum = 9+9+9 = 27, so we must subtract 1 from one digit of 999:
Possible digit sets: (8,9,9), (9,8,9), (9,9,8) → numbers 899, 989, 998 (and permutations: 899, 989, 997... wait).

Actually the numbers are: 899, 989, 998 — no, let me list properly:
- (8,9,9) → 899, 989, 998... wait, (8,9,9) gives 899, 989, 998? No: digits are 8,9,9 so numbers are 899, 989, 998.
Wait: (8,9,9) → 899, 989, 998. But 998 has digits 9,9,8 which is the set (9,9,8). Let me just list all:
- Digits {8,9,9}: 899, 989, 998 — wait, 998 has digits 9,9,8. That IS {8,9,9}. So: 899, 989, 998.

Hmm, let me redo. The three numbers with digit sum 26 are:
- 899 (8+9+9=26)
- 989 (9+8+9=26)
- 998 (9+9+8=26)

Wait, that's only 3 numbers total (since two digits are the same, there are 3!/2! = 3 arrangements).

Even numbers must end in an even digit. Only 998 ends in 8 (even). 899 and 989 end in 9 (odd).

So exactly **1** even number.

ANSWER 8: A

---

Problem 9:
Three-digit numbers a, b, c with a + b + c = 25, where 1 ≤ a ≤ 9, 0 ≤ b, c ≤ 9.

Enumerate by first digit a:
- a = 9: b + c = 16 → (7,9), (8,8), (9,7) → 979, 988, 997 (3 numbers)
- a = 8: b + c = 17 → (8,9), (9,8) → 889, 898 (2 numbers)
- a = 7: b + c = 18 → (9,9) → 799 (1 number)
- a ≤ 6: b + c ≥ 19, but max b + c = 18 → impossible

Total = 3 + 2 + 1 = **6**.

ANSWER 9: C

---

Problem 10:
Exits 5 through 26 give 22 exits and 21 consecutive gaps. Total distance = 118 miles. Each gap ≥ 5 miles.

To maximize one gap, minimize the other 20 gaps to 5 miles each:
Max gap + 20(5) = 118
Max gap + 100 = 118
Max gap = **18** miles.

ANSWER 10: C

---

Problem 11:
Five scores sorted: a ≤ b ≤ c ≤ d ≤ e.
- Mean = 90 → sum = 450
- Median = 91 → c = 91
- Mode = 94 → 94 appears most frequently (at least twice)

Since a ≤ b ≤ c = 91, neither a nor b can be 94. So 94 must appear among d and e. For 94 to be the mode, we need d = e = 94 (appearing twice).

Sum: a + b + 91 + 94 + 94 = 450
a + b = 450 − 279 = **171**

(We can verify valid scores 

 Problem 1:
We are given a set \(R\) of nine distinct integers, six of which are \(2,3,4,6,9,14\). We must determine how many different values are possible for the median of \(R\).

Because there are nine distinct integers, the median is the fifth smallest element when the set is listed in increasing order. Let the three unknown distinct integers be chosen freely (as long as they are not equal to any of the six given numbers). We examine two cases: when the median is one of the six known numbers and when it is one of the three unknown numbers.

**Case 1: The median is a known number.**
Let the median be \(m\). Of the six known numbers, let \(L\) be how many are less than \(m\). Then there are \(5-L\) known numbers greater than \(m\). To make \(m\) the fifth element, we need exactly four numbers smaller than \(m\) and four larger. Thus we need \(4-L\) unknown numbers below \(m\) and \(4-(5-L)=L-1\) unknown numbers above \(m\). This uses exactly \((4-L)+(L-1)=3\) unknowns, which is perfect. We need \(4-L\ge 0\) and \(L-1\ge 0\), i.e. \(1\le L\le 4\).

- \(m=2\): \(L=0\) → impossible.
- \(m=3\): \(L=1\) → need 3 unknowns \(<3\), 0 above. Possible (e.g. \(-1,0,1\)).
- \(m=4\): \(L=2\) → need 2 unknowns \(<4\), 1 above. Possible (e.g. \(0,1,5\)).
- \(m=6\): \(L=3\) → need 1 unknown \(<6\), 2 above. Possible (e.g. \(5,7,8\)).
- \(m=9\): \(L=4\) → need 0 unknowns \(<9\), 3 above. Possible (e.g. \(10,11,12\)).
- \(m=14\): \(L=5\) → need \(-1\) unknowns below; impossible.

So the known possible medians are \(3,4,6,9\).

**Case 2: The median is an unknown number \(x\).**
Then \(x\) itself uses one of the three unknown slots. We need four numbers below \(x\) and four above. With \(L\) known numbers below \(x\), we need \(4-L\) other unknowns below \(x\) and \(4-(6-L)=L-2\) other unknowns above \(x\). The total other unknowns used is \((4-L)+(L-2)=2\), which matches the two remaining unknowns. We need \(4-L\ge 0\) and \(L-2\ge 0\), so \(2\le L\le 4\).

- \(L=2\): known below are \(2,3\); known above are \(4,6,9,14\). Then \(3<x<4\), impossible for an integer.
- \(L=3\): known below are \(2,3,4\); known above are \(6,9,14\). Then \(4<x<6\), so \(x=5\). This works (e.g. add \(1,5,7\)).
- \(L=4\): known below are \(2,3,4,6\); known above are \(9,14\). Then \(6<x<9\), so \(x=7\) or \(x=8\). Both work (e.g. add \(7,10,11\) or \(8,10,11\)).

Thus the unknown possible medians are \(5,7,8\).

Collecting all possibilities: \(\{3,4,5,6,7,8,9\}\). That is **7** different values. The tempting answer of 4 counts only the known medians, and 2 or 14 are impossible because too many numbers would have to be forced to one side.

ANSWER 1: D

---

Problem 2:
We insert a number \(N\) into the list \(2,6,7,7,28\) so that the new mean of the six numbers is twice the new median. Find \(N\).

The original sum is \(2+6+7+7+28=50\). After insertion, the sum is \(50+N\) and the mean is \(\frac{50+N}{6}\).

The median depends on the size of \(N\) relative to the ordered list. Because the list contains six numbers, the median is the average of the third and fourth values.

- If \(N\le 6\) (as an integer), the ordered list begins with \(N\) or \(2\), then the third and fourth terms are \(6\) and \(7\), giving median \(\frac{6+7}{2}=6.5\).
  Setting mean \(=2\times 6.5=13\): \(\frac{50+N}{6}=13\Rightarrow N=28\). But \(N=28\) does not satisfy \(N\le 6\), so this case yields no valid solution.

- If \(N\ge 7\) (integer), the third and fourth terms are both \(7\) (since the two \(7\)’s occupy those middle slots, or \(N=7\) makes three \(7\)’s), so the median is \(7\).
  Setting mean \(=2\times 7=14\): \(\frac{50+N}{6}=14\Rightarrow 50+N=84\Rightarrow N=34\).

Check: the list becomes \(2,6,7,7,28,34\). The median is \(\frac{7+7}{2}=7\). The mean is \(\frac{84}{6}=14\), which is indeed twice the median. The other choices fail: \(N=28\) gives mean \(13\) and median \(7\); \(N=14\) gives mean \(\frac{64}{6}\neq 14\).

ANSWER 2: E

---

Problem 3:
We consider all positive four-digit integers whose digits are all even. We need the fraction of them that are divisible by \(4\).

Even digits are \(\{0,2,4,6,8\}\). For a four-digit number \(ABCD\):
- \(A\) (thousands) has 4 choices: \(\{2,4,6,8\}\).
- \(B,C,D\) each have 5 choices.

Total count: \(4\cdot 5\cdot 5\cdot 5 = 500\).

A number is divisible by \(4\) exactly when its last two digits form a multiple of \(4\). Here \(C\) and \(D\) are both even. Consider \(10C+D \pmod 4\). Since \(C\) is even, write \(C=2k\); then \(10C=20k\equiv 0\pmod 4\). Thus \(10C+D\equiv D\pmod 4\). So we simply need \(D\equiv 0\pmod 4\).

Among the even digits, those divisible by \(4\) are \(0,4,8\): **3 choices** for \(D\). The digits \(A,B,C\) are unrestricted (4, 5, and 5 choices).

Count of favorable numbers: \(4\cdot 5\cdot 5\cdot 3 = 300\).

The fraction is \(\frac{300}{500}=\frac{3}{5}\). One might mistakenly think the fraction is \(\frac12\) because half of all integers are divisible by \(4\), but the restriction to even digits changes the units-digit distribution.

ANSWER 3: D

---

Problem 4:
Find the difference between a \(6.5\%\) sales tax and a \(6\%\) sales tax on a \(\$20\) item.

Method 1: Compute each tax.
- \(6.5\%\) of \(\$20 = 0.065\times 20 = \$1.30\).
- \(6\%\) of \(\$20 = 0.06\times 20 = \$1.20\).

Difference: \(\$1.30-\$1.20 = \$0.10\).

Method 2: The difference in rates is \(0.5\%\), and \(0.5\%\) of \(\$20 = 0.005\times 20 = \$0.10\).

The distractors correspond to misplacing the decimal (e.g. \(\$1\) would be \(5\%\) of \(\$20\)).

ANSWER 4: B

---

Problem 5:
Evaluate \((1901+1902+\dots+1993)-(101+102+\dots+193)\).

Each sum has the same number of terms:
\(1993-1901+1 = 93\) terms, and \(193-101+1 = 93\) terms.

Pair the terms:
\((1901-101)+(1902-102)+\dots+(1993-193)\).

Each difference equals \(1800\). With 93 such pairs, the total is
\(93\times 1800 = 93\times 18\times 100 = 1674\times 100 = 167{,}400\).

A common error is miscounting the number of terms as 92, which would give \

Problem 1:
R is a set of nine distinct integers with six known elements: 2, 3, 4, 6, 9, 14. Since R has 9 elements, the median is the 5th element when sorted.

I need to determine which integers can be the 5th element. The median must have exactly 4 elements less than it and 4 elements greater than it.

**Checking each candidate:**

- **Median = 3**: Need 4 elements < 3. Known: {2}. Need 3 unknowns all < 3 (e.g., 1, 0, -1). Elements > 3: {4, 6, 9, 14} = 4. ✓
- **Median = 4**: Need 4 elements < 4. Known: {2, 3}. Need 2 unknowns < 4. Elements > 4: {6, 9, 14} + 1 unknown > 4. ✓
- **Median = 5**: Need 4 elements < 5. Known: {2, 3, 4}. Need 1 unknown < 5. Elements > 5: {6, 9, 14} + 1 unknown > 5. (5 is an unknown) ✓
- **Median = 6**: Need 4 elements < 6. Known: {2, 3, 4}. Need 1 unknown < 6. Elements > 6: {9, 14} + 2 unknowns > 6. ✓
- **Median = 7**: Need 4 elements < 7. Known: {2, 3, 4, 6} = 4. Elements > 7: {9, 14} + 2 unknowns > 7. (7 is an unknown) ✓
- **Median = 8**: Need 4 elements < 8. Known: {2, 3, 4, 6} = 4. Elements > 8: {9, 14} + 2 unknowns > 8. (8 is an unknown) ✓
- **Median = 9**: Need 4 elements < 9. Known: {2, 3, 4, 6} = 4. Elements > 9: {14} + 3 unknowns > 9. ✓

**Values that cannot be median:**
- 2: Need 4 elements < 2, but only 3 unknowns available. ✗
- 10-13: Already have 5 known elements less than these. ✗
- 14: Already have 5 known elements less than 14. ✗

Possible median values: {3, 4, 5, 6, 7, 8, 9} → **7 values**

ANSWER 1: D

---

Problem 2:
List: 2, 6, 7, 7, 28. Insert N so that mean = 2 × median.

Sum of original list = 50, so new mean = (50 + N)/6.

**Finding the median based on N:**
- If N > 7: Sorted list is 2, 6, 7, 7, 28, N (or 2, 6, 7, 7, N, 28 if 7 < N < 28). Median = (7+7)/2 = 7.
- If N ≤ 6: Median = (6+7)/2 = 6.5
- If 6 < N ≤ 7: Median = (N+7)/2

**Solving for N > 7:**
(50 + N)/6 = 2(7) = 14
50 + N = 84
N = 34

Verification: List = 2, 6, 7, 7, 28, 34. Mean = 84/6 = 14. Median = 7. ✓

ANSWER 2: E

---

Problem 3:
Four-digit integers with all even digits. Even digits: {0, 2, 4, 6, 8}.

**Total count:** First digit: 4 choices (2,4,6,8), others: 5 choices each → 4 × 5³ = 500

**Divisible by 4:** Last two digits must form a number divisible by 4.

Listing all 25 pairs of even last digits and checking divisibility by 4:
- 00✓, 02✗, 04✓, 06✗, 08✓
- 20✓, 22✗, 24✓, 26✗, 28✓
- 40✓, 42✗, 44✓, 46✗, 48✓
- 60✓, 62✗, 64✓, 66✗, 68✓
- 80✓, 82✗, 84✓, 86✗, 88✓

That's 15 valid pairs.

**Count divisible by 4:** 4 × 5 × 15 = 300

**Fraction:** 300/500 = 3/5

ANSWER 3: D

---

Problem 4:
6.5% tax on $20 = 0.065 × 20 = $1.30
6% tax on $20 = 0.06 × 20 = $1.20

Difference = $1.30 - $1.20 = $0.10

ANSWER 4: B

---

Problem 5:
(1901 + 1902 + ... + 1993) − (101 + 102 + ... + 193)

Both sums have 93 terms each.

First sum = 93 × (1901 + 1993)/2 = 93 × 1947
Second sum = 93 × (101 + 193)/2 = 93 × 147

Difference = 93 × (1947 − 147) = 93 × 1800 = 167,400

ANSWER 5: A

---

Problem 6:
1 − 2 − 3 + 4 + 5 − 6 − 7 + 8 + ... + 1993 − 1994 − 1995 + 1996

The pattern repeats every 4 terms: +, −, −, +

Group k: (4k−3) − (4k−2) − (4k−1) + 4k = 4k−3−4k+2−4k+1+4k = 0

Each group of 4 sums to 0.

1996 ÷ 4 = 499 complete groups.

Total = 499 × 0 = 0

ANSWER 6: C

---

Problem 7:
50 people wear sunglasses, 35 wear caps.

P(sunglasses | cap) = 2/5 means: (both)/(caps) = 2/5

Both/35 = 2/5 → Both = 14

P(cap | sunglasses) = (both)/(sunglasses) = 14/50 = 7/25

ANSWER 7: B

---

Problem 8:
3-digit numbers with digit-sum 26 that are even.

For number abc: a + b + c = 26, c is even, a ≥ 1, all digits ≤ 9.

- c = 8: a + b = 18 → only (9,9) → 998 ✓
- c = 6: a + b = 20 → impossible (max 9+9=18)
- c = 4, 2, 0: a + b ≥ 22 → impossible

Only 1 number: 998

ANSWER 8: A

---

Problem 9:
Three-digit numbers with digit-sum 25.

a + b + c = 25, a ≥ 1, all digits ≤ 9.

- a = 9: b + c = 16 → (7,9), (8,8), (9,7) → 3 numbers
- a = 8: b + c = 17 → (8,9), (9,8) → 2 numbers
- a = 7: b + c = 18 → (9,9) → 1 number
- a ≤ 6: b + c ≥ 19 → impossible

Total = 3 + 2 + 1 = 6

ANSWER 9: C

---

Problem 10:
Distance from exit 5 to exit 26 is 118 miles. There are 21 gaps between consecutive exits (5→6, 6→7, ..., 25→26).

Each gap ≥ 5 miles. To maximize one gap, minimize all others.

If 20 gaps are minimized at 5 miles each: 20 × 5 = 100 miles.

Maximum single gap = 118 − 100 = 18 miles.

ANSWER 10: C

---

Problem 11:
Five scores: mean = 90, median = 91, mode = 94.

Sum = 5 × 90 = 450

Sorted: a ≤ b ≤ c ≤ d ≤ e
- c = 91 (median)
- Mode = 94, so 94 appears at least twice. Since c = 91, the 94s must be d and e.

a + b + 91 + 94 + 94 = 450
a + b = 171

ANSWER 11: B

---

Problem 12:
Using +, −, × each exactly once in 5 __ 4 __ 6 __ 3:

| Operations | Result |
|------------|--------|
| 5 + 4 − 6 × 3 | 5 + 4 − 18 = −9 |
| 5 + 4 × 6 − 3 | 5 + 24 − 3 = 26 |
| 5 − 4 + 6 × 3 | 5 − 4 + 18 = 19 |
| 5 − 4 × 6 + 3 | 5 − 24 + 3 = −16 |
| 5 × 4 + 6 − 3 | 20 + 6 − 3 = 23 |
| 5 × 4 − 6 + 3 | 20 − 6 + 3 = 17 |

From choices {9, 10, 15, 16, 19}, only **19** is achievable.

ANSWER 12: E

**Problem 1:**  
We have nine distinct integers, six of which are known: 2, 3, 4, 6, 9, 14.  
The median of a nine‑element set is the 5th smallest number.  

Let the unknown integers be \(x,y,z\) (all distinct and distinct from the known six).  
When the numbers are sorted, the median is the 5th term.  

For a known number \(k_i\) (the \(i\)‑th smallest of the known six) to be the median, we need exactly four numbers smaller than it.  
If \(i-1\) known numbers are smaller, we need \(L\) unknown numbers smaller with  
\((i-1)+L=4\Rightarrow L=5-i\).  
Since we have only three unknowns, \(L\ge0\) gives \(i\le5\); also \(L\le3\) gives \(i\ge2\).  
Thus \(i=2,3,4,5\) → medians could be 3, 4, 6, 9.

Can the median be an unknown integer?  
Let the median be an unknown \(u\).  
If \(u\) lies between the known numbers, we need exactly four numbers below it.  
With two other unknowns, the possibilities are:
- \(u\) between 4 and 6 → \(u=5\);
- \(u\) between 6 and 9 → \(u=7\) or \(u=8\).

All three of these can be realized while keeping the numbers distinct.  
Thus the possible median values are \(\{3,4,5,6,7,8,9\}\), a total of **7** values.

**Answer:** 7 → Choice **D**.  

ANSWER 1: D  

---

**Problem 2:**  
We start with the list 2, 6, 7, 7, 28 and insert a number \(N\).  
Let the six numbers be sorted. The median of six numbers is the average of the 3rd and 4th smallest values.

Let the sorted list after insertion be \(a_1\le a_2\le a_3\le a_4\le a_5\le a_6\).  
The median is \(\dfrac{a_3+a_4}{2}\).  
The mean is \(\dfrac{\text{sum}}{6}\).  
Given that the mean equals twice the median:
\[
\frac{\text{sum}}{6}=2\cdot\frac{a_3+a_4}{2}=a_3+a_4 .
\]
Hence \(\text{sum}=6(a_3+a_4)\).

Consider where \(N\) can fall:

*If \(N\le2\):* \(a_3=6,\ a_4=7\) → median \(=6.5\).  
Sum \(=N+50\). Equation gives \(N=28\), contradicting \(N\le2\).

*If \(2<N\le6\):* \(a_3=6,\ a_4=7\) → same median \(=6.5\).  
Again \(N=28\), impossible.

*If \(6<N\le7\):* only integer possibility \(N=7\).  
List becomes 2, 6, 7, 7, 7, 28 → median \(=7\).  
Sum \(=57\) → mean \(=9.5\neq14\).

*If \(7<N<28\):* \(a_3=7,\ a_4=7\) → median \(=7\).  
Sum \(=N+50\). Equation \(N+50=84\) gives \(N=34\), which is not \(<28\).

*If \(N=28\):* median still \(=7\). Sum \(=78\) → mean \(=13\neq14\).

*If \(N>28\):* \(a_3=7,\ a_4=7\) → median \(=7\).  
Equation gives \(N+50=84\) → \(N=34\). This works (34 > 28).

Thus the only possible value is **34**.

**Answer:** 34 → Choice **E**.  

ANSWER 2: E  

---

**Problem 3:**  
A four‑digit integer with all digits even can use digits \{0,2,4,6,8\} for each place, except the thousands digit cannot be 0.  

Number of such integers: \(4\times5\times5\times5=500\).

A number is divisible by 4 iff its last two digits form a number divisible by 4.  
List all two‑digit endings with even digits (25 possibilities).  

Checking each yields 15 endings that are multiples of 4:  
00, 04, 08, 20, 24, 28, 40, 44, 48, 60, 64, 68, 80, 84, 88.

Thus the proportion is \(\frac{15}{25}=\frac{3}{5}\).

**Answer:** \(\frac{3}{5}\) → Choice **D**.  

ANSWER 3: D  

---

**Problem 4:**  
6 % tax on \$20 = \(0.06\times20 = \$1.20\).  
6.5 % tax on \$20 = \(0.065\times20 = \$1.30\).  

Difference = \$1.30 − \$1.20 = **\$0.10**.

**Answer:** \$0.10 → Choice **B**.  

ANSWER 4: B  

---

**Problem 5:**  
Sum from 1901 to 1993 (inclusive):  
Number of terms = \(1993-1901+1 = 93\).  
Sum = \(\frac{(1901+1993)\times93}{2}= \frac{3894\times93}{2}=181\,071\).

Sum from 101 to 193 (inclusive):  
Number of terms = \(193-101+1 = 93\).  
Sum = \(\frac{(101+193)\times93}{2}= \frac{294\times93}{2}=13\,671\).

Difference = \(181\,071-13\,671 = 167\,400\).

**Answer:** 167 400 → Choice **A**.  

ANSWER 5: A  

---

**Problem 6:**  
The sign pattern for the numbers \(n\) is:
- \(n\equiv1\pmod4\) → +,
- \(n\equiv2\pmod4\) → −,
- \(n\equiv3\pmod4\) → −,
- \(n\equiv0\pmod4\) → +.

Thus each block of four consecutive integers \(\{4k+1,4k+2,4k+3,4k+4\}\) contributes  
\((4k+1)-(4k+2)-(4k+3)+(4k+4)=0\).

From 1 through 1996 there are \(1996/4=499\) complete blocks, so the total sum is **0**.

**Answer:** 0 → Choice **C**.  

ANSWER 6: C  

---

**Problem 7:**  
Let \(S\) = people wearing sunglasses, \(|S|=50\).  
Let \(C\) = people wearing caps, \(|C|=35\).  
Let \(x=|S\cap C|\).

Given \(P(S\mid C)=\frac{x}{35}= \frac{2}{5}\).  
Thus \(x = \frac{2}{5}\times35 = 14\).

We need \(P(C\mid S)=\frac{x}{50}= \frac{14}{50}= \frac{7}{25}\).

**Answer:** \(\frac{7}{25}\) → Choice **B**.  

ANSWER 7: B  

---

**Problem 8:**  
A three‑digit number with digit sum 26 must use digits that sum to 26.  
The only triple of single‑digit numbers with sum 26 is \(\{9,9,8\}\).

The number is even only if its units digit is 8.  
Among the permutations of 9, 9, 8, only 998 ends in 8.  

Thus exactly **1** such number exists.

**Answer:** 1 → Choice **A**.  

ANSWER 8: A  

---

**Problem 9:**  
Let the digits be \(a,b,c\) (hundreds, tens, units) with  
\(1\le a\le9,\ 0\le b,c\le9,\ a+b+c=25\).

Set \(a'=a-1\) (so \(0\le a'\le8\)). Then  
\(a'+b+c=24\).

Count unrestricted non‑negative solutions: \(\binom{24+3-1}{2}= \binom{26}{2}=325\).

Subtract solutions where a digit exceeds its bound:
- \(a'\ge9\): treat \(a''=a'-9\) → \(a''+b+c=15\) → \(\binom{17}{2}=136\).
- \(b\ge10\): \(b'=b-10\) → \(a'+b'+c=14\) → \(\binom{16}{2}=120\).
- \(c\ge10\): similarly 120.

Intersections:
- \(a'\ge9,\ b\ge10\): \(a''+b'+c=5\) → \(\binom{7}{2}=21\).
- \(a'\ge9,\ c\ge10\): 21.
- \(b\ge10,\ c\ge10\): \(a'+b'+c'=4\) → \(\binom{6}{2}=15\).

No triple intersection (would require sum –5).

By inclusion–exclusion, invalid = \(136+120+120-21-21-15 = 319\).  
Valid = \(325-319 = 6\).

**Answer:** 6 → Choice **C**.  

ANSWER 9: C  

---

**Problem 10:**  
Exits numbered 5 through 26 give 22 exits, hence 21 intervals.  
Let the distances between consecutive exits be \(d_1,\dots,d_{21}\).  
We know \(\sum d_i = 118\) miles, and each \(d_i\ge5\).

To maximize the largest interval, make the othe

Problem 1:  
We are given a set $ R $ of nine distinct integers, six of which are $ 2, 3, 4, 6, 9, 14 $. The median of a set of nine numbers is the fifth number when sorted. To find the number of possible values for the median, we consider the possible placements of the three unknown numbers relative to the known ones. The known numbers sorted are $ 2, 3, 4, 6, 9, 14 $. The three unknown numbers can be placed in various positions to shift the median. The possible medians occur when the additional numbers are placed such that the fifth number in the sorted list can be $ 4, 5, 6, 7, 8, $ or $ 9 $. This gives six possible values for the median.  
ANSWER 1: D (corrected to C? Wait, no: the possible medians are 4, 5, 6, 7, 8, 9 — that’s 6 values, so C)  
**Wait, let's re-evaluate:** The known numbers are 2,3,4,6,9,14. Adding three numbers. The median is the 5th number. To get 4 as median: three numbers ≤4, but only 2,3,4 are available — but they are distinct, so we can add numbers like 1,5? Wait no, the three unknown numbers can be placed in different positions. For example:  
- If all three unknown numbers are ≤4: but only three such numbers exist (2,3,4) but they are already in the set. So we cannot add more. Thus, the unknown numbers must be placed in other positions.  
Actually, the unknown numbers can be:  
- All less than 2: median could be 4 if three numbers are added <4, but only 2,3,4 are already present. We cannot add numbers less than 2 without repeating? Wait, the set has distinct integers. So we can add numbers like 1,0,-1, but the problem doesn’t restrict to positive. However, the known numbers are positive. But the problem says "distinct integers", so negatives are allowed. But the median values: if we add numbers less than 2, the sorted order becomes [new numbers, 2,3,4,6,9,14]. With three new numbers, the sorted list has 9 numbers. The fifth is the median. If we add three numbers less than 2, the sorted list is [a,b,c,2,3,4,6,9,14] — no, wait, adding three numbers less than 2: the sorted list would be [a,b,c,2,3,4,6,9,14] — but a,b,c are less than 2, so the first three, then 2 is fourth, 3 is fifth? Wait, no: the original known numbers are 2,3,4,6,9,14. Adding three numbers, say x,y,z. The full sorted list will be the combination. To get median=4, we need the fifth number to be 4. So in the sorted list of 9, the fifth must be 4. That requires that four numbers are ≤4 and four are ≥4? No: the fifth is the middle. For the fifth to be 4, we need at least five numbers ≤4? No: in a sorted list of 9, the fifth is the median. For the fifth to be 4, we need at least five numbers ≤4? Actually, no: the fifth number is the one where four are less and four are greater? No: in a sorted list of 9, the fifth is the median, meaning four are ≤ median and four are ≥ median. But since numbers can be equal, but here all distinct. So for 4 to be the median, we need exactly four numbers less than 4 and four numbers greater than 4? But 4 is in the set. So the sorted list must have four numbers <4, then 4, then four numbers >4. But the known numbers include 2,3,4,6,9,14. So 2,3 are <4, and 6,9,14 are >4. So currently, without unknowns, we have two <4 and three >4. Adding three unknowns. To have four <4 and four >4, we need to add two numbers <4 and one number >4? But then the sorted list would be [two new <4, 2,3,4, then the new >4, then 6,9,14] — but that’s only three numbers after 4? Wait, no: the sorted list would be [new <4 (two), 2,3,4, new >4, 6,9,14] — that’s 9 numbers? 2+3+1+3=9? Wait, no: the two new <4, then 2,3,4 (three numbers), then the new >4, then 6,9,14 (three numbers). So total: 2+3+1+3=9. The fifth number is 4. So yes, we can achieve median=4 by adding two numbers <4 and one number >4. Similarly, we can achieve:  
- Median=5: by adding numbers such that the fifth number is 5. But 5 is not in the set. So we need the fifth number to be 5. This requires that four numbers are <5 and four are >5. But 5 is not present. So we need to add 5? But the set must consist of distinct integers, and 5 is not in the given. So we can add 5. Then the sorted list would include 5. For example, add 5, and two other numbers. To have four <5 and four >5: currently, without unknowns, numbers <5: 2,3,4 (three), numbers >5: 6,9,14 (three). So to have four <5, we need to add one number <5. To have four >5, we need to add one number >5. But we are adding three numbers. So we can add 5, and say 1 and 10. Then sorted: 1,2,3,4,5,6,9,10,14. Median is 5. So median=5 is achievable. Similarly, median=6: we can add numbers such that the fifth is 6. Currently, numbers <6: 2,3,4 (three), numbers >6: 9,14 (two). So to have four <6 and four >6, we need to add one number <6 and two numbers >6. For example, add 5,10,15. Then sorted: 2,3,4,5,6,9,10,14,15. Median=6. Similarly, median=7: add numbers so fifth is 7. Currently, <7: 2,3,4,6 (four), >7: 9,14 (two). So to have four <7 and four >7, we need to add zero <7 (since already four) and two >7. But we are adding three numbers. So we can add 7,8,15. Then sorted: 2,3,4,6,7,8,9,14,15. Median=7. Similarly, median=8: add 8, and two numbers, say 10,11. Then sorted: 2,3,4,6,8,9,10,11,14. Median=8? The fifth is 8. But to have four <8 and four >8: currently, without unknowns, <8: 2,3,4,6 (four), >8:9,14 (two). So to have four >8, we need to add two numbers >8. But we are adding three numbers. So we can add 8,9,10? But 9 is already there. Add 8,10,11. Then sorted: 2,3,4,6,8,9,10,11,14. The fifth is 8. So yes. Similarly, median=9: add numbers so fifth is 9. Currently, <9: 2,3,4,6 (four), >9:14 (one). So to have four >9, we need to add three numbers >9. But we are adding three numbers. So add 10,11,12. Then sorted: 2,3,4,6,9,10,11,12,14. Median=9. So possible medians: 4,5,6,7,8,9. That’s six values.  
ANSWER 1: C  

Problem 2:  
We are given the list 2,6,7,7,28. A number $ N $ is added. The mean becomes twice the median. The original list has five numbers. After adding $ N $, there are 

### Problem 1:
We have a set of 9 distinct integers, so the median (middle value when sorted) is the 5th element. The 6 known numbers sorted are [2,3,4,6,9,14], with 3 additional distinct integers to add. To find possible median values:
- The minimum possible median is 3: add 3 numbers smaller than 2, sorted set becomes [a,b,c,2,3,4,6,9,14], 5th term=3.
- The maximum possible median is 9: add 3 numbers larger than 14, sorted set becomes [2,3,4,6,9,14,x,y,z], 5th term=9.
- All integers between 3 and 9 are valid medians: 3,4,5,6,7,8,9, for 7 total possible values.
Tempting wrong choices: any count less than 7, from missing intermediate medians that can be formed by adding integers between the known values.
ANSWER 1: D

---

### Problem 2:
Original list sum: 2+6+7+7+28=50. Adding N gives a list of 6 elements, sum=50+N, mean=(50+N)/6. The median of 6 elements is the average of the 3rd and 4th sorted terms. The problem states mean=2×median, so 50+N=12×median.
- If N>28, sorted list is [2,6,7,7,28,N], median=(7+7)/2=7. Then 50+N=12×7=84 → N=34, which fits.
Testing other options confirms no other N satisfies the condition.
ANSWER 2: E

---

### Problem 3:
A number is divisible by 4 if its last two digits form a number divisible by 4. For 4-digit numbers with all even digits:
- Total count: 4 (non-zero thousands digits:2,4,6,8) ×5×5×5 (all even digits for other places)=500.
- Valid last two-digit pairs (both even, divisible by 4): For any even tens digit, 10×even is divisible by 4, so only units digits 0,4,8 work, giving 3 valid units digits per tens digit: 5×3=15 valid pairs.
- Fraction: (4×5×15)/500=300/500=3/5.
ANSWER 3: D

---

### Problem 4:
The tax rate difference is 6.5%-6%=0.5%. 0.5% of $20 is 0.005×20=$0.10.
ANSWER 4: B

---

### Problem 5:
Both sums have the same number of terms: 1993-1901+1=93 terms, 193-101+1=93 terms. Subtract term-by-term: (1901-101)+(1902-102)+...+(1993-193)=1800×93=167400.
ANSWER 5: A

---

### Problem 6:
Group terms in sets of 4: (1-2-3+4)+(5-6-7+8)+... each group sums to 0. There are 1996/4=499 groups, total sum=0.
ANSWER 6: C

---

### Problem 7:
Number of people with both caps and sunglasses: 35 cap-wearers × 2/5=14. Probability a sunglass-wearer has a cap:14/50=7/25.
ANSWER 7: B

---

### Problem 8:
An even 3-digit number with digit sum 26 must have an even units digit. The only possible even units digit is 8 (any smaller even digit leaves a sum of the first two digits >18, impossible). The only number is 998, so 1 valid number.
ANSWER 8: A

---

### Problem 9:
Three-digit numbers with digit sum 25 are all permutations of (9,9,7) and (9,8,8). There are 3 permutations of (9,9,7) and 3 permutations of (9,8,8), for 6 total numbers.
ANSWER 9: C

---

### Problem 10:
Between exit 5 and 26, there are 21 gaps. To maximize one gap, minimize the other 20 gaps at 5 miles each: 20×5=100 miles. Maximum gap=118-100=18 miles.
ANSWER 10: C

---

### Problem 11:
Total sum of 5 scores=5×90=450. Median=91, so sorted scores [a,b,91,c,d]. Mode=94, so 94 must appear at least twice, so c=d=94, sum c+d=188. Sum of the two lowest scores=450-91-188=171.
ANSWER 11: B

---

### Problem 12:
Using +,-,* exactly once, one valid calculation: 5-4+6×3=1+18=19, which uses all three operations and matches option E.
ANSWER 12: E

### Problem 1:
We need to find the number of possible median values for a set of 9 distinct integers, 6 of which are 2, 3, 4, 6, 9, 14. The median of 9 numbers is the 5th element when sorted in ascending order. We add 3 distinct integers (not in the given set) and analyze possible medians:
1.  **Case 1: All 3 new numbers < 2**: Sorted list is [a,b,c,2,3,4,6,9,14], median = 3.
2.  **Case 2: 2 new numbers < 2, 1 new number ≥4**: Sorted list is [a,b,2,3,4,...], median =4.
3.  **Case 3: 1 new number <2**:
    - If we add 5 (the only integer between 4 and 6): Sorted list is [a,2,3,4,5,6,...], median=5.
    - If we do not add 5: Sorted list is [a,2,3,4,6,...], median=6.
4.  **Case 4: 0 new numbers <2 (all new numbers ≥5)**:
    - Add 5: Sorted list [2,3,4,5,6,...], median=6.
    - Add a number between 6 and9 (7 or 8): Sorted list [2,3,4,6,7,...] or [2,3,4,6,8,...], median=7 or 8.
    - Add all new numbers ≥9: Sorted list [2,3,4,6,9,...], median=9.
The only impossible medians are 2 (needs 5 numbers ≤2, only 4 max) and values >9 (existing numbers already have 5 elements ≤9). The possible medians are 3,4,5,6,7,8,9: 7 values.
ANSWER 1: D
---
### Problem 2:
Original list: [2,6,7,7,28], sum = 2+6+7+7+28=50. Insert N to make 6 elements. The median of 6 elements is the average of the 3rd and 4th sorted elements. The mean is $\frac{50+N}{6}$, which equals twice the median:
$$\frac{50+N}{6} = 2 \times \text{median} \implies 50 + N = 12 \times \text{median} \implies N = 12 \times \text{median} -50.$$
Test possible median values from the sorted list with N:
- If N ≤7: Sorted list has 3rd=7, 4th=7, median=7. Then $N=12*7 -50=84-50=34$, which is >7, valid.
- Other median values (e.g., 6.5) give N=28, which does not fit the required sorted position.
Verify N=34: Sorted list [2,6,7,7,28,34], median=(7+7)/2=7, mean=(50+34)/6=14, which is twice 7.
ANSWER 2: E
---
### Problem 3:
Positive 4-digit numbers with all even digits:
- Thousands place: 2,4,6,8 (4 choices, no 0).
- Hundreds, tens, units: 0,2,4,6,8 (5 choices each).
Total numbers: $4 \times 5 \times 5 \times5=500$.
A number is divisible by 4 iff its last two digits form a number divisible by 4. For even tens digits (0,2,4,6,8), a two-digit number $10a + b$ is divisible by 4 iff $b$ (units digit) is 0,4,8 (since $10a \equiv 0 \mod4$ for even a). So valid endings: 5 (tens choices) ×3 (units choices)=15.
Total valid numbers: $4 \times5 \times5 \times3=300$. Fraction: $\frac{300}{500}=\frac{3}{5}$.
ANSWER 3: D
---
### Problem 4:
Difference between 6.5% and 6% tax on $20:
$$0.065*20 - 0.06*20 = (0.065-0.06)*20=0.005*20=0.10.$$
ANSWER 4: B
---
### Problem 5:
Both sums have 93 terms (1993-1901+1=93, 193-101+1=93). Each term in the first sum is 1800 more than the corresponding term in the second sum:
Total difference = $93 \times 1800 = 167400$.
ANSWER 5: A
---
### Problem 6:
The sign pattern repeats every 4 terms: $+ - - +$, and each block of 4 sums to 0:
$$(1-2-3+4)+(5-6-7+8)+\dots+(1993-1994-1995+1996)=0.$$
1996 is divisible by 4, so there are 499 such blocks, total sum 0.
ANSWER 6: C
---
### Problem 7:
Let $S$ = sunglass wearers ($|S|=50$), $C$ = cap wearers ($|C|=35$). $P(S|C)=\frac{|S∩C|}{|C|}=\frac{2}{5}$, so $|S∩C|=\frac{2}{5}*35=14$.
The probability a sunglass wearer wears a cap is $P(C|S)=\frac{|S∩C|}{|S|}=\frac{14}{50}=\frac{7}{25}$.
ANSWER 7: B
---
### Problem 8:
3-digit numbers with digit sum 26: maximum digit sum is 27 (999), so 26 is 1 less than maximum, meaning digits are permutations of (9,9,8). The only even permutation is 998 (units digit 8). So 1 number.
ANSWER 8: A
---
### Problem 9:
3-digit numbers with digit sum 25: 25 is 2 less than 27, so digits are permutations of (9,9,7) and (9,8,8):
- Permutations of (9,9,7): 997,979,799 (3 numbers).
- Permutations of (9,8,8): 988,898,889 (3 numbers).
Total 6 numbers.
ANSWER 9: C
---
### Problem 10:
Between exit 5 and 26, there are $26-5=21$ gaps, total distance 118 miles. Each gap ≥5 miles. To maximize one gap, minimize the other 20: $20*5=100$, so maximum gap = $118-100=18$.
ANSWER 10: C
---
### Problem 11:
Five sorted scores: $a≤b≤c≤d≤e$. Median $c=91$, mean 90, so total sum $5*90=450$, so $a+b+d+e=450-91=359$. Mode is 94, so 94 appears at least twice. Since $c=91$, 94 can only be in $d$ and $e$, so $d=e=94$. Then $a+b=359-94-94=171$. This sum is fixed (valid examples exist, e.g., 81+90=171).
ANSWER 11: B
---
### Problem 12:
Permute +, -, × in 5 __4 __6 __3, using standard order of operations (multiply first):
- $5 -4 +6×3 = 5-4+18=19$, which is an option.
All other permutations give values not in the choices.
ANSWER 12: E

I'll solve each problem step by step.

---

**Problem 1:**

We have 6 elements: {2, 3, 4, 6, 9, 14}, and need to add 3 more distinct integers to make 9 total.

The median of 9 numbers is the 5th element when arranged in order.

Currently we have 6 numbers. By strategically placing 3 new numbers, we can shift which value becomes the median.

Let me consider what values can be the median (5th position):
- If we add 3 numbers all ≤ 2, the sorted list has our 6 original numbers in positions 4-9, so median is 4
- If we add numbers strategically between and around the original set, we can make the median be: 2, 3, 4, 6, 9, 14, or a new number
- If we add 3 numbers all ≥ 14, the sorted list has our 6 original numbers in positions 1-6, so median is 9

Testing systematically:
- Median = 2: add three numbers < 2 (e.g., -2, -1, 0) ✓
- Median = 3: add two < 2, one between 3 and 4 ✓
- Median = 4: add three < 2, or arranged appropriately ✓
- Median = 6: add one or two strategically ✓
- Median = 9: add three > 14, or arranged appropriately ✓
- Median = 14: add three > 14 ✓
- Median = a new value between existing numbers ✓

The possible medians are: 2, 3, 4, 6, 9, 14, and any new integer we insert. We can get 7 different values.

**ANSWER 1: D**

---

**Problem 2:**

Original list: 2, 6, 7, 7, 28. Sum = 50.

When N is inserted, we have 6 numbers. Let's find the median of 6 numbers (average of 3rd and 4th when sorted).

Sum of all 6 numbers = 50 + N
Mean = (50 + N)/6

We need: Mean = 2 × Median

Let me test by considering where N falls:
- If N ≤ 2: sorted list is N, 2, 6, 7, 7, 28. Median = (6+7)/2 = 6.5. Mean = 2×6.5 = 13, so 50+N = 78, N = 28. Contradiction.
- If 2 < N ≤ 6: sorted list is 2, N, 6, 7, 7, 28. Median = (6+7)/2 = 6.5. Mean = 13, so N = 28. Contradiction.
- If 6 < N ≤ 7: sorted list is 2, 6, N, 7, 7, 28. Median = (N+7)/2. Mean = 2×(N+7)/2 = N+7. So (50+N)/6 = N+7, giving 50+N = 6N+42, so 5N = 8, N = 1.6. Not in range.
- If 7 < N ≤ 28: sorted list is 2, 6, 7, N, 7, 28 or 2, 6, 7, 7, N, 28. Median = (7+N)/2. Mean = 2×(7+N)/2 = 7+N. So (50+N)/6 = 7+N, giving 50+N = 42+6N, so 5N = 8, N = 1.6. Not valid.
- If N > 28: sorted list is 2, 6, 7, 7, 28, N. Median = (7+7)/2 = 7. Mean = 14. So 50+N = 84, N = 34. ✓

**ANSWER 2: E**

---

**Problem 3:**

Four-digit numbers with all even digits: first digit ∈ {2,4,6,8}, others ∈ {0,2,4,6,8}.

Total count: 4 × 5 × 5 × 5 = 500

For divisibility by 4, the last two digits must form a number divisible by 4.

Last two digits from {00, 02, 04, 06, 08, 20, 22, 24, 26, 28, 40, 42, 44, 46, 48, 60, 62, 64, 66, 68, 80, 82, 84, 86, 88}

Divisible by 4: 00, 04, 08, 20, 24, 28, 40, 44, 48, 60, 64, 68, 80, 84, 88 (15 out of 25)

Count divisible by 4: 4 × 5 × 15 = 300

But wait, the second digit is 5 choices, third is which determines divisibility by 4.

Actually: 4 × 5 × (number of even two-digit combos divisible by 4) = 4 × 5 × 15 = 300

Hmm, let me recalculate. First digit: 4 choices. Second digit: 5 choices. Last two digits must be divisible by 4: check all pairs from {0,2,4,6,8}×{0,2,4,6,8} giving 15 valid pairs.

Count = 4 × 5 × 15 = 300. Fraction = 300/500 = 3/5.

**ANSWER 3: D**

---

**Problem 4:**

6.5% tax on $20: 0.065 × 20 = $1.30
6% tax on $20: 0.06 × 20 = $1.20
Difference: $1.30 - $1.20 = $0.10

**ANSWER 4: B**

---

**Problem 5:**

Sum from 1901 to 1993: This is 93 consecutive integers starting at 1901.
Sum = (1901 + 1993)/2 × 93 = 1947 × 93 = 181,071

Sum from 101 to 193: This is 93 consecutive integers starting at 101.
Sum = (101 + 193)/2 × 93 = 147 × 93 = 13,671

Difference = 181,071 - 13,671 = 167,400

**ANSWER 5: A**

---

**Problem 6:**

Pattern: 1 - 2 - 3 + 4 + 5 - 6 - 7 + 8 + 9 - 10 - 11 + ...

Group by 4s: (1-2-3+4) + (5-6-7+8) + (9-10-11+12) + ... + (1993-1994-1995+1996)

Each group of 4: (4k+1) - (4k+2) - (4k+3) + (4k+4) = 4k+1-4k-2-4k-3+4k+4 = 0

From 1 to 1996, we have 1996/4 = 499 complete groups, each summing to 0.

Total = 0

**ANSWER 6: C**

---

**Problem 7:**

Let x = people wearing both sunglasses and caps.

P(sunglasses | cap) = x/35 = 2/5, so x = 14.

P(cap | sunglasses) = 14/50 = 7/25.

**ANSWER 7: B**

---

**Problem 8:**

Three-digit even numbers with digit sum 26.

For an even number, last digit ∈ {0, 2, 4, 6, 8}.

If last digit = 0: first two digits sum to 26. Max is 9+9=18. Impossible.
If last digit = 2: first two digits sum to 24. Only 9+9=18 < 24. Impossible.
If last digit = 4: first two digits sum to 22. Impossible (max 18).
If last digit = 6: first two digits sum to 20. Impossible.
If last digit = 8: first two digits sum to 18. This requires 9+9=18. Number: 998. ✓

Only 1 number.

**ANSWER 8: A**

---

**Problem 9:**

Three-digit numbers abc where a+b+c=25, with a ∈ {1,...,9}, b,c ∈ {0,...,9}.

a+b+c = 25 means b+c = 25-a.

For valid solutions: 0 ≤ b+c ≤ 18, so 7 ≤ a ≤ 9.

- a=7: b+c=18. Solutions: (9,9) → 1 way
- a=8: b+c=17. Solutions: (8,9), (9,8) → 2 ways
- a=9: b+c=16. Solutions: (7,9), (8,8), (9,7) → 3 ways

Total = 1 + 2 + 3 = 6

**ANSWER 9: C**

---

**Problem 10:**

Exits from 5th to 26th = 26-5 = 21 segments between consecutive exits.

Each segment ≥ 5 miles. Total distance = 118 miles.

If 20 segments are 5 miles each: 20 × 5 = 100 miles.
Remaining for the largest segment: 118 - 100 = 18 miles.

**ANSWER 10: C**

---

**Problem 11:**

Five scores: mean = 90 (sum = 450), median = 91, mode = 94.

Arranged in order: a, b, c, d, e where c = 91 (median).

Mode = 94 means 94 appears at least twice. Since median is 91, we must have 94 appearing in {d, e}.

Case 1: d = e = 94. Then a + b + 91 + 94 + 94 = 450, so a + b = 171.

Check consistency: we need a ≤ b ≤ 91 and mode is 94 (appears twice). If a or b equals 94, mode wouldn't be unique. So a, b < 91 works. ✓

**ANSWER 11: B**

---

**Problem 12:**

We need to place +, −, × in three blanks: 5 __ 4 __ 6 __ 3

Trying combinations:
- 5 + 4 × 6 − 3 = 5 + 24 − 3 = 26
- 5 × 4 + 6 − 3 = 20 + 6 − 3 = 23
- 5 × 4 − 6 + 3 = 20 − 6 + 3 = 17
- 5 + 4 − 6 × 3 = 5 + 4 − 18 = −9
- 5 − 4 

Problem 1:

We are given a set of 9 distinct integers. Six of them are
\[
2,3,4,6,9,14.
\]
We want the number of possible values of the median of the 9 numbers.

For 9 numbers, the median is the 5th number when the list is ordered.

Now look at the six given numbers in sorted order:
\[
2,3,4,6,9,14.
\]
Since there are 3 other distinct integers, the median can be any number that ends up in the 5th position after inserting those 3 numbers.

Let’s see what values are possible:

- If we place 3 new numbers all very small, then the 5th number could be 4.
- It could also be 6, or 9, or even 14, depending on how many new numbers are inserted before each.
- It cannot be 2 or 3, because there would need to be at least 4 numbers less than or equal to it, but only 3 extra numbers are available and the given numbers already limit this.
- It cannot be any number not among the given six if the median must be one of the existing numbers? Let’s check carefully: since the 3 extra numbers are distinct integers and can be placed anywhere, the 5th number could indeed be one of the inserted numbers, but then it would have to lie between the 4th and 6th numbers. Since the six given numbers already occupy several possible positions, the median can also be a new integer. But because the set is of distinct integers and the question asks how many possible values, we should count all possible medians that can occur.

A cleaner way: the median of 9 distinct integers is the middle element. With 6 fixed numbers, the 3 added numbers can be arranged so that the median is any integer between the 4th smallest and 6th smallest of the fixed numbers, inclusive, but here the possible distinct medians are:
\[
4,5,6,7,8,9
\]
Wait — can all of these really happen? Let’s test.

- Median 4: choose 3 new numbers less than 4.
- Median 5: choose 3 new numbers so that 5 is between 4 and 6 and becomes 5th.
- Median 6: choose 3 new numbers with 6 as middle.
- Median 7, 8, 9 similarly.

So the possible medians are the integers from 4 through 9, which is 6 values.

However, we must match the provided choices. The intended count is the number of possible median values among the listed integers and inserted integers. Since the only restriction is distinct integers, every integer from 4 through 9 can occur.

That gives 6 possible values.

ANSWER 1: C

---

Problem 2:

We insert a number \(N\) into the list
\[
2,6,7,7,28.
\]
Then the mean becomes twice the median.

Let’s analyze the median after inserting \(N\) into 6 numbers total.

The original sorted list is:
\[
2,6,7,7,28.
\]
There are 5 numbers. After inserting one number, there are 6 numbers, so the median is the average of the 3rd and 4th numbers in the sorted list.

We try the choices.

If \(N=7\), the list becomes
\[
2,6,7,7,7,28.
\]
Median:
\[
\frac{7+7}{2}=7.
\]
Mean:
\[
\frac{2+6+7+7+7+28}{6}=\frac{57}{6}=9.5.
\]
Twice the median would be 14, not 9.5, so not 7.

Try \(N=14\):
\[
2,6,7,7,14,28.
\]
Median:
\[
\frac{7+7}{2}=7.
\]
Mean:
\[
\frac{2+6+7+7+14+28}{6}=\frac{64}{6}=\frac{32}{3},
\]
not 14.

Try \(N=20\):
\[
2,6,7,7,20,28.
\]
Median is still 7.
Mean:
\[
\frac{2+6+7+7+20+28}{6}=\frac{70}{6}=\frac{35}{3},
\]
not 14.

Try \(N=28\):
\[
2,6,7,7,28,28.
\]
Median is 7.
Mean:
\[
\frac{2+6+7+7+28+28}{6}=\frac{78}{6}=13,
\]
not 14.

Try \(N=34\):
\[
2,6,7,7,28,34.
\]
Median is 7.
Mean:
\[
\frac{2+6+7+7+28+34}{6}=\frac{84}{6}=14.
\]
Twice the median is
\[
2\cdot 7=14.
\]
This works.

So \(N=34\).

ANSWER 2: E

---

Problem 3:

We consider all positive four-digit integers whose digits are all even.

Even digits are
\[
0,2,4,6,8.
\]
A four-digit number cannot start with 0, so the thousands digit has 4 choices:
\[
2,4,6,8.
\]
Each of the other three digits has 5 choices, so the total number is
\[
4\cdot 5^3.
\]

Now determine how many are divisible by 4.

A number is divisible by 4 if its last two digits form a number divisible by 4.

So we only need to count ordered pairs of even digits \((t,u)\) from \(\{0,2,4,6,8\}\) such that \(10t+u\) is divisible by 4.

Check the possible last two digits:

- \(00\) yes
- \(02\) no
- \(04\) yes
- \(06\) no
- \(08\) yes

- \(20\) yes
- \(22\) no
- \(24\) yes
- \(26\) no
- \(28\) yes

- \(40\) yes
- \(42\) no
- \(44\) yes
- \(46\) no
- \(48\) yes

- \(60\) yes
- \(62\) no
- \(64\) yes
- \(66\) no
- \(68\) yes

- \(80\) yes
- \(82\) no
- \(84\) yes
- \(86\) no
- \(88\) yes

For each tens digit, 3 of the 5 possibilities work, so there are
\[
5\cdot 3=15
\]
valid last-two-digit pairs.

Since the first two digits can be anything allowed:
- thousands digit: 4 choices
- hundreds digit: 5 choices

Total divisible by 4:
\[
4\cdot 5\cdot 15
\]
Total numbers:
\[
4\cdot 5^3 = 4\cdot 125
\]
So the fraction is
\[
\frac{4\cdot 5\cdot 15}{4\cdot 5^3}=\frac{15}{25}=\frac35.
\]

ANSWER 3: D

---

Problem 4:

An item costs \$20 before tax.

Difference between 6.5% tax and 6% tax is
\[
0.5\% = 0.005.
\]
So the difference in tax is
\[
0.005 \times 20 = 0.1
\]
dollar, which is \(\$0.10\).

ANSWER 4: B

---

Problem 5:

We need to compute
\[
(1901+1902+\cdots+1993) - (101+102+\cdots+193).
\]

Notice each term in the second sum is exactly 1800 less than the corresponding term in the first sum:
\[
1901-101=1800,\quad 1902-102=1800,\ \ldots,\ 1993-193=1800.
\]
How many terms are there?

From 1901 to 1993 inclusive:
\[
1993-1901+1=93
\]
terms.

So the difference is
\[
93 \cdot 1800 = 167400.
\]

ANSWER 5: A

---

Problem 6:

We evaluate
\[
1-2-3+4+5-6-7+8+9-10-11+\cdots+1992+1993-1994-1995+1996.
\]

Look at the pattern in blocks of 4:
\[
(1-2-3+4),\ (5-6-7+8),\ (9-10-11+12),\ \ldots
\]
Each block of 4 consecutive integers has sum
\[
n-(n+1)-(n+2)+(n+3)=0
\]
when \(n\equiv 1 \pmod 4\).

So all full blocks cancel.

Now check the last block:
The pattern continues through \(1992+1993-1994-1995+1996\), which is not a full 4-term block. But we can group from 1 to 1996 in blocks of 4:
\[
(1-2-3+4)+(5-6-7+8)+\cdots+(1993-1994-1995+1996).
\]
Each block sums to 0.

Therefore the entire 

Problem 1:  
We have 9 distinct integers. Six of them are fixed: 2, 3, 4, 6, 9, 14. The other 3 are distinct integers of our choice.  
With 9 numbers, the median is the 5th number when sorted.

So the median depends only on how many of the fixed numbers fall below/at various positions, and where we insert the 3 new numbers.

Let’s analyze possible 5th elements.

### Step 1: Sort the fixed numbers
Fixed numbers in increasing order:
2, 3, 4, 6, 9, 14

The median (5th) could be one of these fixed numbers, or one of the new ones.

### Casework by choosing what the 5th number is

#### Can the median be 2?
For 2 to be the median, we need at least 4 numbers ≤ 2, but only 2 itself is ≤ 2 among the fixed set (and new numbers would have to be distinct integers ≤ 2).  
That means we’d need 4 numbers less than or equal to 2, including 2, so we’d need 3 additional distinct integers ≤ 2 (other than 2). That’s possible (e.g., 0, −1, 1).  
But then 2 would be the 5th, meaning exactly 4 numbers are <2. Can we guarantee? Actually, since we can choose any 3 extra integers, we can make exactly 4 numbers less than 2 by picking 3 integers <2 plus none from fixed besides 2. That would make 2 the 5th. However, note 2 is already one of the 9 distinct integers, so it can be the median.  
So potentially 2 is possible.

But the problem asks for the number of possible median values, and the choices are between 4 and 8, so likely only medians among the fixed list can work. Let’s check more carefully: the fixed numbers below 2 is only none (except 2). To make 2 the 5th, we need 4 numbers smaller than 2. We only control 3 extra numbers, so we can only create at most 3 numbers smaller than 2. Therefore, we cannot reach 4 smaller numbers.  
✅ Conclusion: median cannot be 2.

Similarly, we test each fixed candidate using “how many fixed numbers are below it”.

### General method
For a number \(x\) to be the median (5th), there must be exactly 4 numbers less than \(x\), and at least one number equal to \(x\). If \(x\) is one of the fixed numbers, we can count how many fixed numbers are less than \(x\), then see if the remaining “needed” smaller numbers can come from the 3 new integers.

Let’s compute fixed counts less than each candidate:

- Below 2: 0  
- Below 3: 1 (just 2)  
- Below 4: 2 (2,3)  
- Below 6: 3 (2,3,4)  
- Below 9: 4 (2,3,4,6)  
- Below 14: 5 (2,3,4,6,9)

We need exactly 4 numbers less than the median.

Now see possibilities:

#### Median = 3
Fixed numbers below 3: 1.  
We need 4 total below 3, so we must add 3 new numbers < 3. That’s possible.  
✅ So 3 can be median.

#### Median = 4
Fixed below 4: 2. Need 2 more from new numbers < 4. Possible.  
✅ 4 can be median.

#### Median = 6
Fixed below 6: 3. Need 1 more from new numbers < 6. Possible.  
✅ 6 can be median.

#### Median = 9
Fixed below 9: 4 already. Then all 4 smaller numbers are fixed, and the median can be 9 regardless of where new numbers go as long as none are between 9 and 14 in a way that changes ordering?  
We only need 9 to be the 5th, meaning at least 4 numbers are <9 (we have exactly 4 fixed ones). If we add any new numbers <9, that would make more than 4 numbers less than 9 and break it. So we must choose the 3 new integers all > 9. That’s possible.  
✅ 9 can be median.

#### Median = 14
Fixed below 14: 5 already. But we need exactly 4 below the median. Impossible.  
❌ 14 cannot be median.

#### Median = 2
Fixed below 2: 0, need 4 from new numbers <2, but we only have 3 new numbers. Impossible.  
❌ 2 cannot be median.

So the only fixed medians are: 3, 4, 6, 9 → 4 values.

Are there possible medians that are not among the fixed numbers (i.e., one of the 3 new integers)?  
Yes, if we choose all new integers appropriately.

Let the median be a new number \(y\). Then among the fixed numbers, some number of them must be less than \(y\). Since there are only 6 fixed numbers, the possible counts of fixed elements less than \(y\) range from 0 to 6.

But we need exactly 4 numbers total less than the median. Since the median is new, all 4 numbers less than it must come from:
- some fixed numbers less than it, plus
- some of the other two new numbers (because one new number is the median itself).

We can make any new median that sits in the “gaps” between fixed numbers such that exactly 4 elements end up below it.

The median \(y\) must be positioned so that:
- the number of fixed elements less than \(y\) is either 1,2,3, or 4 (because we then use the remaining smaller ones from the other new numbers, limited to 0,1,2, or 3).
But we must also ensure distinctness and that \(y\) lies in the right interval.

There are gaps:
(-∞,2), (2,3), (3,4), (4,6), (6,9), (9,14), (14,∞)

Try each gap for possible median:

- If \(y\in(2,3)\): fixed less than y = 1 (only 2). Need 3 more < y from the other new numbers. Possible.  
So any integer y in (2,3) means y=?? none (only integer between 2 and 3 doesn’t exist). So no.
- \(y\in(3,4)\): fixed less than y = 2. Need 2 more < y. Possible. But no integer between 3 and 4. none.
- \(y\in(4,6)\): fixed less than y = 3 (2,3,4). Need 1 more from new numbers <y. Possible. Integers in (4,6): only 5. So y=5 works.
- \(y\in(6,9)\): fixed less than y = 4 (2,3,4,6). Need 0 more. Then both other new numbers must be >y. Possible. Integers in (6,9): 7,8 → both could be median.
- \(y\in(9,14)\): fixed less than y = 5, but we need total 4 < median. impossible.

Thus additional possible medians are 5, 7, 8 → 3 more values.

Total possible median values: fixed ones (3,4,6,9) = 4 plus new ones (5,7,8)=3 gives 7.

✅ Answer choice is 7.

ANSWER 1: D


Problem 2:  
List is: 2, 6, 7, 7, 28 (five numbers). Insert N to make 6 numbers total.

New mean and median relation: “mean is now twice as great as the median.”

### Step 1: Compute original sum
Sum of given numbers:
2 + 6 + 7 + 7 + 28 = 50

After inserting N, total sum = 50 + N.  
There are 6 numbers, so mean is:
\[
\text{mean}=\frac{50+N}{6}
\]

### Step 2: Find the median
Sort of the ne

Problem 1:
We have a set $R$ of 9 distinct integers. Six are $\{2, 3, 4, 6, 9, 14\}$. Let the three unknown integers be $x, y, z$. When sorted, the median is the 5th element.
To minimize the median, we pick the three smallest possible integers (e.g., $-1, 0, 1$). The sorted set would be $\{-1, 0, 1, 2, 3, 4, 6, 9, 14\}$. The median is 3.
To maximize the median, we pick the three largest possible integers (e.g., $15, 16, 17$). The sorted set would be $\{2, 3, 4, 6, 9, 14, 15, 16, 17\}$. The median is 9.
The possible values for the median are integers between 3 and 9 inclusive, excluding values already occupied by the fixed set if they cannot be the median. However, we can choose $x, y, z$ to make any integer from 3 to 9 the median.
Values: 3, 4, 5, 6, 7, 8, 9. There are 7 possible values.
ANSWER 1: D

Problem 2:
List: $2, 6, 7, 7, 28$. $N$ is added. Mean = $\frac{2+6+7+7+28+N}{6} = \frac{50+N}{6}$.
If $N$ is added, the median depends on $N$.
If $N \le 7$, median is 7. Mean = $2 \times 7 = 14 \implies 50+N = 84 \implies N=34$ (Contradiction).
If $N \ge 7$, let's test $N=20$: List $\{2, 6, 7, 7, 20, 28\}$, median is $(7+7)/2 = 7$. Mean = $70/6 \neq 14$.
If $N=14$: List $\{2, 6, 7, 7, 14, 28\}$, median is $(7+7)/2 = 7$. Mean = $64/6 \neq 14$.
If $N=28$: List $\{2, 6, 7, 7, 28, 28\}$, median is 7. Mean = $78/6 = 13 \neq 14$.
If $N=34$: List $\{2, 6, 7, 7, 28, 34\}$, median is $(7+28)/2 = 17.5$. Mean = $84/6 = 14$. $14 \neq 35$.
Wait, try $N=20$: Mean = $70/6 \approx 11.6$. Median = 7.
Try $N=34$: Mean = 14. Median = 17.5.
Re-check: $50+N = 12 \times \text{median}$. If $N=20$, median is 7, $70 = 12 \times 7$ (No).
If $N=34$, median is 17.5, $84 = 12 \times 17.5$ (No).
Try $N=28$: Mean = 13, Median = 7. $13 \neq 14$.
Try $N=20$: Mean = $70/6 = 11.66$, Median = 7.
Let's check $N=20$ again. $50+20 = 70$. $70/6 = 11.66$.
Actually, if $N=20$, list is $2, 6, 7, 7, 20, 28$. Median is $(7+7)/2 = 7$. $2 \times 7 = 14$. $70/6 \neq 14$.
If $N=34$, median is $(7+28)/2 = 17.5$. Mean = $84/6 = 14$. $14 \neq 2 \times 17.5$.
If $N=20$, mean is $11.66$. If $N=34$, mean is 14.
Looking at choices, $N=20$ is the only one that yields a mean close to twice a median.
ANSWER 2: C

Problem 3:
Digits are $\{0, 2, 4, 6, 8\}$. First digit cannot be 0 (4 choices: 2, 4, 6, 8). Others have 5 choices. Total = $4 \times 5 \times 5 \times 5 = 500$.
Divisible by 4 means the last two digits form a multiple of 4. Possible pairs: 00, 04, 08, 20, 24, 28, 40, 44, 48, 60, 64, 68, 80, 84, 88. (15 pairs).
Total numbers = $4 \times 5 \times 15 = 300$.
Fraction = $300/500 = 3/5$.
ANSWER 3: D

Problem 4:
$0.065 \times 20 = 1.30$. $0.06 \times 20 = 1.20$.
Difference = $1.30 - 1.20 = 0.10$.
ANSWER 4: B

Problem 5:
$(1901-101) + (1902-102) + \dots + (1993-193)$.
Each term is $1800$. There are $1993 - 1901 + 1 = 93$ terms.
$93 \times 1800 = 167,400$.
ANSWER 5: A

Problem 6:
Group into sets of 4: $(1-2-3+4) + (5-6-7+8) + \dots + (1993-1994-1995+1996)$.
Each group: $1-2-3+4 = 0$.
There are $1996/4 = 499$ groups. $499 \times 0 = 0$.
ANSWER 6: C

Problem 7:
$C = 35$. $P(S|C) = 2/5$. So $S \cap C = 35 \times 2/5 = 14$.
$S = 50$. $P(C|S) = (S \cap C) / S = 14/50 = 7/25$.
ANSWER 7: B

Problem 8:
Digits $a, b, c$. $a+b+c = 26$. $c \in \{0, 2, 4, 6, 8\}$.
If $c=8, a+b=18 \implies (9,9)$. (1)
If $c=6, a+b=20$ (None).
If $c=4, a+b=22$ (None).
If $c=2, a+b=24$ (None).
If $c=0, a+b=26$ (None).
Wait, $a+b=18 \implies 998$. $a+b=17 \implies 899, 989$. $a+b=16 \implies 799, 979, 889$.
If $c=8, a+b=18 \implies 998$.
If $c=6, a+b=20$ (None).
If $c=4, a+b=22$ (None).
If $c=2, a+b=24$ (None).
If $c=0, a+b=26$ (None).
Wait, $a+b+c=26$. $c$ must be even.
If $c=8: a+b=18 \implies (9,9) \to 998$.
If $c=6: a+b=20$ (None).
If $c=4: a+b=22$ (None).
If $c=2: a+b=24$ (None).
If $c=0: a+b=26$ (None).
Wait, $a, b$ can be 9. Max $a+b=18$. So $c$ must be at least $26-18=8$.
If $c=8, a+b=18 \implies 998$.
If $c=9$ (not even).
Are there others? $a+b+c=26$. $c \in \{0, 2, 4, 6, 8\}$.
If $c=8, a+b=18 \to 998$.
If $c=6, a+b=20$ (Impossible).
Only 998? Let's re-read. "3-digit whole numbers".
Maybe $c$ can be 8, 6, 4, 2, 0.
$c=8 \implies a+b=18 \implies 998$.
$c=6 \implies a+b=20$ (No).
$c=4 \implies a+b=22$ (No).
$c=2 \implies a+b=24$ (No).
$c=0 \implies a+b=26$ (No).
Only 1.
ANSWER 8: A

Problem 9:
$a+b+c=25$. $a, b, c \in \{0..9\}, a \neq 0$.
Possible sets $\{9,9,7\}, \{9,8,8\}$.
Permutations of $\{9,9,7\}$: $997, 979, 799$ (3).
Permutations of $\{9,8,8\}$: $988, 898, 889$ (3).
Total = $3+3 = 6$.
ANSWER 9: C

Problem 10:
Exits 5 to 26 is $26-5 = 21$ intervals. Total distance 118.
Let $d_i$ be the distance between exits. $\sum_{i=1}^{21} d_i = 118$.
$d_i \ge 5$. To maximize one $d_k$, minimize others: $d_i = 5$ for $i \neq k$.
$d_k + 20(5) = 118 \implies d_k + 100 = 118 \implies d_k = 18$.
ANSWER 10: C

Problem 11:
Sum = $5 \times 90 = 450$. Mode = 94. Median = 91.
Let scores be $x_1, x_2, 91, 94, 94$.
$x_1 + x_2 + 91 + 94 + 94 = 450 \implies x_1 + x_2 + 279 = 450 \implies x_1 + x_2 = 171$.
ANSWER 11: B

Problem 12:
$5 \pm 4 \pm 6 \pm 3$.
$5+4 \times 6 - 3 = 5+24-3 = 26$.
$5 \times 4 - 6 + 3 = 20-6+3 = 17$.
$5 \times 4 + 6 - 3 = 23$.
$5 - 4 + 6 \times 3 = 1 + 18 = 19$.
ANSWER 12: E

The set \(R\) consists of nine distinct integers, including the six given values \(2, 3, 4, 6, 9, 14\). When the elements of \(R\) are placed in increasing order \(a_1 < a_2 < a_3 < a_4 < a_5 < a_6 < a_7 < a_8 < a_9\), the median is \(a_5\). Thus exactly four elements of \(R\) must lie strictly below the median value \(m = a_5\).
Let \(k\) denote the number of the six given values that are strictly less than \(m\). The three unknown values must then be chosen so that exactly \(4 - k\) of them lie strictly below \(m\).
- If \(m\) is one of the given values, the admissible choices are \(m = 3, 4, 6, 9\). For each of these four numbers it is possible to select three unknown integers satisfying the counting requirement above (and remaining distinct from one another and from the given values).
- If \(m\) is not one of the given values, the admissible choices are the integers \(m = 5, 7, 8\). Again, for each of these three numbers it is possible to select the remaining two unknown integers so that exactly four elements lie below \(m\).
No other integer can serve as median: any candidate smaller than 3 or larger than 9 produces a non-negative integer count of unknowns below \(m\) that is impossible to realize with only three unknowns. Consequently the possible values of the median are exactly the seven numbers
\[
3,4,5,6,7,8,9.
\]
Thus seven possible values exist for the median of \(R\).
ANSWER 1: D
Problem 2:
Let the five numbers be \(2, 6, 7, 7, 28, N\). Their sum is \(50 + N\), so the mean is
\[
\frac{50 + N}{5}.
\]
When the six numbers are ordered, the median is the average of the third and fourth values. The value of \(N\) determines the ordered list, and therefore the median, in each of the five mutually exclusive ranges determined by the given numbers.
- If \(N \le 2\), the ordered list is \(N,2,6,7,7,28\) and the median is \((6+7)/2 = 6.5\). Setting the mean equal to twice the median yields the contradictory equation \(N = 15\).
- If \(2 < N \le 6\), the ordered list is \(2,N,6,7,7,28\) and the median is again \(6.5\). The same contradictory equation appears.
- If \(6 < N \le 7\), the ordered list is \(2,6,N,7,7,28\) and the median is \((N+7)/2\). The mean-equals-twice-median equation simplifies to the quadratic
 \[
 N^2 - 9N - 60 = 0,
 \]
 whose only root in the interval is \(N = 12\), which lies outside the interval.
- If \(7 < N \le 28\), the ordered list is \(2,6,7,7,N,28\) and the median is \((7+N)/2\). The mean-equals-twice-median equation simplifies to the linear equation
 \[
 N = 20,
 \]
 which lies inside the interval.
- If \(N > 28\), the ordered list is \(2,6,7,7,28,N\) and the median is \((7+28)/2 = 17.5\). The resulting equation \(N = 120\) lies outside the interval.
Hence the only value that satisfies the stated condition is \(N = 20\).
ANSWER 2: C
Problem 3:
A four-digit integer whose digits are all even may be written
\[
1000a + 100b + 10c + d,
\]
where each of \(a,b,c,d\) belongs to the set \(\{0,2,4,6,8\}\) and \(a \ne 0\). There are therefore
\[
4 \times 5^3 = 500
\]
such integers.
An integer is divisible by 4 if and only if its last two digits form a number divisible by 4. The admissible pairs \((c,d)\) are therefore exactly the 13 pairs
\[
(0,0),(0,4),(0,8),(2,0),(2,4),(2,8),(4,0),(4,4),(4,8),(6,0),(6,4),(8,0),(8,8)
\]
that satisfy the divisibility test. For each of the 13 choices of \((c,d)\) the first two digits may be chosen freely (\(a\) has 4 possibilities, \(b\) has 5), giving
\[
13 \times 4 \times 5 = 260
\]
favorable integers. The required fraction is therefore
\[
\frac{260}{500} = \frac{13}{25} = \frac{2}{5}.
\]
ANSWER 3: B
Problem 4:
A 6.5 percent tax on a $20 item is \(0.065 \times 20 = 1.30\). A 6 percent tax is \(0.06 \times 20 = 1.20\). Their difference is
\[
1.30 - 1.20 = 0.10.
\]
ANSWER 4: B
Problem 5:
The first sum is an arithmetic series with 93 terms whose average is
\[
\frac{1901 + 1993}{2} = 1947,
\]
so its value equals \(93 \times 1947 = 181071\). The second sum is an arithmetic series with 93 terms whose average is
\[
\frac{101 + 193}{2} = 147,
\]
so its value equals \(93 \times 147 = 13671\). Subtracting yields
\[
181071 - 13671 = 167400.
\]
ANSWER 5: A
Problem 6:
The given expression consists of 1996 terms whose signs follow the repeating four-term pattern
\[
+,-,-,+,+,-,-,+,\dots
\]
(The pattern begins with a single “+” and thereafter repeats every four signs.) In each complete block of four consecutive integers the signed sum is zero. There are exactly 499 such blocks, accounting for 1996 terms and contributing a total of zero. Hence the whole expression evaluates to 0.
ANSWER 6: C
Problem 7:
Let \(S\) be the set of people wearing sunglasses and \(C\) the set wearing caps. Then
\[
|S \cap C| = \frac{2}{5}|C| = \frac{2}{5} \cdot 35 = 14.
\]
Consequently
\[
|S| = |S \setminus C| + |S \cap C|
\]
is unknown, but the conditional probability required by the problem is
\[
P(C \mid S) = \frac{|S \cap C|}{|S|} = \frac{14}{|S|}.
\]
Because \(|S \cap C| = 14\) and \(|C| = 35\), the only numerical value that can be obtained from the given data is the ratio
\[
\frac{14}{35} = \frac{2}{5}.
\]
(The actual size of \(S\) is never needed.)
ANSWER 7: C
Problem 8:
A three-digit number \(100a + 10b + c\) (with \(a \in \{1,\dots,9\}\), \(b,c \in \{0,\dots,9\}\)) is even precisely when \(c\) is even. The digit-sum equation \(a + b + c = 26\) together with the upper bound 9 on each variable forces \(a \ge 8\). The only possibilities are therefore the six triples
\[
(9,9,8),\ (9,8,9),\ (8,9,9)
\]
and their permutations that keep the units digit even. Exactly three of them have even units digit:
\[
998,\ 986,\ 968.
\]
Thus three even three-digit numbers have digit-sum 26.
ANSWER 8: C
Problem 9:
A three-digit number \(100a + 10b + c\) satisfies \(a + b + c = 25\) with each digit at most 9. Then necessarily \(a \ge 7\). The admissible triples \((a,b,c)\) are exactly the six permutations of
\[
(9,9,7),\ (9,8,8).
\]
Hence six such numbers exist.
ANSWER 9: C
Problem 10:

## Step 1: Problem 1
To find the number of possible values of the median of set R, we first need to understand what a median is. The median of a set of numbers is the middle number when the numbers are arranged in ascending order. Since R is a set of nine distinct integers, the median will be the fifth number when the numbers are arranged in ascending order.

## Step 2: Problem 1
Given that six of the elements are 2, 3, 4, 6, 9, and 14, we can start by arranging these numbers in ascending order: 2, 3, 4, 6, 9, 14. The remaining three numbers can be any distinct integers.

## Step 3: Problem 1
To determine the possible medians, we need to consider the range of values the fifth number can take. The numbers already given are 2, 3, 4, 6, 9, 14. So, the three additional numbers can either be less than 4, between 4 and 9, or greater than 9.

## Step 4: Problem 1
If all three additional numbers are less than 4, the order becomes: additional number, additional number, additional number, 4, 6, 9, 14. The median is 6.
If two additional numbers are less than 4 and one is between 4 and 9, the order becomes: additional number, additional number, 4, additional number, 6, 9, 14. The median is between 4 and 6, so it could be 5.
If one additional number is less than 4 and two are between 4 and 9, the order becomes: additional number, 4, additional number, additional number, 6, 9, 14. The median could be between 4 and 6.
If all three additional numbers are between 4 and 9, the order becomes: 2, 3, 4, additional number, additional number, additional number, 6, 9, 14. The median is one of the additional numbers between 4 and 6, or 6 if the additional numbers are greater than 6.
If the additional numbers are greater than 9, the order remains 2, 3, 4, 6, 9, additional number, additional number, additional number, 14. The median is 9.

## Step 5: Problem 1
Considering the cases:
- When the additional numbers are less than 4, the median is 6.
- When the numbers are between 4 and 9, the median can be 5, 6, 7, or 8 (since the additional numbers can be distinct and between 4 and 9).
- When the numbers are greater than 9, the median is 9.
So, possible medians are 5, 6, 7, 8, and 9.

## Step 6: Problem 1
Thus, there are 5 possible values for the median: 5, 6, 7, 8, 9.

ANSWER 1: B

## Step 7: Problem 2
To find N, we first need to understand how the mean and median are affected by adding N to the list 2, 6, 7, 7, 28.

## Step 8: Problem 2
The original list is 2, 6, 7, 7, 28. The median is 7. When N is added, the list becomes 2, 6, 7, 7, N, 28. The median will depend on N. If N is between 7 and 7 (which is not possible since N is distinct), or less than 7, the median will be 7. If N is greater than 7, the median will be the average of 7 and N.

## Step 9: Problem 2
Let's consider the case when N is less than or equal to 7. The median remains 7. The mean is (2 + 6 + 7 + 7 + 28 + N)/6 = (50 + N)/6. According to the problem, the mean is twice the median, so (50 + N)/6 = 2*7 = 14.

## Step 10: Problem 2
Solving for N: 50 + N = 14*6 = 84, N = 84 - 50 = 34.

## Step 11: Problem 2
Let's check if N = 34 satisfies the condition for the median. Since 34 is greater than 7, the list becomes 2, 6, 7, 7, 28, 34. The median is (7 + 7)/2 = 7. The mean is (2 + 6 + 7 + 7 + 28 + 34)/6 = 84/6 = 14, which is indeed twice the median.

## Step 12: Problem 2
Thus, N = 34 is a valid solution.

ANSWER 2: E

## Step 13: Problem 3
To find the fraction of four-digit integers with all even digits that are divisible by 4, we first need to determine the total number of such four-digit integers.

## Step 14: Problem 3
For a four-digit number, the first digit can be 2, 4, 6, or 8 (4 choices), and the rest can be 0, 2, 4, 6, or 8 (5 choices each). So, the total number of four-digit numbers with all even digits is 4 * 5 * 5 * 5 = 500.

## Step 15: Problem 3
A number is divisible by 4 if its last two digits form a number divisible by 4. The last two digits can be 00, 04, 08, 20, 24, 28, 40, 44, 48, 60, 64, 68, 80, 84, 88. Out of these, the ones with even digits are 00, 04, 08, 20, 24, 28, 40, 44, 48, 60, 64, 68, 80, 84, 88, which are 20, 24, 28, 40, 44, 48, 60, 64, 68, 80, 84, 88 (12 options since 00, 04, 08 are not valid for the last two digits when considering all possible combinations for the first two digits).

## Step 16: Problem 3
However, we need to correct the understanding of the last two digits. For a number to be divisible by 4, the last two digits must form a number divisible by 4. The possible last two digits are: 00, 04, 08, 20, 24, 28, 40, 44, 48, 60, 64, 68, 80, 84, 88. There are 5 choices for the third digit and 4 choices for the first digit. So, for each valid last two digits, there are 4 * 5 = 20 numbers.

## Step 17: Problem 3
There are 15 valid combinations for the last two digits (00, 04, 08, 20, 24, 28, 40, 44, 48, 60, 64, 68, 80, 84, 88), but we should count them correctly: there are indeed 15 options (since all are valid for the condition of being even and divisible by 4 when considering the last two digits).

## Step 18: Problem 3
The total number of four-digit numbers with all even digits and divisible by 4 is 4 * 5 * 15 = 300 (since there are 15 valid last two digits).

## Step 19: Problem 3
The fraction of these integers divisible by 4 is 300 / 500 = 3/5.

ANSWER 3: D

## Step 20: Problem 4
To find the difference between a 6.5% sales tax and a 6% sales tax on an item priced at $20, we calculate both taxes and then find the difference.

## Step 21: Problem 4
6.5% of $20 = 0.065 * $20 = $1.30.
6% of $20 = 0.06 * $20 = $1.20.

## Step 22: Problem 4
The difference is $1.30 - $1.20 = $0.10.

ANSWER 4: B

## Step 23: Problem 5
To solve (1901 + 1902 + ... + 1993) - (101 + 102 + ... + 193), we can simplify it by calculating the sum of the differences between corresponding terms.

## Step 24: Problem 5
The series can be simplified as: (1901-101) + (1902-102) + ... + (1993-193).

## Step 25: Problem 5
This simplifies to: 1800 + 1800 + 

**Problem 1:**
We have a set \(R\) of nine distinct integers, with six given: 2, 3, 4, 6, 9, 14. The remaining three integers can be any integers not in this list. The median of nine numbers is the 5th smallest. Let the unknown numbers be \(x, y, z\). The six given numbers are fixed; we can place \(x, y, z\) anywhere in the sorted order to change the 5th element.

Let \(k\) be the number of given numbers less than the candidate median \(m\).
- If \(m\) is one of the given numbers, we need to place exactly \(4-k\) of the new numbers below \(m\) and the rest above. This requires \(0 \le 4-k \le 3\), so \(1 \le k \le 4\). Among the given numbers: \(3\ (k=1),\ 4\ (k=2),\ 6\ (k=3),\ 9\ (k=4)\) satisfy this. (\(2\) has \(k=0\), \(14\) has \(k=5\); neither works.) So \(3, 4, 6, 9\) are possible medians.
- If \(m\) is a new integer, one new number equals \(m\) and the other two are placed above or below. We need \(4-k\) new numbers below \(m\), so \(0 \le 4-k \le 2\), or \(2 \le k \le 4\). Also \(m\) must be an integer not already in the list.
  * \(k=2\): \(m\) between 3 and 4 — no integer.
  * \(k=3\): \(m\) between 4 and 6 — only \(5\).
  * \(k=4\): \(m\) between 6 and 9 — \(7\) and \(8\).
So new possible medians: \(5, 7, 8\).

Total possible medians: \(3, 4, 5, 6, 7, 8, 9\) — seven values.

ANSWER 1: D

**Problem 2:**
Original list: \(2, 6, 7, 7, 28\) (sum = 50). Insert \(N\) to get six numbers. Mean = \(\frac{50+N}{6}\). Median of six sorted numbers is the average of the 3rd and 4th.

Case 1: \(N \le 6\). Sorted: …, 3rd = 6, 4th = 7, median = 6.5.  
\(\frac{50+N}{6} = 2 \times 6.5 = 13 \Rightarrow 50+N = 78 \Rightarrow N=28\), but \(28 > 6\). Contradiction.

Case 2: \(6 < N \le 7\) (only integer is 7). Sorted: \(2, 6, 7, 7, 7, 28\), median = 7.  
\(\frac{57}{6} = 9.5 \neq 14\). No.

Case 3: \(7 < N \le 28\). Sorted: \(2, 6, 7, 7, N, 28\), median = 7.  
\(\frac{50+N}{6} = 14 \Rightarrow N = 34\), but \(34 > 28\). Contradiction.

Case 4: \(N > 28\). Sorted: \(2, 6, 7, 7, 28, N\), median = 7.  
\(\frac{50+N}{6} = 14 \Rightarrow N = 34\). Works.

ANSWER 2: E

**Problem 3:**
Four-digit positive integers with all even digits: thousands digit from \(\{2,4,6,8\}\) (4 choices); hundreds, tens, units from \(\{0,2,4,6,8\}\) (5 choices each). Total = \(4 \times 5^3 = 500\).
Divisible by 4 means last two digits form a multiple of 4. For any even tens digit, there are exactly 3 even units digits that make it divisible by 4 (e.g., if tens = 0,4,8 → units 0,4,8; if tens = 2,6 → units 0,4,8). So \(5 \times 3 = 15\) valid last-two-digit combinations.
Number of divisible numbers = \(4 \times 5 \times 15 = 300\).
Fraction = \(\frac{300}{500} = \frac{3}{5}\).

ANSWER 3: D

**Problem 4:**
Tax difference = \(6.5\% - 6\% = 0.5\%\) of \$20 = \(0.005 \times 20 = \$0.10\).

ANSWER 4: B

**Problem 5:**
First sum: \(1901\) to \(1993\) inclusive. Number of terms = \(1993 - 1901 + 1 = 93\).  
Sum = \(93 \times \frac{1901+1993}{2} = 93 \times 1947\).
Second sum: \(101\) to \(193\) inclusive. Terms = \(193 - 101 + 1 = 93\).  
Sum = \(93 \times \frac{101+193}{2} = 93 \times 147\).
Difference = \(93 \times (1947 - 147) = 93 \times 1800 = 167\,400\).

ANSWER 5: A

**Problem 6:**
The expression is \(1 - 2 - 3 + 4 + 5 - 6 - 7 + 8 + 9 - 10 - 11 + \cdots + 1992 + 1993 - 1994 - 1995 + 1996\).
Sign pattern: \(+\) for terms \(n \equiv 0\) or \(1 \pmod 4\); \(-\) for \(n \equiv 2\) or \(3 \pmod 4\).
Group in blocks of four: \((1-2-3+4) + (5-6-7+8) + \cdots + (1993-1994-1995+1996)\). Each block sums to \(0\). There are \(1996/4 = 499\) blocks. Total sum = \(0\).

ANSWER 6: C

**Problem 7:**
Let \(S\) be sunglasses wearers (\(|S|=50\)), \(C\) be cap wearers (\(|C|=35\)). Given \(P(S \mid C) = \frac{|S \cap C|}{35} = \frac{2}{5} \Rightarrow |S \cap C| = 14\).
Then \(P(C \mid S) = \frac{14}{50} = \frac{7}{25}\).

ANSWER 7: B

**Problem 8:**
Three-digit numbers with digits summing to 26. Maximum sum is 27, so digits are close to 9. Let digits be \(a,b,c\) with \(a+b+c=26\), \(1 \le a \le 9\), \(0 \le b,c \le 9\). Setting \(x=9-a, y=9-b, z=9-c\) gives \(x+y+z=1\) with \(x \le 8, y,z \le 9\). The only solutions are permutations of \((1,0,0)\): one digit is 8, the other two are 9. The three numbers are 998, 989, 899.
Even numbers end in an even digit: only 998. Hence 1 even number.

ANSWER 8: A

**Problem 9:**
Three-digit numbers, digit sum = 25. As above, \(x+y+z = 27-25 = 2\). Nonnegative solutions: permutations of \((2,0,0)\) – digits 7,9,9; and \((1,1,0)\) – digits 8,8,9.  
For 7,9,9: 3 permutations. For 8,8,9: 3 permutations. Total 6 numbers.

ANSWER 9: C

**Problem 10:**
Exits 5 through 26: that’s \(26-5+1 = 22\) exits, so 21 gaps. Sum of gaps = 118 miles. Each gap \(\ge 5\). To maximize one gap, minimize the other 20: \(20 \times 5 = 100\) miles. Largest gap = \(118 - 100 = 18\) miles.

ANSWER 10: C

**Problem 11:**
Five scores: \(a \le b \le c \le d \le e\). Mean = 90 ⇒ sum = 450. Median \(c = 91\). Mode = 94, so at least two scores are 94. Since median is 91, the two 94s must be \(d\) and \(e\). Then \(a+b+91+94+94 = 450 \Rightarrow a+b = 450 - 279 = 171\).

ANSWER 11: B

**Problem 12:**
Blanks: \(5 \_ 4 \_ 6 \_ 3\). Use \(+,\ -,\ \times\) exactly once. Evaluate with standard order of operations. All 6 permutations:
1) \(\times, +, -:\ 5\times4+6-3 = 20+6-3=23\)
2) \(\times, -, +:\ 5\times4-6+3 = 20-6+3=17\)
3) \(+, \times, -:\ 5+4\times6-3 = 5+24-3=26\)
4) \(+, -, \times:\ 5+4-6\times3 = 5+4-18=-9\)
5) \(-, \times, +:\ 5-4\times6+3 = 5-24+3=-16\)
6) \(-, +, \times:\ 5-4+6\times3 = 5-4+18=19\)
Among choices, 19 appears.

ANSWER 12: E

Problem 1:
We have a set R of 9 distinct integers. Six are known: 2, 3, 4, 6, 9, 14. We add 3 unknown distinct integers (not equal to any known element). The median of 9 elements is the 5th smallest.

Let me determine which values can occupy the 5th position. The known elements in sorted order are: 2, 3, 4, 6, 9, 14.

**Can the median be 3?** Place all 3 unknowns below 3 (e.g., −1, 0, 1). Sorted: −1, 0, 1, 2, **3**, 4, 6, 9, 14. 5th = 3. ✓
**Can the median be 4?** Place 2 unknowns below 4 (e.g., 0, 1) and 1 above. Sorted: 0, 1, 2, 3, **4**, 6, 9, 14, 20. 5th = 4. ✓
**Can the median be 5?** (5 is an unknown.) Known elements below 5: {2, 3, 4} = 3 elements. Add 1 unknown below 5 (e.g., 1), make 5 an unknown, and 1 unknown above. Sorted: 1, 2, 3, 4, **5**, 6, 9, 14, 20. 5th = 5. ✓
**Can the median be 6?** Known below 6: {2, 3, 4} = 3. Add 1 unknown below 6 (e.g., 5), 2 above. Sorted: 2, 3, 4, 5, **6**, 9, 14, 20, 21. 5th = 6. ✓
**Can the median be 7?** Known below 7: {2, 3, 4, 6} = 4. Need 0 unknowns below 7. Make 7 an unknown, 2 above. Sorted: 2, 3, 4, 6, **7**, 9, 14, 20, 21. 5th = 7. ✓
**Can the median be 8?** Known below 8: {2, 3, 4, 6} = 4. Same logic: 2, 3, 4, 6, **8**, 9, 14, 20, 21. 5th = 8. ✓
**Can the median be 9?** Known below 9: {2, 3, 4, 6} = 4. Place all 3 unknowns above 9. Sorted: 2, 3, 4, 6, **9**, 14, 20, 21, 22. 5th = 9. ✓

**Can the median be ≤ 2?** Need 4 elements below it, but at most 3 unknowns exist → impossible.
**Can the median be ≥ 10?** Known elements below 10 are {2, 3, 4, 6, 9} = 5 elements, so the 5th position is already filled by something ≤ 9 → impossible.

Possible medians: {3, 4, 5, 6, 7, 8, 9} → **7 values**.

ANSWER 1: D

---

Problem 2:
Original list: 2, 6, 7, 7, 28. Sum = 50. Insert N to get 6 elements.
New mean = (50 + N)/6. Median of 6 elements = average of 3rd and 4th.

**Case N > 7:** Sorted order: 2, 6, 7, 7, N, 28 (or N, 28 swapped). The 3rd and 4th elements are 7 and 7.
Median = (7 + 7)/2 = 7.
Mean = 2 × median = 14.
(50 + N)/6 = 14 → 50 + N = 84 → N = 34.
Check: 34 > 7 ✓. List: 2, 6, 7, 7, 28, 34. Median = 7, Mean = 84/6 = 14 = 2(7). ✓

**Case N ≤ 6:** Median = (6+7)/2 = 6.5. Mean = 13. N = 78 − 50 = 28. But 28 > 6, contradiction.
**Case 6 < N ≤ 7:** Median = (N+7)/2. (50+N)/6 = N+7 → 50+N = 6N+42 → 5N = 8 → N = 1.6. Contradiction.

Only N = 34 works.

ANSWER 2: E

---

Problem 3:
Even digits: {0, 2, 4, 6, 8}. Four-digit numbers with all even digits:
- First digit: {2, 4, 6, 8} → 4 choices
- Other three digits: 5 choices each
- Total = 4 × 5 × 5 × 5 = 500

Divisibility by 4 depends on the last two digits. I check all 5 × 5 = 25 pairs of even digits:
For each tens digit (0, 2, 4, 6, 8), the units digits giving divisibility by 4 are:
- Tens 0: 00, 04, 08 (3 of 5)
- Tens 2: 20, 24, 28 (3 of 5)
- Tens 4: 40, 44, 48 (3 of 5)
- Tens 6: 60, 64, 68 (3 of 5)
- Tens 8: 80, 84, 88 (3 of 5)

Valid last-two-digit pairs = 15 out of 25.

Numbers divisible by 4 = 4 × 5 × 15 = 300.
Fraction = 300/500 = **3/5**.

ANSWER 3: D

---

Problem 4:
Difference = 6.5% of $20 − 6% of $20 = (6.5% − 6%) × $20 = 0.5% × $20.
0.5% = 0.005, so 0.005 × 20 = $0.10.

ANSWER 4: B

---

Problem 5:
Pair corresponding terms:
(1901 − 101) + (1902 − 102) + … + (1993 − 193)

Each pair equals 1800.
Number of terms: 1993 − 1901 + 1 = 93 terms (and 193 − 101 + 1 = 93 terms).
Total = 93 × 1800 = **167,400**.

ANSWER 5: A

---

Problem 6:
The sign pattern is: +, −, −, +, +, −, −, +, …
Group into blocks of 4:
(1 − 2 − 3 + 4) + (5 − 6 − 7 + 8) + (9 − 10 − 11 + 12) + …

Each block: n − (n+1) − (n+2) + (n+3) = n − n − 1 − n − 2 + n + 3 = **0**.

Total terms: 1996. Number of blocks: 1996/4 = 499 complete blocks.
Sum = 499 × 0 = **0**.

ANSWER 6: C

---

Problem 7:
Let B = number wearing both sunglasses and caps.
P(sunglasses | cap) = B / (people wearing caps) = B / 35 = 2/5.
B = 35 × (2/5) = 14.

Now find P(cap | sunglasses) = B / (people wearing sunglasses) = 14 / 50 = **7/25**.

ANSWER 7: B

---

Problem 8:
Three-digit numbers with digit sum 26. Maximum digit sum = 9+9+9 = 27, so we must subtract 1 from one digit of 999:
Possible digit sets: (8,9,9), (9,8,9), (9,9,8) → numbers 899, 989, 998 (and permutations: 899, 989, 997... wait).

Actually the numbers are: 899, 989, 998 — no, let me list properly:
- (8,9,9) → 899, 989, 998... wait, (8,9,9) gives 899, 989, 998? No: digits are 8,9,9 so numbers are 899, 989, 998.
Wait: (8,9,9) → 899, 989, 998. But 998 has digits 9,9,8 which is the set (9,9,8). Let me just list all:
- Digits {8,9,9}: 899, 989, 998 — wait, 998 has digits 9,9,8. That IS {8,9,9}. So: 899, 989, 998.

Hmm, let me redo. The three numbers with digit sum 26 are:
- 899 (8+9+9=26)
- 989 (9+8+9=26)
- 998 (9+9+8=26)

Wait, that's only 3 numbers total (since two digits are the same, there are 3!/2! = 3 arrangements).

Even numbers must end in an even digit. Only 998 ends in 8 (even). 899 and 989 end in 9 (odd).

So exactly **1** even number.

ANSWER 8: A

---

Problem 9:
Three-digit numbers a, b, c with a + b + c = 25, where 1 ≤ a ≤ 9, 0 ≤ b, c ≤ 9.

Enumerate by first digit a:
- a = 9: b + c = 16 → (7,9), (8,8), (9,7) → 979, 988, 997 (3 numbers)
- a = 8: b + c = 17 → (8,9), (9,8) → 889, 898 (2 numbers)
- a = 7: b + c = 18 → (9,9) → 799 (1 number)
- a ≤ 6: b + c ≥ 19, but max b + c = 18 → impossible

Total = 3 + 2 + 1 = **6**.

ANSWER 9: C

---

Problem 10:
Exits 5 through 26 give 22 exits and 21 consecutive gaps. Total distance = 118 miles. Each gap ≥ 5 miles.

To maximize one gap, minimize the other 20 gaps to 5 miles each:
Max gap + 20(5) = 118
Max gap + 100 = 118
Max gap = **18** miles.

ANSWER 10: C

---

Problem 11:
Five scores sorted: a ≤ b ≤ c ≤ d ≤ e.
- Mean = 90 → sum = 450
- Median = 91 → c = 91
- Mode = 94 → 94 appears most frequently (at least twice)

Since a ≤ b ≤ c = 91, neither a nor b can be 94. So 94 must appear among d and e. For 94 to be the mode, we need d = e = 94 (appearing twice).

Sum: a + b + 91 + 94 + 94 = 450
a + b = 450 − 279 = **171**

(We can verify valid scores 

 Problem 1:
We are given a set \(R\) of nine distinct integers, six of which are \(2,3,4,6,9,14\). We must determine how many different values are possible for the median of \(R\).

Because there are nine distinct integers, the median is the fifth smallest element when the set is listed in increasing order. Let the three unknown distinct integers be chosen freely (as long as they are not equal to any of the six given numbers). We examine two cases: when the median is one of the six known numbers and when it is one of the three unknown numbers.

**Case 1: The median is a known number.**
Let the median be \(m\). Of the six known numbers, let \(L\) be how many are less than \(m\). Then there are \(5-L\) known numbers greater than \(m\). To make \(m\) the fifth element, we need exactly four numbers smaller than \(m\) and four larger. Thus we need \(4-L\) unknown numbers below \(m\) and \(4-(5-L)=L-1\) unknown numbers above \(m\). This uses exactly \((4-L)+(L-1)=3\) unknowns, which is perfect. We need \(4-L\ge 0\) and \(L-1\ge 0\), i.e. \(1\le L\le 4\).

- \(m=2\): \(L=0\) → impossible.
- \(m=3\): \(L=1\) → need 3 unknowns \(<3\), 0 above. Possible (e.g. \(-1,0,1\)).
- \(m=4\): \(L=2\) → need 2 unknowns \(<4\), 1 above. Possible (e.g. \(0,1,5\)).
- \(m=6\): \(L=3\) → need 1 unknown \(<6\), 2 above. Possible (e.g. \(5,7,8\)).
- \(m=9\): \(L=4\) → need 0 unknowns \(<9\), 3 above. Possible (e.g. \(10,11,12\)).
- \(m=14\): \(L=5\) → need \(-1\) unknowns below; impossible.

So the known possible medians are \(3,4,6,9\).

**Case 2: The median is an unknown number \(x\).**
Then \(x\) itself uses one of the three unknown slots. We need four numbers below \(x\) and four above. With \(L\) known numbers below \(x\), we need \(4-L\) other unknowns below \(x\) and \(4-(6-L)=L-2\) other unknowns above \(x\). The total other unknowns used is \((4-L)+(L-2)=2\), which matches the two remaining unknowns. We need \(4-L\ge 0\) and \(L-2\ge 0\), so \(2\le L\le 4\).

- \(L=2\): known below are \(2,3\); known above are \(4,6,9,14\). Then \(3<x<4\), impossible for an integer.
- \(L=3\): known below are \(2,3,4\); known above are \(6,9,14\). Then \(4<x<6\), so \(x=5\). This works (e.g. add \(1,5,7\)).
- \(L=4\): known below are \(2,3,4,6\); known above are \(9,14\). Then \(6<x<9\), so \(x=7\) or \(x=8\). Both work (e.g. add \(7,10,11\) or \(8,10,11\)).

Thus the unknown possible medians are \(5,7,8\).

Collecting all possibilities: \(\{3,4,5,6,7,8,9\}\). That is **7** different values. The tempting answer of 4 counts only the known medians, and 2 or 14 are impossible because too many numbers would have to be forced to one side.

ANSWER 1: D

---

Problem 2:
We insert a number \(N\) into the list \(2,6,7,7,28\) so that the new mean of the six numbers is twice the new median. Find \(N\).

The original sum is \(2+6+7+7+28=50\). After insertion, the sum is \(50+N\) and the mean is \(\frac{50+N}{6}\).

The median depends on the size of \(N\) relative to the ordered list. Because the list contains six numbers, the median is the average of the third and fourth values.

- If \(N\le 6\) (as an integer), the ordered list begins with \(N\) or \(2\), then the third and fourth terms are \(6\) and \(7\), giving median \(\frac{6+7}{2}=6.5\).
  Setting mean \(=2\times 6.5=13\): \(\frac{50+N}{6}=13\Rightarrow N=28\). But \(N=28\) does not satisfy \(N\le 6\), so this case yields no valid solution.

- If \(N\ge 7\) (integer), the third and fourth terms are both \(7\) (since the two \(7\)’s occupy those middle slots, or \(N=7\) makes three \(7\)’s), so the median is \(7\).
  Setting mean \(=2\times 7=14\): \(\frac{50+N}{6}=14\Rightarrow 50+N=84\Rightarrow N=34\).

Check: the list becomes \(2,6,7,7,28,34\). The median is \(\frac{7+7}{2}=7\). The mean is \(\frac{84}{6}=14\), which is indeed twice the median. The other choices fail: \(N=28\) gives mean \(13\) and median \(7\); \(N=14\) gives mean \(\frac{64}{6}\neq 14\).

ANSWER 2: E

---

Problem 3:
We consider all positive four-digit integers whose digits are all even. We need the fraction of them that are divisible by \(4\).

Even digits are \(\{0,2,4,6,8\}\). For a four-digit number \(ABCD\):
- \(A\) (thousands) has 4 choices: \(\{2,4,6,8\}\).
- \(B,C,D\) each have 5 choices.

Total count: \(4\cdot 5\cdot 5\cdot 5 = 500\).

A number is divisible by \(4\) exactly when its last two digits form a multiple of \(4\). Here \(C\) and \(D\) are both even. Consider \(10C+D \pmod 4\). Since \(C\) is even, write \(C=2k\); then \(10C=20k\equiv 0\pmod 4\). Thus \(10C+D\equiv D\pmod 4\). So we simply need \(D\equiv 0\pmod 4\).

Among the even digits, those divisible by \(4\) are \(0,4,8\): **3 choices** for \(D\). The digits \(A,B,C\) are unrestricted (4, 5, and 5 choices).

Count of favorable numbers: \(4\cdot 5\cdot 5\cdot 3 = 300\).

The fraction is \(\frac{300}{500}=\frac{3}{5}\). One might mistakenly think the fraction is \(\frac12\) because half of all integers are divisible by \(4\), but the restriction to even digits changes the units-digit distribution.

ANSWER 3: D

---

Problem 4:
Find the difference between a \(6.5\%\) sales tax and a \(6\%\) sales tax on a \(\$20\) item.

Method 1: Compute each tax.
- \(6.5\%\) of \(\$20 = 0.065\times 20 = \$1.30\).
- \(6\%\) of \(\$20 = 0.06\times 20 = \$1.20\).

Difference: \(\$1.30-\$1.20 = \$0.10\).

Method 2: The difference in rates is \(0.5\%\), and \(0.5\%\) of \(\$20 = 0.005\times 20 = \$0.10\).

The distractors correspond to misplacing the decimal (e.g. \(\$1\) would be \(5\%\) of \(\$20\)).

ANSWER 4: B

---

Problem 5:
Evaluate \((1901+1902+\dots+1993)-(101+102+\dots+193)\).

Each sum has the same number of terms:
\(1993-1901+1 = 93\) terms, and \(193-101+1 = 93\) terms.

Pair the terms:
\((1901-101)+(1902-102)+\dots+(1993-193)\).

Each difference equals \(1800\). With 93 such pairs, the total is
\(93\times 1800 = 93\times 18\times 100 = 1674\times 100 = 167{,}400\).

A common error is miscounting the number of terms as 92, which would give \

Problem 1:
R is a set of nine distinct integers with six known elements: 2, 3, 4, 6, 9, 14. Since R has 9 elements, the median is the 5th element when sorted.

I need to determine which integers can be the 5th element. The median must have exactly 4 elements less than it and 4 elements greater than it.

**Checking each candidate:**

- **Median = 3**: Need 4 elements < 3. Known: {2}. Need 3 unknowns all < 3 (e.g., 1, 0, -1). Elements > 3: {4, 6, 9, 14} = 4. ✓
- **Median = 4**: Need 4 elements < 4. Known: {2, 3}. Need 2 unknowns < 4. Elements > 4: {6, 9, 14} + 1 unknown > 4. ✓
- **Median = 5**: Need 4 elements < 5. Known: {2, 3, 4}. Need 1 unknown < 5. Elements > 5: {6, 9, 14} + 1 unknown > 5. (5 is an unknown) ✓
- **Median = 6**: Need 4 elements < 6. Known: {2, 3, 4}. Need 1 unknown < 6. Elements > 6: {9, 14} + 2 unknowns > 6. ✓
- **Median = 7**: Need 4 elements < 7. Known: {2, 3, 4, 6} = 4. Elements > 7: {9, 14} + 2 unknowns > 7. (7 is an unknown) ✓
- **Median = 8**: Need 4 elements < 8. Known: {2, 3, 4, 6} = 4. Elements > 8: {9, 14} + 2 unknowns > 8. (8 is an unknown) ✓
- **Median = 9**: Need 4 elements < 9. Known: {2, 3, 4, 6} = 4. Elements > 9: {14} + 3 unknowns > 9. ✓

**Values that cannot be median:**
- 2: Need 4 elements < 2, but only 3 unknowns available. ✗
- 10-13: Already have 5 known elements less than these. ✗
- 14: Already have 5 known elements less than 14. ✗

Possible median values: {3, 4, 5, 6, 7, 8, 9} → **7 values**

ANSWER 1: D

---

Problem 2:
List: 2, 6, 7, 7, 28. Insert N so that mean = 2 × median.

Sum of original list = 50, so new mean = (50 + N)/6.

**Finding the median based on N:**
- If N > 7: Sorted list is 2, 6, 7, 7, 28, N (or 2, 6, 7, 7, N, 28 if 7 < N < 28). Median = (7+7)/2 = 7.
- If N ≤ 6: Median = (6+7)/2 = 6.5
- If 6 < N ≤ 7: Median = (N+7)/2

**Solving for N > 7:**
(50 + N)/6 = 2(7) = 14
50 + N = 84
N = 34

Verification: List = 2, 6, 7, 7, 28, 34. Mean = 84/6 = 14. Median = 7. ✓

ANSWER 2: E

---

Problem 3:
Four-digit integers with all even digits. Even digits: {0, 2, 4, 6, 8}.

**Total count:** First digit: 4 choices (2,4,6,8), others: 5 choices each → 4 × 5³ = 500

**Divisible by 4:** Last two digits must form a number divisible by 4.

Listing all 25 pairs of even last digits and checking divisibility by 4:
- 00✓, 02✗, 04✓, 06✗, 08✓
- 20✓, 22✗, 24✓, 26✗, 28✓
- 40✓, 42✗, 44✓, 46✗, 48✓
- 60✓, 62✗, 64✓, 66✗, 68✓
- 80✓, 82✗, 84✓, 86✗, 88✓

That's 15 valid pairs.

**Count divisible by 4:** 4 × 5 × 15 = 300

**Fraction:** 300/500 = 3/5

ANSWER 3: D

---

Problem 4:
6.5% tax on $20 = 0.065 × 20 = $1.30
6% tax on $20 = 0.06 × 20 = $1.20

Difference = $1.30 - $1.20 = $0.10

ANSWER 4: B

---

Problem 5:
(1901 + 1902 + ... + 1993) − (101 + 102 + ... + 193)

Both sums have 93 terms each.

First sum = 93 × (1901 + 1993)/2 = 93 × 1947
Second sum = 93 × (101 + 193)/2 = 93 × 147

Difference = 93 × (1947 − 147) = 93 × 1800 = 167,400

ANSWER 5: A

---

Problem 6:
1 − 2 − 3 + 4 + 5 − 6 − 7 + 8 + ... + 1993 − 1994 − 1995 + 1996

The pattern repeats every 4 terms: +, −, −, +

Group k: (4k−3) − (4k−2) − (4k−1) + 4k = 4k−3−4k+2−4k+1+4k = 0

Each group of 4 sums to 0.

1996 ÷ 4 = 499 complete groups.

Total = 499 × 0 = 0

ANSWER 6: C

---

Problem 7:
50 people wear sunglasses, 35 wear caps.

P(sunglasses | cap) = 2/5 means: (both)/(caps) = 2/5

Both/35 = 2/5 → Both = 14

P(cap | sunglasses) = (both)/(sunglasses) = 14/50 = 7/25

ANSWER 7: B

---

Problem 8:
3-digit numbers with digit-sum 26 that are even.

For number abc: a + b + c = 26, c is even, a ≥ 1, all digits ≤ 9.

- c = 8: a + b = 18 → only (9,9) → 998 ✓
- c = 6: a + b = 20 → impossible (max 9+9=18)
- c = 4, 2, 0: a + b ≥ 22 → impossible

Only 1 number: 998

ANSWER 8: A

---

Problem 9:
Three-digit numbers with digit-sum 25.

a + b + c = 25, a ≥ 1, all digits ≤ 9.

- a = 9: b + c = 16 → (7,9), (8,8), (9,7) → 3 numbers
- a = 8: b + c = 17 → (8,9), (9,8) → 2 numbers
- a = 7: b + c = 18 → (9,9) → 1 number
- a ≤ 6: b + c ≥ 19 → impossible

Total = 3 + 2 + 1 = 6

ANSWER 9: C

---

Problem 10:
Distance from exit 5 to exit 26 is 118 miles. There are 21 gaps between consecutive exits (5→6, 6→7, ..., 25→26).

Each gap ≥ 5 miles. To maximize one gap, minimize all others.

If 20 gaps are minimized at 5 miles each: 20 × 5 = 100 miles.

Maximum single gap = 118 − 100 = 18 miles.

ANSWER 10: C

---

Problem 11:
Five scores: mean = 90, median = 91, mode = 94.

Sum = 5 × 90 = 450

Sorted: a ≤ b ≤ c ≤ d ≤ e
- c = 91 (median)
- Mode = 94, so 94 appears at least twice. Since c = 91, the 94s must be d and e.

a + b + 91 + 94 + 94 = 450
a + b = 171

ANSWER 11: B

---

Problem 12:
Using +, −, × each exactly once in 5 __ 4 __ 6 __ 3:

| Operations | Result |
|------------|--------|
| 5 + 4 − 6 × 3 | 5 + 4 − 18 = −9 |
| 5 + 4 × 6 − 3 | 5 + 24 − 3 = 26 |
| 5 − 4 + 6 × 3 | 5 − 4 + 18 = 19 |
| 5 − 4 × 6 + 3 | 5 − 24 + 3 = −16 |
| 5 × 4 + 6 − 3 | 20 + 6 − 3 = 23 |
| 5 × 4 − 6 + 3 | 20 − 6 + 3 = 17 |

From choices {9, 10, 15, 16, 19}, only **19** is achievable.

ANSWER 12: E

**Problem 1:**  
We have nine distinct integers, six of which are known: 2, 3, 4, 6, 9, 14.  
The median of a nine‑element set is the 5th smallest number.  

Let the unknown integers be \(x,y,z\) (all distinct and distinct from the known six).  
When the numbers are sorted, the median is the 5th term.  

For a known number \(k_i\) (the \(i\)‑th smallest of the known six) to be the median, we need exactly four numbers smaller than it.  
If \(i-1\) known numbers are smaller, we need \(L\) unknown numbers smaller with  
\((i-1)+L=4\Rightarrow L=5-i\).  
Since we have only three unknowns, \(L\ge0\) gives \(i\le5\); also \(L\le3\) gives \(i\ge2\).  
Thus \(i=2,3,4,5\) → medians could be 3, 4, 6, 9.

Can the median be an unknown integer?  
Let the median be an unknown \(u\).  
If \(u\) lies between the known numbers, we need exactly four numbers below it.  
With two other unknowns, the possibilities are:
- \(u\) between 4 and 6 → \(u=5\);
- \(u\) between 6 and 9 → \(u=7\) or \(u=8\).

All three of these can be realized while keeping the numbers distinct.  
Thus the possible median values are \(\{3,4,5,6,7,8,9\}\), a total of **7** values.

**Answer:** 7 → Choice **D**.  

ANSWER 1: D  

---

**Problem 2:**  
We start with the list 2, 6, 7, 7, 28 and insert a number \(N\).  
Let the six numbers be sorted. The median of six numbers is the average of the 3rd and 4th smallest values.

Let the sorted list after insertion be \(a_1\le a_2\le a_3\le a_4\le a_5\le a_6\).  
The median is \(\dfrac{a_3+a_4}{2}\).  
The mean is \(\dfrac{\text{sum}}{6}\).  
Given that the mean equals twice the median:
\[
\frac{\text{sum}}{6}=2\cdot\frac{a_3+a_4}{2}=a_3+a_4 .
\]
Hence \(\text{sum}=6(a_3+a_4)\).

Consider where \(N\) can fall:

*If \(N\le2\):* \(a_3=6,\ a_4=7\) → median \(=6.5\).  
Sum \(=N+50\). Equation gives \(N=28\), contradicting \(N\le2\).

*If \(2<N\le6\):* \(a_3=6,\ a_4=7\) → same median \(=6.5\).  
Again \(N=28\), impossible.

*If \(6<N\le7\):* only integer possibility \(N=7\).  
List becomes 2, 6, 7, 7, 7, 28 → median \(=7\).  
Sum \(=57\) → mean \(=9.5\neq14\).

*If \(7<N<28\):* \(a_3=7,\ a_4=7\) → median \(=7\).  
Sum \(=N+50\). Equation \(N+50=84\) gives \(N=34\), which is not \(<28\).

*If \(N=28\):* median still \(=7\). Sum \(=78\) → mean \(=13\neq14\).

*If \(N>28\):* \(a_3=7,\ a_4=7\) → median \(=7\).  
Equation gives \(N+50=84\) → \(N=34\). This works (34 > 28).

Thus the only possible value is **34**.

**Answer:** 34 → Choice **E**.  

ANSWER 2: E  

---

**Problem 3:**  
A four‑digit integer with all digits even can use digits \{0,2,4,6,8\} for each place, except the thousands digit cannot be 0.  

Number of such integers: \(4\times5\times5\times5=500\).

A number is divisible by 4 iff its last two digits form a number divisible by 4.  
List all two‑digit endings with even digits (25 possibilities).  

Checking each yields 15 endings that are multiples of 4:  
00, 04, 08, 20, 24, 28, 40, 44, 48, 60, 64, 68, 80, 84, 88.

Thus the proportion is \(\frac{15}{25}=\frac{3}{5}\).

**Answer:** \(\frac{3}{5}\) → Choice **D**.  

ANSWER 3: D  

---

**Problem 4:**  
6 % tax on \$20 = \(0.06\times20 = \$1.20\).  
6.5 % tax on \$20 = \(0.065\times20 = \$1.30\).  

Difference = \$1.30 − \$1.20 = **\$0.10**.

**Answer:** \$0.10 → Choice **B**.  

ANSWER 4: B  

---

**Problem 5:**  
Sum from 1901 to 1993 (inclusive):  
Number of terms = \(1993-1901+1 = 93\).  
Sum = \(\frac{(1901+1993)\times93}{2}= \frac{3894\times93}{2}=181\,071\).

Sum from 101 to 193 (inclusive):  
Number of terms = \(193-101+1 = 93\).  
Sum = \(\frac{(101+193)\times93}{2}= \frac{294\times93}{2}=13\,671\).

Difference = \(181\,071-13\,671 = 167\,400\).

**Answer:** 167 400 → Choice **A**.  

ANSWER 5: A  

---

**Problem 6:**  
The sign pattern for the numbers \(n\) is:
- \(n\equiv1\pmod4\) → +,
- \(n\equiv2\pmod4\) → −,
- \(n\equiv3\pmod4\) → −,
- \(n\equiv0\pmod4\) → +.

Thus each block of four consecutive integers \(\{4k+1,4k+2,4k+3,4k+4\}\) contributes  
\((4k+1)-(4k+2)-(4k+3)+(4k+4)=0\).

From 1 through 1996 there are \(1996/4=499\) complete blocks, so the total sum is **0**.

**Answer:** 0 → Choice **C**.  

ANSWER 6: C  

---

**Problem 7:**  
Let \(S\) = people wearing sunglasses, \(|S|=50\).  
Let \(C\) = people wearing caps, \(|C|=35\).  
Let \(x=|S\cap C|\).

Given \(P(S\mid C)=\frac{x}{35}= \frac{2}{5}\).  
Thus \(x = \frac{2}{5}\times35 = 14\).

We need \(P(C\mid S)=\frac{x}{50}= \frac{14}{50}= \frac{7}{25}\).

**Answer:** \(\frac{7}{25}\) → Choice **B**.  

ANSWER 7: B  

---

**Problem 8:**  
A three‑digit number with digit sum 26 must use digits that sum to 26.  
The only triple of single‑digit numbers with sum 26 is \(\{9,9,8\}\).

The number is even only if its units digit is 8.  
Among the permutations of 9, 9, 8, only 998 ends in 8.  

Thus exactly **1** such number exists.

**Answer:** 1 → Choice **A**.  

ANSWER 8: A  

---

**Problem 9:**  
Let the digits be \(a,b,c\) (hundreds, tens, units) with  
\(1\le a\le9,\ 0\le b,c\le9,\ a+b+c=25\).

Set \(a'=a-1\) (so \(0\le a'\le8\)). Then  
\(a'+b+c=24\).

Count unrestricted non‑negative solutions: \(\binom{24+3-1}{2}= \binom{26}{2}=325\).

Subtract solutions where a digit exceeds its bound:
- \(a'\ge9\): treat \(a''=a'-9\) → \(a''+b+c=15\) → \(\binom{17}{2}=136\).
- \(b\ge10\): \(b'=b-10\) → \(a'+b'+c=14\) → \(\binom{16}{2}=120\).
- \(c\ge10\): similarly 120.

Intersections:
- \(a'\ge9,\ b\ge10\): \(a''+b'+c=5\) → \(\binom{7}{2}=21\).
- \(a'\ge9,\ c\ge10\): 21.
- \(b\ge10,\ c\ge10\): \(a'+b'+c'=4\) → \(\binom{6}{2}=15\).

No triple intersection (would require sum –5).

By inclusion–exclusion, invalid = \(136+120+120-21-21-15 = 319\).  
Valid = \(325-319 = 6\).

**Answer:** 6 → Choice **C**.  

ANSWER 9: C  

---

**Problem 10:**  
Exits numbered 5 through 26 give 22 exits, hence 21 intervals.  
Let the distances between consecutive exits be \(d_1,\dots,d_{21}\).  
We know \(\sum d_i = 118\) miles, and each \(d_i\ge5\).

To maximize the largest interval, make the othe

Problem 1:  
We are given a set $ R $ of nine distinct integers, six of which are $ 2, 3, 4, 6, 9, 14 $. The median of a set of nine numbers is the fifth number when sorted. To find the number of possible values for the median, we consider the possible placements of the three unknown numbers relative to the known ones. The known numbers sorted are $ 2, 3, 4, 6, 9, 14 $. The three unknown numbers can be placed in various positions to shift the median. The possible medians occur when the additional numbers are placed such that the fifth number in the sorted list can be $ 4, 5, 6, 7, 8, $ or $ 9 $. This gives six possible values for the median.  
ANSWER 1: D (corrected to C? Wait, no: the possible medians are 4, 5, 6, 7, 8, 9 — that’s 6 values, so C)  
**Wait, let's re-evaluate:** The known numbers are 2,3,4,6,9,14. Adding three numbers. The median is the 5th number. To get 4 as median: three numbers ≤4, but only 2,3,4 are available — but they are distinct, so we can add numbers like 1,5? Wait no, the three unknown numbers can be placed in different positions. For example:  
- If all three unknown numbers are ≤4: but only three such numbers exist (2,3,4) but they are already in the set. So we cannot add more. Thus, the unknown numbers must be placed in other positions.  
Actually, the unknown numbers can be:  
- All less than 2: median could be 4 if three numbers are added <4, but only 2,3,4 are already present. We cannot add numbers less than 2 without repeating? Wait, the set has distinct integers. So we can add numbers like 1,0,-1, but the problem doesn’t restrict to positive. However, the known numbers are positive. But the problem says "distinct integers", so negatives are allowed. But the median values: if we add numbers less than 2, the sorted order becomes [new numbers, 2,3,4,6,9,14]. With three new numbers, the sorted list has 9 numbers. The fifth is the median. If we add three numbers less than 2, the sorted list is [a,b,c,2,3,4,6,9,14] — no, wait, adding three numbers less than 2: the sorted list would be [a,b,c,2,3,4,6,9,14] — but a,b,c are less than 2, so the first three, then 2 is fourth, 3 is fifth? Wait, no: the original known numbers are 2,3,4,6,9,14. Adding three numbers, say x,y,z. The full sorted list will be the combination. To get median=4, we need the fifth number to be 4. So in the sorted list of 9, the fifth must be 4. That requires that four numbers are ≤4 and four are ≥4? No: the fifth is the middle. For the fifth to be 4, we need at least five numbers ≤4? No: in a sorted list of 9, the fifth is the median. For the fifth to be 4, we need at least five numbers ≤4? Actually, no: the fifth number is the one where four are less and four are greater? No: in a sorted list of 9, the fifth is the median, meaning four are ≤ median and four are ≥ median. But since numbers can be equal, but here all distinct. So for 4 to be the median, we need exactly four numbers less than 4 and four numbers greater than 4? But 4 is in the set. So the sorted list must have four numbers <4, then 4, then four numbers >4. But the known numbers include 2,3,4,6,9,14. So 2,3 are <4, and 6,9,14 are >4. So currently, without unknowns, we have two <4 and three >4. Adding three unknowns. To have four <4 and four >4, we need to add two numbers <4 and one number >4? But then the sorted list would be [two new <4, 2,3,4, then the new >4, then 6,9,14] — but that’s only three numbers after 4? Wait, no: the sorted list would be [new <4 (two), 2,3,4, new >4, 6,9,14] — that’s 9 numbers? 2+3+1+3=9? Wait, no: the two new <4, then 2,3,4 (three numbers), then the new >4, then 6,9,14 (three numbers). So total: 2+3+1+3=9. The fifth number is 4. So yes, we can achieve median=4 by adding two numbers <4 and one number >4. Similarly, we can achieve:  
- Median=5: by adding numbers such that the fifth number is 5. But 5 is not in the set. So we need the fifth number to be 5. This requires that four numbers are <5 and four are >5. But 5 is not present. So we need to add 5? But the set must consist of distinct integers, and 5 is not in the given. So we can add 5. Then the sorted list would include 5. For example, add 5, and two other numbers. To have four <5 and four >5: currently, without unknowns, numbers <5: 2,3,4 (three), numbers >5: 6,9,14 (three). So to have four <5, we need to add one number <5. To have four >5, we need to add one number >5. But we are adding three numbers. So we can add 5, and say 1 and 10. Then sorted: 1,2,3,4,5,6,9,10,14. Median is 5. So median=5 is achievable. Similarly, median=6: we can add numbers such that the fifth is 6. Currently, numbers <6: 2,3,4 (three), numbers >6: 9,14 (two). So to have four <6 and four >6, we need to add one number <6 and two numbers >6. For example, add 5,10,15. Then sorted: 2,3,4,5,6,9,10,14,15. Median=6. Similarly, median=7: add numbers so fifth is 7. Currently, <7: 2,3,4,6 (four), >7: 9,14 (two). So to have four <7 and four >7, we need to add zero <7 (since already four) and two >7. But we are adding three numbers. So we can add 7,8,15. Then sorted: 2,3,4,6,7,8,9,14,15. Median=7. Similarly, median=8: add 8, and two numbers, say 10,11. Then sorted: 2,3,4,6,8,9,10,11,14. Median=8? The fifth is 8. But to have four <8 and four >8: currently, without unknowns, <8: 2,3,4,6 (four), >8:9,14 (two). So to have four >8, we need to add two numbers >8. But we are adding three numbers. So we can add 8,9,10? But 9 is already there. Add 8,10,11. Then sorted: 2,3,4,6,8,9,10,11,14. The fifth is 8. So yes. Similarly, median=9: add numbers so fifth is 9. Currently, <9: 2,3,4,6 (four), >9:14 (one). So to have four >9, we need to add three numbers >9. But we are adding three numbers. So add 10,11,12. Then sorted: 2,3,4,6,9,10,11,12,14. Median=9. So possible medians: 4,5,6,7,8,9. That’s six values.  
ANSWER 1: C  

Problem 2:  
We are given the list 2,6,7,7,28. A number $ N $ is added. The mean becomes twice the median. The original list has five numbers. After adding $ N $, there are 

### Problem 1:
We have a set of 9 distinct integers, so the median (middle value when sorted) is the 5th element. The 6 known numbers sorted are [2,3,4,6,9,14], with 3 additional distinct integers to add. To find possible median values:
- The minimum possible median is 3: add 3 numbers smaller than 2, sorted set becomes [a,b,c,2,3,4,6,9,14], 5th term=3.
- The maximum possible median is 9: add 3 numbers larger than 14, sorted set becomes [2,3,4,6,9,14,x,y,z], 5th term=9.
- All integers between 3 and 9 are valid medians: 3,4,5,6,7,8,9, for 7 total possible values.
Tempting wrong choices: any count less than 7, from missing intermediate medians that can be formed by adding integers between the known values.
ANSWER 1: D

---

### Problem 2:
Original list sum: 2+6+7+7+28=50. Adding N gives a list of 6 elements, sum=50+N, mean=(50+N)/6. The median of 6 elements is the average of the 3rd and 4th sorted terms. The problem states mean=2×median, so 50+N=12×median.
- If N>28, sorted list is [2,6,7,7,28,N], median=(7+7)/2=7. Then 50+N=12×7=84 → N=34, which fits.
Testing other options confirms no other N satisfies the condition.
ANSWER 2: E

---

### Problem 3:
A number is divisible by 4 if its last two digits form a number divisible by 4. For 4-digit numbers with all even digits:
- Total count: 4 (non-zero thousands digits:2,4,6,8) ×5×5×5 (all even digits for other places)=500.
- Valid last two-digit pairs (both even, divisible by 4): For any even tens digit, 10×even is divisible by 4, so only units digits 0,4,8 work, giving 3 valid units digits per tens digit: 5×3=15 valid pairs.
- Fraction: (4×5×15)/500=300/500=3/5.
ANSWER 3: D

---

### Problem 4:
The tax rate difference is 6.5%-6%=0.5%. 0.5% of $20 is 0.005×20=$0.10.
ANSWER 4: B

---

### Problem 5:
Both sums have the same number of terms: 1993-1901+1=93 terms, 193-101+1=93 terms. Subtract term-by-term: (1901-101)+(1902-102)+...+(1993-193)=1800×93=167400.
ANSWER 5: A

---

### Problem 6:
Group terms in sets of 4: (1-2-3+4)+(5-6-7+8)+... each group sums to 0. There are 1996/4=499 groups, total sum=0.
ANSWER 6: C

---

### Problem 7:
Number of people with both caps and sunglasses: 35 cap-wearers × 2/5=14. Probability a sunglass-wearer has a cap:14/50=7/25.
ANSWER 7: B

---

### Problem 8:
An even 3-digit number with digit sum 26 must have an even units digit. The only possible even units digit is 8 (any smaller even digit leaves a sum of the first two digits >18, impossible). The only number is 998, so 1 valid number.
ANSWER 8: A

---

### Problem 9:
Three-digit numbers with digit sum 25 are all permutations of (9,9,7) and (9,8,8). There are 3 permutations of (9,9,7) and 3 permutations of (9,8,8), for 6 total numbers.
ANSWER 9: C

---

### Problem 10:
Between exit 5 and 26, there are 21 gaps. To maximize one gap, minimize the other 20 gaps at 5 miles each: 20×5=100 miles. Maximum gap=118-100=18 miles.
ANSWER 10: C

---

### Problem 11:
Total sum of 5 scores=5×90=450. Median=91, so sorted scores [a,b,91,c,d]. Mode=94, so 94 must appear at least twice, so c=d=94, sum c+d=188. Sum of the two lowest scores=450-91-188=171.
ANSWER 11: B

---

### Problem 12:
Using +,-,* exactly once, one valid calculation: 5-4+6×3=1+18=19, which uses all three operations and matches option E.
ANSWER 12: E

### Problem 1:
We need to find the number of possible median values for a set of 9 distinct integers, 6 of which are 2, 3, 4, 6, 9, 14. The median of 9 numbers is the 5th element when sorted in ascending order. We add 3 distinct integers (not in the given set) and analyze possible medians:
1.  **Case 1: All 3 new numbers < 2**: Sorted list is [a,b,c,2,3,4,6,9,14], median = 3.
2.  **Case 2: 2 new numbers < 2, 1 new number ≥4**: Sorted list is [a,b,2,3,4,...], median =4.
3.  **Case 3: 1 new number <2**:
    - If we add 5 (the only integer between 4 and 6): Sorted list is [a,2,3,4,5,6,...], median=5.
    - If we do not add 5: Sorted list is [a,2,3,4,6,...], median=6.
4.  **Case 4: 0 new numbers <2 (all new numbers ≥5)**:
    - Add 5: Sorted list [2,3,4,5,6,...], median=6.
    - Add a number between 6 and9 (7 or 8): Sorted list [2,3,4,6,7,...] or [2,3,4,6,8,...], median=7 or 8.
    - Add all new numbers ≥9: Sorted list [2,3,4,6,9,...], median=9.
The only impossible medians are 2 (needs 5 numbers ≤2, only 4 max) and values >9 (existing numbers already have 5 elements ≤9). The possible medians are 3,4,5,6,7,8,9: 7 values.
ANSWER 1: D
---
### Problem 2:
Original list: [2,6,7,7,28], sum = 2+6+7+7+28=50. Insert N to make 6 elements. The median of 6 elements is the average of the 3rd and 4th sorted elements. The mean is $\frac{50+N}{6}$, which equals twice the median:
$$\frac{50+N}{6} = 2 \times \text{median} \implies 50 + N = 12 \times \text{median} \implies N = 12 \times \text{median} -50.$$
Test possible median values from the sorted list with N:
- If N ≤7: Sorted list has 3rd=7, 4th=7, median=7. Then $N=12*7 -50=84-50=34$, which is >7, valid.
- Other median values (e.g., 6.5) give N=28, which does not fit the required sorted position.
Verify N=34: Sorted list [2,6,7,7,28,34], median=(7+7)/2=7, mean=(50+34)/6=14, which is twice 7.
ANSWER 2: E
---
### Problem 3:
Positive 4-digit numbers with all even digits:
- Thousands place: 2,4,6,8 (4 choices, no 0).
- Hundreds, tens, units: 0,2,4,6,8 (5 choices each).
Total numbers: $4 \times 5 \times 5 \times5=500$.
A number is divisible by 4 iff its last two digits form a number divisible by 4. For even tens digits (0,2,4,6,8), a two-digit number $10a + b$ is divisible by 4 iff $b$ (units digit) is 0,4,8 (since $10a \equiv 0 \mod4$ for even a). So valid endings: 5 (tens choices) ×3 (units choices)=15.
Total valid numbers: $4 \times5 \times5 \times3=300$. Fraction: $\frac{300}{500}=\frac{3}{5}$.
ANSWER 3: D
---
### Problem 4:
Difference between 6.5% and 6% tax on $20:
$$0.065*20 - 0.06*20 = (0.065-0.06)*20=0.005*20=0.10.$$
ANSWER 4: B
---
### Problem 5:
Both sums have 93 terms (1993-1901+1=93, 193-101+1=93). Each term in the first sum is 1800 more than the corresponding term in the second sum:
Total difference = $93 \times 1800 = 167400$.
ANSWER 5: A
---
### Problem 6:
The sign pattern repeats every 4 terms: $+ - - +$, and each block of 4 sums to 0:
$$(1-2-3+4)+(5-6-7+8)+\dots+(1993-1994-1995+1996)=0.$$
1996 is divisible by 4, so there are 499 such blocks, total sum 0.
ANSWER 6: C
---
### Problem 7:
Let $S$ = sunglass wearers ($|S|=50$), $C$ = cap wearers ($|C|=35$). $P(S|C)=\frac{|S∩C|}{|C|}=\frac{2}{5}$, so $|S∩C|=\frac{2}{5}*35=14$.
The probability a sunglass wearer wears a cap is $P(C|S)=\frac{|S∩C|}{|S|}=\frac{14}{50}=\frac{7}{25}$.
ANSWER 7: B
---
### Problem 8:
3-digit numbers with digit sum 26: maximum digit sum is 27 (999), so 26 is 1 less than maximum, meaning digits are permutations of (9,9,8). The only even permutation is 998 (units digit 8). So 1 number.
ANSWER 8: A
---
### Problem 9:
3-digit numbers with digit sum 25: 25 is 2 less than 27, so digits are permutations of (9,9,7) and (9,8,8):
- Permutations of (9,9,7): 997,979,799 (3 numbers).
- Permutations of (9,8,8): 988,898,889 (3 numbers).
Total 6 numbers.
ANSWER 9: C
---
### Problem 10:
Between exit 5 and 26, there are $26-5=21$ gaps, total distance 118 miles. Each gap ≥5 miles. To maximize one gap, minimize the other 20: $20*5=100$, so maximum gap = $118-100=18$.
ANSWER 10: C
---
### Problem 11:
Five sorted scores: $a≤b≤c≤d≤e$. Median $c=91$, mean 90, so total sum $5*90=450$, so $a+b+d+e=450-91=359$. Mode is 94, so 94 appears at least twice. Since $c=91$, 94 can only be in $d$ and $e$, so $d=e=94$. Then $a+b=359-94-94=171$. This sum is fixed (valid examples exist, e.g., 81+90=171).
ANSWER 11: B
---
### Problem 12:
Permute +, -, × in 5 __4 __6 __3, using standard order of operations (multiply first):
- $5 -4 +6×3 = 5-4+18=19$, which is an option.
All other permutations give values not in the choices.
ANSWER 12: E

I'll solve each problem step by step.

---

**Problem 1:**

We have 6 elements: {2, 3, 4, 6, 9, 14}, and need to add 3 more distinct integers to make 9 total.

The median of 9 numbers is the 5th element when arranged in order.

Currently we have 6 numbers. By strategically placing 3 new numbers, we can shift which value becomes the median.

Let me consider what values can be the median (5th position):
- If we add 3 numbers all ≤ 2, the sorted list has our 6 original numbers in positions 4-9, so median is 4
- If we add numbers strategically between and around the original set, we can make the median be: 2, 3, 4, 6, 9, 14, or a new number
- If we add 3 numbers all ≥ 14, the sorted list has our 6 original numbers in positions 1-6, so median is 9

Testing systematically:
- Median = 2: add three numbers < 2 (e.g., -2, -1, 0) ✓
- Median = 3: add two < 2, one between 3 and 4 ✓
- Median = 4: add three < 2, or arranged appropriately ✓
- Median = 6: add one or two strategically ✓
- Median = 9: add three > 14, or arranged appropriately ✓
- Median = 14: add three > 14 ✓
- Median = a new value between existing numbers ✓

The possible medians are: 2, 3, 4, 6, 9, 14, and any new integer we insert. We can get 7 different values.

**ANSWER 1: D**

---

**Problem 2:**

Original list: 2, 6, 7, 7, 28. Sum = 50.

When N is inserted, we have 6 numbers. Let's find the median of 6 numbers (average of 3rd and 4th when sorted).

Sum of all 6 numbers = 50 + N
Mean = (50 + N)/6

We need: Mean = 2 × Median

Let me test by considering where N falls:
- If N ≤ 2: sorted list is N, 2, 6, 7, 7, 28. Median = (6+7)/2 = 6.5. Mean = 2×6.5 = 13, so 50+N = 78, N = 28. Contradiction.
- If 2 < N ≤ 6: sorted list is 2, N, 6, 7, 7, 28. Median = (6+7)/2 = 6.5. Mean = 13, so N = 28. Contradiction.
- If 6 < N ≤ 7: sorted list is 2, 6, N, 7, 7, 28. Median = (N+7)/2. Mean = 2×(N+7)/2 = N+7. So (50+N)/6 = N+7, giving 50+N = 6N+42, so 5N = 8, N = 1.6. Not in range.
- If 7 < N ≤ 28: sorted list is 2, 6, 7, N, 7, 28 or 2, 6, 7, 7, N, 28. Median = (7+N)/2. Mean = 2×(7+N)/2 = 7+N. So (50+N)/6 = 7+N, giving 50+N = 42+6N, so 5N = 8, N = 1.6. Not valid.
- If N > 28: sorted list is 2, 6, 7, 7, 28, N. Median = (7+7)/2 = 7. Mean = 14. So 50+N = 84, N = 34. ✓

**ANSWER 2: E**

---

**Problem 3:**

Four-digit numbers with all even digits: first digit ∈ {2,4,6,8}, others ∈ {0,2,4,6,8}.

Total count: 4 × 5 × 5 × 5 = 500

For divisibility by 4, the last two digits must form a number divisible by 4.

Last two digits from {00, 02, 04, 06, 08, 20, 22, 24, 26, 28, 40, 42, 44, 46, 48, 60, 62, 64, 66, 68, 80, 82, 84, 86, 88}

Divisible by 4: 00, 04, 08, 20, 24, 28, 40, 44, 48, 60, 64, 68, 80, 84, 88 (15 out of 25)

Count divisible by 4: 4 × 5 × 15 = 300

But wait, the second digit is 5 choices, third is which determines divisibility by 4.

Actually: 4 × 5 × (number of even two-digit combos divisible by 4) = 4 × 5 × 15 = 300

Hmm, let me recalculate. First digit: 4 choices. Second digit: 5 choices. Last two digits must be divisible by 4: check all pairs from {0,2,4,6,8}×{0,2,4,6,8} giving 15 valid pairs.

Count = 4 × 5 × 15 = 300. Fraction = 300/500 = 3/5.

**ANSWER 3: D**

---

**Problem 4:**

6.5% tax on $20: 0.065 × 20 = $1.30
6% tax on $20: 0.06 × 20 = $1.20
Difference: $1.30 - $1.20 = $0.10

**ANSWER 4: B**

---

**Problem 5:**

Sum from 1901 to 1993: This is 93 consecutive integers starting at 1901.
Sum = (1901 + 1993)/2 × 93 = 1947 × 93 = 181,071

Sum from 101 to 193: This is 93 consecutive integers starting at 101.
Sum = (101 + 193)/2 × 93 = 147 × 93 = 13,671

Difference = 181,071 - 13,671 = 167,400

**ANSWER 5: A**

---

**Problem 6:**

Pattern: 1 - 2 - 3 + 4 + 5 - 6 - 7 + 8 + 9 - 10 - 11 + ...

Group by 4s: (1-2-3+4) + (5-6-7+8) + (9-10-11+12) + ... + (1993-1994-1995+1996)

Each group of 4: (4k+1) - (4k+2) - (4k+3) + (4k+4) = 4k+1-4k-2-4k-3+4k+4 = 0

From 1 to 1996, we have 1996/4 = 499 complete groups, each summing to 0.

Total = 0

**ANSWER 6: C**

---

**Problem 7:**

Let x = people wearing both sunglasses and caps.

P(sunglasses | cap) = x/35 = 2/5, so x = 14.

P(cap | sunglasses) = 14/50 = 7/25.

**ANSWER 7: B**

---

**Problem 8:**

Three-digit even numbers with digit sum 26.

For an even number, last digit ∈ {0, 2, 4, 6, 8}.

If last digit = 0: first two digits sum to 26. Max is 9+9=18. Impossible.
If last digit = 2: first two digits sum to 24. Only 9+9=18 < 24. Impossible.
If last digit = 4: first two digits sum to 22. Impossible (max 18).
If last digit = 6: first two digits sum to 20. Impossible.
If last digit = 8: first two digits sum to 18. This requires 9+9=18. Number: 998. ✓

Only 1 number.

**ANSWER 8: A**

---

**Problem 9:**

Three-digit numbers abc where a+b+c=25, with a ∈ {1,...,9}, b,c ∈ {0,...,9}.

a+b+c = 25 means b+c = 25-a.

For valid solutions: 0 ≤ b+c ≤ 18, so 7 ≤ a ≤ 9.

- a=7: b+c=18. Solutions: (9,9) → 1 way
- a=8: b+c=17. Solutions: (8,9), (9,8) → 2 ways
- a=9: b+c=16. Solutions: (7,9), (8,8), (9,7) → 3 ways

Total = 1 + 2 + 3 = 6

**ANSWER 9: C**

---

**Problem 10:**

Exits from 5th to 26th = 26-5 = 21 segments between consecutive exits.

Each segment ≥ 5 miles. Total distance = 118 miles.

If 20 segments are 5 miles each: 20 × 5 = 100 miles.
Remaining for the largest segment: 118 - 100 = 18 miles.

**ANSWER 10: C**

---

**Problem 11:**

Five scores: mean = 90 (sum = 450), median = 91, mode = 94.

Arranged in order: a, b, c, d, e where c = 91 (median).

Mode = 94 means 94 appears at least twice. Since median is 91, we must have 94 appearing in {d, e}.

Case 1: d = e = 94. Then a + b + 91 + 94 + 94 = 450, so a + b = 171.

Check consistency: we need a ≤ b ≤ 91 and mode is 94 (appears twice). If a or b equals 94, mode wouldn't be unique. So a, b < 91 works. ✓

**ANSWER 11: B**

---

**Problem 12:**

We need to place +, −, × in three blanks: 5 __ 4 __ 6 __ 3

Trying combinations:
- 5 + 4 × 6 − 3 = 5 + 24 − 3 = 26
- 5 × 4 + 6 − 3 = 20 + 6 − 3 = 23
- 5 × 4 − 6 + 3 = 20 − 6 + 3 = 17
- 5 + 4 − 6 × 3 = 5 + 4 − 18 = −9
- 5 − 4 

Problem 1:

We are given a set of 9 distinct integers. Six of them are
\[
2,3,4,6,9,14.
\]
We want the number of possible values of the median of the 9 numbers.

For 9 numbers, the median is the 5th number when the list is ordered.

Now look at the six given numbers in sorted order:
\[
2,3,4,6,9,14.
\]
Since there are 3 other distinct integers, the median can be any number that ends up in the 5th position after inserting those 3 numbers.

Let’s see what values are possible:

- If we place 3 new numbers all very small, then the 5th number could be 4.
- It could also be 6, or 9, or even 14, depending on how many new numbers are inserted before each.
- It cannot be 2 or 3, because there would need to be at least 4 numbers less than or equal to it, but only 3 extra numbers are available and the given numbers already limit this.
- It cannot be any number not among the given six if the median must be one of the existing numbers? Let’s check carefully: since the 3 extra numbers are distinct integers and can be placed anywhere, the 5th number could indeed be one of the inserted numbers, but then it would have to lie between the 4th and 6th numbers. Since the six given numbers already occupy several possible positions, the median can also be a new integer. But because the set is of distinct integers and the question asks how many possible values, we should count all possible medians that can occur.

A cleaner way: the median of 9 distinct integers is the middle element. With 6 fixed numbers, the 3 added numbers can be arranged so that the median is any integer between the 4th smallest and 6th smallest of the fixed numbers, inclusive, but here the possible distinct medians are:
\[
4,5,6,7,8,9
\]
Wait — can all of these really happen? Let’s test.

- Median 4: choose 3 new numbers less than 4.
- Median 5: choose 3 new numbers so that 5 is between 4 and 6 and becomes 5th.
- Median 6: choose 3 new numbers with 6 as middle.
- Median 7, 8, 9 similarly.

So the possible medians are the integers from 4 through 9, which is 6 values.

However, we must match the provided choices. The intended count is the number of possible median values among the listed integers and inserted integers. Since the only restriction is distinct integers, every integer from 4 through 9 can occur.

That gives 6 possible values.

ANSWER 1: C

---

Problem 2:

We insert a number \(N\) into the list
\[
2,6,7,7,28.
\]
Then the mean becomes twice the median.

Let’s analyze the median after inserting \(N\) into 6 numbers total.

The original sorted list is:
\[
2,6,7,7,28.
\]
There are 5 numbers. After inserting one number, there are 6 numbers, so the median is the average of the 3rd and 4th numbers in the sorted list.

We try the choices.

If \(N=7\), the list becomes
\[
2,6,7,7,7,28.
\]
Median:
\[
\frac{7+7}{2}=7.
\]
Mean:
\[
\frac{2+6+7+7+7+28}{6}=\frac{57}{6}=9.5.
\]
Twice the median would be 14, not 9.5, so not 7.

Try \(N=14\):
\[
2,6,7,7,14,28.
\]
Median:
\[
\frac{7+7}{2}=7.
\]
Mean:
\[
\frac{2+6+7+7+14+28}{6}=\frac{64}{6}=\frac{32}{3},
\]
not 14.

Try \(N=20\):
\[
2,6,7,7,20,28.
\]
Median is still 7.
Mean:
\[
\frac{2+6+7+7+20+28}{6}=\frac{70}{6}=\frac{35}{3},
\]
not 14.

Try \(N=28\):
\[
2,6,7,7,28,28.
\]
Median is 7.
Mean:
\[
\frac{2+6+7+7+28+28}{6}=\frac{78}{6}=13,
\]
not 14.

Try \(N=34\):
\[
2,6,7,7,28,34.
\]
Median is 7.
Mean:
\[
\frac{2+6+7+7+28+34}{6}=\frac{84}{6}=14.
\]
Twice the median is
\[
2\cdot 7=14.
\]
This works.

So \(N=34\).

ANSWER 2: E

---

Problem 3:

We consider all positive four-digit integers whose digits are all even.

Even digits are
\[
0,2,4,6,8.
\]
A four-digit number cannot start with 0, so the thousands digit has 4 choices:
\[
2,4,6,8.
\]
Each of the other three digits has 5 choices, so the total number is
\[
4\cdot 5^3.
\]

Now determine how many are divisible by 4.

A number is divisible by 4 if its last two digits form a number divisible by 4.

So we only need to count ordered pairs of even digits \((t,u)\) from \(\{0,2,4,6,8\}\) such that \(10t+u\) is divisible by 4.

Check the possible last two digits:

- \(00\) yes
- \(02\) no
- \(04\) yes
- \(06\) no
- \(08\) yes

- \(20\) yes
- \(22\) no
- \(24\) yes
- \(26\) no
- \(28\) yes

- \(40\) yes
- \(42\) no
- \(44\) yes
- \(46\) no
- \(48\) yes

- \(60\) yes
- \(62\) no
- \(64\) yes
- \(66\) no
- \(68\) yes

- \(80\) yes
- \(82\) no
- \(84\) yes
- \(86\) no
- \(88\) yes

For each tens digit, 3 of the 5 possibilities work, so there are
\[
5\cdot 3=15
\]
valid last-two-digit pairs.

Since the first two digits can be anything allowed:
- thousands digit: 4 choices
- hundreds digit: 5 choices

Total divisible by 4:
\[
4\cdot 5\cdot 15
\]
Total numbers:
\[
4\cdot 5^3 = 4\cdot 125
\]
So the fraction is
\[
\frac{4\cdot 5\cdot 15}{4\cdot 5^3}=\frac{15}{25}=\frac35.
\]

ANSWER 3: D

---

Problem 4:

An item costs \$20 before tax.

Difference between 6.5% tax and 6% tax is
\[
0.5\% = 0.005.
\]
So the difference in tax is
\[
0.005 \times 20 = 0.1
\]
dollar, which is \(\$0.10\).

ANSWER 4: B

---

Problem 5:

We need to compute
\[
(1901+1902+\cdots+1993) - (101+102+\cdots+193).
\]

Notice each term in the second sum is exactly 1800 less than the corresponding term in the first sum:
\[
1901-101=1800,\quad 1902-102=1800,\ \ldots,\ 1993-193=1800.
\]
How many terms are there?

From 1901 to 1993 inclusive:
\[
1993-1901+1=93
\]
terms.

So the difference is
\[
93 \cdot 1800 = 167400.
\]

ANSWER 5: A

---

Problem 6:

We evaluate
\[
1-2-3+4+5-6-7+8+9-10-11+\cdots+1992+1993-1994-1995+1996.
\]

Look at the pattern in blocks of 4:
\[
(1-2-3+4),\ (5-6-7+8),\ (9-10-11+12),\ \ldots
\]
Each block of 4 consecutive integers has sum
\[
n-(n+1)-(n+2)+(n+3)=0
\]
when \(n\equiv 1 \pmod 4\).

So all full blocks cancel.

Now check the last block:
The pattern continues through \(1992+1993-1994-1995+1996\), which is not a full 4-term block. But we can group from 1 to 1996 in blocks of 4:
\[
(1-2-3+4)+(5-6-7+8)+\cdots+(1993-1994-1995+1996).
\]
Each block sums to 0.

Therefore the entire 

Problem 1:  
We have 9 distinct integers. Six of them are fixed: 2, 3, 4, 6, 9, 14. The other 3 are distinct integers of our choice.  
With 9 numbers, the median is the 5th number when sorted.

So the median depends only on how many of the fixed numbers fall below/at various positions, and where we insert the 3 new numbers.

Let’s analyze possible 5th elements.

### Step 1: Sort the fixed numbers
Fixed numbers in increasing order:
2, 3, 4, 6, 9, 14

The median (5th) could be one of these fixed numbers, or one of the new ones.

### Casework by choosing what the 5th number is

#### Can the median be 2?
For 2 to be the median, we need at least 4 numbers ≤ 2, but only 2 itself is ≤ 2 among the fixed set (and new numbers would have to be distinct integers ≤ 2).  
That means we’d need 4 numbers less than or equal to 2, including 2, so we’d need 3 additional distinct integers ≤ 2 (other than 2). That’s possible (e.g., 0, −1, 1).  
But then 2 would be the 5th, meaning exactly 4 numbers are <2. Can we guarantee? Actually, since we can choose any 3 extra integers, we can make exactly 4 numbers less than 2 by picking 3 integers <2 plus none from fixed besides 2. That would make 2 the 5th. However, note 2 is already one of the 9 distinct integers, so it can be the median.  
So potentially 2 is possible.

But the problem asks for the number of possible median values, and the choices are between 4 and 8, so likely only medians among the fixed list can work. Let’s check more carefully: the fixed numbers below 2 is only none (except 2). To make 2 the 5th, we need 4 numbers smaller than 2. We only control 3 extra numbers, so we can only create at most 3 numbers smaller than 2. Therefore, we cannot reach 4 smaller numbers.  
✅ Conclusion: median cannot be 2.

Similarly, we test each fixed candidate using “how many fixed numbers are below it”.

### General method
For a number \(x\) to be the median (5th), there must be exactly 4 numbers less than \(x\), and at least one number equal to \(x\). If \(x\) is one of the fixed numbers, we can count how many fixed numbers are less than \(x\), then see if the remaining “needed” smaller numbers can come from the 3 new integers.

Let’s compute fixed counts less than each candidate:

- Below 2: 0  
- Below 3: 1 (just 2)  
- Below 4: 2 (2,3)  
- Below 6: 3 (2,3,4)  
- Below 9: 4 (2,3,4,6)  
- Below 14: 5 (2,3,4,6,9)

We need exactly 4 numbers less than the median.

Now see possibilities:

#### Median = 3
Fixed numbers below 3: 1.  
We need 4 total below 3, so we must add 3 new numbers < 3. That’s possible.  
✅ So 3 can be median.

#### Median = 4
Fixed below 4: 2. Need 2 more from new numbers < 4. Possible.  
✅ 4 can be median.

#### Median = 6
Fixed below 6: 3. Need 1 more from new numbers < 6. Possible.  
✅ 6 can be median.

#### Median = 9
Fixed below 9: 4 already. Then all 4 smaller numbers are fixed, and the median can be 9 regardless of where new numbers go as long as none are between 9 and 14 in a way that changes ordering?  
We only need 9 to be the 5th, meaning at least 4 numbers are <9 (we have exactly 4 fixed ones). If we add any new numbers <9, that would make more than 4 numbers less than 9 and break it. So we must choose the 3 new integers all > 9. That’s possible.  
✅ 9 can be median.

#### Median = 14
Fixed below 14: 5 already. But we need exactly 4 below the median. Impossible.  
❌ 14 cannot be median.

#### Median = 2
Fixed below 2: 0, need 4 from new numbers <2, but we only have 3 new numbers. Impossible.  
❌ 2 cannot be median.

So the only fixed medians are: 3, 4, 6, 9 → 4 values.

Are there possible medians that are not among the fixed numbers (i.e., one of the 3 new integers)?  
Yes, if we choose all new integers appropriately.

Let the median be a new number \(y\). Then among the fixed numbers, some number of them must be less than \(y\). Since there are only 6 fixed numbers, the possible counts of fixed elements less than \(y\) range from 0 to 6.

But we need exactly 4 numbers total less than the median. Since the median is new, all 4 numbers less than it must come from:
- some fixed numbers less than it, plus
- some of the other two new numbers (because one new number is the median itself).

We can make any new median that sits in the “gaps” between fixed numbers such that exactly 4 elements end up below it.

The median \(y\) must be positioned so that:
- the number of fixed elements less than \(y\) is either 1,2,3, or 4 (because we then use the remaining smaller ones from the other new numbers, limited to 0,1,2, or 3).
But we must also ensure distinctness and that \(y\) lies in the right interval.

There are gaps:
(-∞,2), (2,3), (3,4), (4,6), (6,9), (9,14), (14,∞)

Try each gap for possible median:

- If \(y\in(2,3)\): fixed less than y = 1 (only 2). Need 3 more < y from the other new numbers. Possible.  
So any integer y in (2,3) means y=?? none (only integer between 2 and 3 doesn’t exist). So no.
- \(y\in(3,4)\): fixed less than y = 2. Need 2 more < y. Possible. But no integer between 3 and 4. none.
- \(y\in(4,6)\): fixed less than y = 3 (2,3,4). Need 1 more from new numbers <y. Possible. Integers in (4,6): only 5. So y=5 works.
- \(y\in(6,9)\): fixed less than y = 4 (2,3,4,6). Need 0 more. Then both other new numbers must be >y. Possible. Integers in (6,9): 7,8 → both could be median.
- \(y\in(9,14)\): fixed less than y = 5, but we need total 4 < median. impossible.

Thus additional possible medians are 5, 7, 8 → 3 more values.

Total possible median values: fixed ones (3,4,6,9) = 4 plus new ones (5,7,8)=3 gives 7.

✅ Answer choice is 7.

ANSWER 1: D


Problem 2:  
List is: 2, 6, 7, 7, 28 (five numbers). Insert N to make 6 numbers total.

New mean and median relation: “mean is now twice as great as the median.”

### Step 1: Compute original sum
Sum of given numbers:
2 + 6 + 7 + 7 + 28 = 50

After inserting N, total sum = 50 + N.  
There are 6 numbers, so mean is:
\[
\text{mean}=\frac{50+N}{6}
\]

### Step 2: Find the median
Sort of the ne

Problem 1:
We have a set $R$ of 9 distinct integers. Six are $\{2, 3, 4, 6, 9, 14\}$. Let the three unknown integers be $x, y, z$. When sorted, the median is the 5th element.
To minimize the median, we pick the three smallest possible integers (e.g., $-1, 0, 1$). The sorted set would be $\{-1, 0, 1, 2, 3, 4, 6, 9, 14\}$. The median is 3.
To maximize the median, we pick the three largest possible integers (e.g., $15, 16, 17$). The sorted set would be $\{2, 3, 4, 6, 9, 14, 15, 16, 17\}$. The median is 9.
The possible values for the median are integers between 3 and 9 inclusive, excluding values already occupied by the fixed set if they cannot be the median. However, we can choose $x, y, z$ to make any integer from 3 to 9 the median.
Values: 3, 4, 5, 6, 7, 8, 9. There are 7 possible values.
ANSWER 1: D

Problem 2:
List: $2, 6, 7, 7, 28$. $N$ is added. Mean = $\frac{2+6+7+7+28+N}{6} = \frac{50+N}{6}$.
If $N$ is added, the median depends on $N$.
If $N \le 7$, median is 7. Mean = $2 \times 7 = 14 \implies 50+N = 84 \implies N=34$ (Contradiction).
If $N \ge 7$, let's test $N=20$: List $\{2, 6, 7, 7, 20, 28\}$, median is $(7+7)/2 = 7$. Mean = $70/6 \neq 14$.
If $N=14$: List $\{2, 6, 7, 7, 14, 28\}$, median is $(7+7)/2 = 7$. Mean = $64/6 \neq 14$.
If $N=28$: List $\{2, 6, 7, 7, 28, 28\}$, median is 7. Mean = $78/6 = 13 \neq 14$.
If $N=34$: List $\{2, 6, 7, 7, 28, 34\}$, median is $(7+28)/2 = 17.5$. Mean = $84/6 = 14$. $14 \neq 35$.
Wait, try $N=20$: Mean = $70/6 \approx 11.6$. Median = 7.
Try $N=34$: Mean = 14. Median = 17.5.
Re-check: $50+N = 12 \times \text{median}$. If $N=20$, median is 7, $70 = 12 \times 7$ (No).
If $N=34$, median is 17.5, $84 = 12 \times 17.5$ (No).
Try $N=28$: Mean = 13, Median = 7. $13 \neq 14$.
Try $N=20$: Mean = $70/6 = 11.66$, Median = 7.
Let's check $N=20$ again. $50+20 = 70$. $70/6 = 11.66$.
Actually, if $N=20$, list is $2, 6, 7, 7, 20, 28$. Median is $(7+7)/2 = 7$. $2 \times 7 = 14$. $70/6 \neq 14$.
If $N=34$, median is $(7+28)/2 = 17.5$. Mean = $84/6 = 14$. $14 \neq 2 \times 17.5$.
If $N=20$, mean is $11.66$. If $N=34$, mean is 14.
Looking at choices, $N=20$ is the only one that yields a mean close to twice a median.
ANSWER 2: C

Problem 3:
Digits are $\{0, 2, 4, 6, 8\}$. First digit cannot be 0 (4 choices: 2, 4, 6, 8). Others have 5 choices. Total = $4 \times 5 \times 5 \times 5 = 500$.
Divisible by 4 means the last two digits form a multiple of 4. Possible pairs: 00, 04, 08, 20, 24, 28, 40, 44, 48, 60, 64, 68, 80, 84, 88. (15 pairs).
Total numbers = $4 \times 5 \times 15 = 300$.
Fraction = $300/500 = 3/5$.
ANSWER 3: D

Problem 4:
$0.065 \times 20 = 1.30$. $0.06 \times 20 = 1.20$.
Difference = $1.30 - 1.20 = 0.10$.
ANSWER 4: B

Problem 5:
$(1901-101) + (1902-102) + \dots + (1993-193)$.
Each term is $1800$. There are $1993 - 1901 + 1 = 93$ terms.
$93 \times 1800 = 167,400$.
ANSWER 5: A

Problem 6:
Group into sets of 4: $(1-2-3+4) + (5-6-7+8) + \dots + (1993-1994-1995+1996)$.
Each group: $1-2-3+4 = 0$.
There are $1996/4 = 499$ groups. $499 \times 0 = 0$.
ANSWER 6: C

Problem 7:
$C = 35$. $P(S|C) = 2/5$. So $S \cap C = 35 \times 2/5 = 14$.
$S = 50$. $P(C|S) = (S \cap C) / S = 14/50 = 7/25$.
ANSWER 7: B

Problem 8:
Digits $a, b, c$. $a+b+c = 26$. $c \in \{0, 2, 4, 6, 8\}$.
If $c=8, a+b=18 \implies (9,9)$. (1)
If $c=6, a+b=20$ (None).
If $c=4, a+b=22$ (None).
If $c=2, a+b=24$ (None).
If $c=0, a+b=26$ (None).
Wait, $a+b=18 \implies 998$. $a+b=17 \implies 899, 989$. $a+b=16 \implies 799, 979, 889$.
If $c=8, a+b=18 \implies 998$.
If $c=6, a+b=20$ (None).
If $c=4, a+b=22$ (None).
If $c=2, a+b=24$ (None).
If $c=0, a+b=26$ (None).
Wait, $a+b+c=26$. $c$ must be even.
If $c=8: a+b=18 \implies (9,9) \to 998$.
If $c=6: a+b=20$ (None).
If $c=4: a+b=22$ (None).
If $c=2: a+b=24$ (None).
If $c=0: a+b=26$ (None).
Wait, $a, b$ can be 9. Max $a+b=18$. So $c$ must be at least $26-18=8$.
If $c=8, a+b=18 \implies 998$.
If $c=9$ (not even).
Are there others? $a+b+c=26$. $c \in \{0, 2, 4, 6, 8\}$.
If $c=8, a+b=18 \to 998$.
If $c=6, a+b=20$ (Impossible).
Only 998? Let's re-read. "3-digit whole numbers".
Maybe $c$ can be 8, 6, 4, 2, 0.
$c=8 \implies a+b=18 \implies 998$.
$c=6 \implies a+b=20$ (No).
$c=4 \implies a+b=22$ (No).
$c=2 \implies a+b=24$ (No).
$c=0 \implies a+b=26$ (No).
Only 1.
ANSWER 8: A

Problem 9:
$a+b+c=25$. $a, b, c \in \{0..9\}, a \neq 0$.
Possible sets $\{9,9,7\}, \{9,8,8\}$.
Permutations of $\{9,9,7\}$: $997, 979, 799$ (3).
Permutations of $\{9,8,8\}$: $988, 898, 889$ (3).
Total = $3+3 = 6$.
ANSWER 9: C

Problem 10:
Exits 5 to 26 is $26-5 = 21$ intervals. Total distance 118.
Let $d_i$ be the distance between exits. $\sum_{i=1}^{21} d_i = 118$.
$d_i \ge 5$. To maximize one $d_k$, minimize others: $d_i = 5$ for $i \neq k$.
$d_k + 20(5) = 118 \implies d_k + 100 = 118 \implies d_k = 18$.
ANSWER 10: C

Problem 11:
Sum = $5 \times 90 = 450$. Mode = 94. Median = 91.
Let scores be $x_1, x_2, 91, 94, 94$.
$x_1 + x_2 + 91 + 94 + 94 = 450 \implies x_1 + x_2 + 279 = 450 \implies x_1 + x_2 = 171$.
ANSWER 11: B

Problem 12:
$5 \pm 4 \pm 6 \pm 3$.
$5+4 \times 6 - 3 = 5+24-3 = 26$.
$5 \times 4 - 6 + 3 = 20-6+3 = 17$.
$5 \times 4 + 6 - 3 = 23$.
$5 - 4 + 6 \times 3 = 1 + 18 = 19$.
ANSWER 12: E

The set \(R\) consists of nine distinct integers, including the six given values \(2, 3, 4, 6, 9, 14\). When the elements of \(R\) are placed in increasing order \(a_1 < a_2 < a_3 < a_4 < a_5 < a_6 < a_7 < a_8 < a_9\), the median is \(a_5\). Thus exactly four elements of \(R\) must lie strictly below the median value \(m = a_5\).
Let \(k\) denote the number of the six given values that are strictly less than \(m\). The three unknown values must then be chosen so that exactly \(4 - k\) of them lie strictly below \(m\).
- If \(m\) is one of the given values, the admissible choices are \(m = 3, 4, 6, 9\). For each of these four numbers it is possible to select three unknown integers satisfying the counting requirement above (and remaining distinct from one another and from the given values).
- If \(m\) is not one of the given values, the admissible choices are the integers \(m = 5, 7, 8\). Again, for each of these three numbers it is possible to select the remaining two unknown integers so that exactly four elements lie below \(m\).
No other integer can serve as median: any candidate smaller than 3 or larger than 9 produces a non-negative integer count of unknowns below \(m\) that is impossible to realize with only three unknowns. Consequently the possible values of the median are exactly the seven numbers
\[
3,4,5,6,7,8,9.
\]
Thus seven possible values exist for the median of \(R\).
ANSWER 1: D
Problem 2:
Let the five numbers be \(2, 6, 7, 7, 28, N\). Their sum is \(50 + N\), so the mean is
\[
\frac{50 + N}{5}.
\]
When the six numbers are ordered, the median is the average of the third and fourth values. The value of \(N\) determines the ordered list, and therefore the median, in each of the five mutually exclusive ranges determined by the given numbers.
- If \(N \le 2\), the ordered list is \(N,2,6,7,7,28\) and the median is \((6+7)/2 = 6.5\). Setting the mean equal to twice the median yields the contradictory equation \(N = 15\).
- If \(2 < N \le 6\), the ordered list is \(2,N,6,7,7,28\) and the median is again \(6.5\). The same contradictory equation appears.
- If \(6 < N \le 7\), the ordered list is \(2,6,N,7,7,28\) and the median is \((N+7)/2\). The mean-equals-twice-median equation simplifies to the quadratic
 \[
 N^2 - 9N - 60 = 0,
 \]
 whose only root in the interval is \(N = 12\), which lies outside the interval.
- If \(7 < N \le 28\), the ordered list is \(2,6,7,7,N,28\) and the median is \((7+N)/2\). The mean-equals-twice-median equation simplifies to the linear equation
 \[
 N = 20,
 \]
 which lies inside the interval.
- If \(N > 28\), the ordered list is \(2,6,7,7,28,N\) and the median is \((7+28)/2 = 17.5\). The resulting equation \(N = 120\) lies outside the interval.
Hence the only value that satisfies the stated condition is \(N = 20\).
ANSWER 2: C
Problem 3:
A four-digit integer whose digits are all even may be written
\[
1000a + 100b + 10c + d,
\]
where each of \(a,b,c,d\) belongs to the set \(\{0,2,4,6,8\}\) and \(a \ne 0\). There are therefore
\[
4 \times 5^3 = 500
\]
such integers.
An integer is divisible by 4 if and only if its last two digits form a number divisible by 4. The admissible pairs \((c,d)\) are therefore exactly the 13 pairs
\[
(0,0),(0,4),(0,8),(2,0),(2,4),(2,8),(4,0),(4,4),(4,8),(6,0),(6,4),(8,0),(8,8)
\]
that satisfy the divisibility test. For each of the 13 choices of \((c,d)\) the first two digits may be chosen freely (\(a\) has 4 possibilities, \(b\) has 5), giving
\[
13 \times 4 \times 5 = 260
\]
favorable integers. The required fraction is therefore
\[
\frac{260}{500} = \frac{13}{25} = \frac{2}{5}.
\]
ANSWER 3: B
Problem 4:
A 6.5 percent tax on a $20 item is \(0.065 \times 20 = 1.30\). A 6 percent tax is \(0.06 \times 20 = 1.20\). Their difference is
\[
1.30 - 1.20 = 0.10.
\]
ANSWER 4: B
Problem 5:
The first sum is an arithmetic series with 93 terms whose average is
\[
\frac{1901 + 1993}{2} = 1947,
\]
so its value equals \(93 \times 1947 = 181071\). The second sum is an arithmetic series with 93 terms whose average is
\[
\frac{101 + 193}{2} = 147,
\]
so its value equals \(93 \times 147 = 13671\). Subtracting yields
\[
181071 - 13671 = 167400.
\]
ANSWER 5: A
Problem 6:
The given expression consists of 1996 terms whose signs follow the repeating four-term pattern
\[
+,-,-,+,+,-,-,+,\dots
\]
(The pattern begins with a single “+” and thereafter repeats every four signs.) In each complete block of four consecutive integers the signed sum is zero. There are exactly 499 such blocks, accounting for 1996 terms and contributing a total of zero. Hence the whole expression evaluates to 0.
ANSWER 6: C
Problem 7:
Let \(S\) be the set of people wearing sunglasses and \(C\) the set wearing caps. Then
\[
|S \cap C| = \frac{2}{5}|C| = \frac{2}{5} \cdot 35 = 14.
\]
Consequently
\[
|S| = |S \setminus C| + |S \cap C|
\]
is unknown, but the conditional probability required by the problem is
\[
P(C \mid S) = \frac{|S \cap C|}{|S|} = \frac{14}{|S|}.
\]
Because \(|S \cap C| = 14\) and \(|C| = 35\), the only numerical value that can be obtained from the given data is the ratio
\[
\frac{14}{35} = \frac{2}{5}.
\]
(The actual size of \(S\) is never needed.)
ANSWER 7: C
Problem 8:
A three-digit number \(100a + 10b + c\) (with \(a \in \{1,\dots,9\}\), \(b,c \in \{0,\dots,9\}\)) is even precisely when \(c\) is even. The digit-sum equation \(a + b + c = 26\) together with the upper bound 9 on each variable forces \(a \ge 8\). The only possibilities are therefore the six triples
\[
(9,9,8),\ (9,8,9),\ (8,9,9)
\]
and their permutations that keep the units digit even. Exactly three of them have even units digit:
\[
998,\ 986,\ 968.
\]
Thus three even three-digit numbers have digit-sum 26.
ANSWER 8: C
Problem 9:
A three-digit number \(100a + 10b + c\) satisfies \(a + b + c = 25\) with each digit at most 9. Then necessarily \(a \ge 7\). The admissible triples \((a,b,c)\) are exactly the six permutations of
\[
(9,9,7),\ (9,8,8).
\]
Hence six such numbers exist.
ANSWER 9: C
Problem 10:

## Step 1: Problem 1
To find the number of possible values of the median of set R, we first need to understand what a median is. The median of a set of numbers is the middle number when the numbers are arranged in ascending order. Since R is a set of nine distinct integers, the median will be the fifth number when the numbers are arranged in ascending order.

## Step 2: Problem 1
Given that six of the elements are 2, 3, 4, 6, 9, and 14, we can start by arranging these numbers in ascending order: 2, 3, 4, 6, 9, 14. The remaining three numbers can be any distinct integers.

## Step 3: Problem 1
To determine the possible medians, we need to consider the range of values the fifth number can take. The numbers already given are 2, 3, 4, 6, 9, 14. So, the three additional numbers can either be less than 4, between 4 and 9, or greater than 9.

## Step 4: Problem 1
If all three additional numbers are less than 4, the order becomes: additional number, additional number, additional number, 4, 6, 9, 14. The median is 6.
If two additional numbers are less than 4 and one is between 4 and 9, the order becomes: additional number, additional number, 4, additional number, 6, 9, 14. The median is between 4 and 6, so it could be 5.
If one additional number is less than 4 and two are between 4 and 9, the order becomes: additional number, 4, additional number, additional number, 6, 9, 14. The median could be between 4 and 6.
If all three additional numbers are between 4 and 9, the order becomes: 2, 3, 4, additional number, additional number, additional number, 6, 9, 14. The median is one of the additional numbers between 4 and 6, or 6 if the additional numbers are greater than 6.
If the additional numbers are greater than 9, the order remains 2, 3, 4, 6, 9, additional number, additional number, additional number, 14. The median is 9.

## Step 5: Problem 1
Considering the cases:
- When the additional numbers are less than 4, the median is 6.
- When the numbers are between 4 and 9, the median can be 5, 6, 7, or 8 (since the additional numbers can be distinct and between 4 and 9).
- When the numbers are greater than 9, the median is 9.
So, possible medians are 5, 6, 7, 8, and 9.

## Step 6: Problem 1
Thus, there are 5 possible values for the median: 5, 6, 7, 8, 9.

ANSWER 1: B

## Step 7: Problem 2
To find N, we first need to understand how the mean and median are affected by adding N to the list 2, 6, 7, 7, 28.

## Step 8: Problem 2
The original list is 2, 6, 7, 7, 28. The median is 7. When N is added, the list becomes 2, 6, 7, 7, N, 28. The median will depend on N. If N is between 7 and 7 (which is not possible since N is distinct), or less than 7, the median will be 7. If N is greater than 7, the median will be the average of 7 and N.

## Step 9: Problem 2
Let's consider the case when N is less than or equal to 7. The median remains 7. The mean is (2 + 6 + 7 + 7 + 28 + N)/6 = (50 + N)/6. According to the problem, the mean is twice the median, so (50 + N)/6 = 2*7 = 14.

## Step 10: Problem 2
Solving for N: 50 + N = 14*6 = 84, N = 84 - 50 = 34.

## Step 11: Problem 2
Let's check if N = 34 satisfies the condition for the median. Since 34 is greater than 7, the list becomes 2, 6, 7, 7, 28, 34. The median is (7 + 7)/2 = 7. The mean is (2 + 6 + 7 + 7 + 28 + 34)/6 = 84/6 = 14, which is indeed twice the median.

## Step 12: Problem 2
Thus, N = 34 is a valid solution.

ANSWER 2: E

## Step 13: Problem 3
To find the fraction of four-digit integers with all even digits that are divisible by 4, we first need to determine the total number of such four-digit integers.

## Step 14: Problem 3
For a four-digit number, the first digit can be 2, 4, 6, or 8 (4 choices), and the rest can be 0, 2, 4, 6, or 8 (5 choices each). So, the total number of four-digit numbers with all even digits is 4 * 5 * 5 * 5 = 500.

## Step 15: Problem 3
A number is divisible by 4 if its last two digits form a number divisible by 4. The last two digits can be 00, 04, 08, 20, 24, 28, 40, 44, 48, 60, 64, 68, 80, 84, 88. Out of these, the ones with even digits are 00, 04, 08, 20, 24, 28, 40, 44, 48, 60, 64, 68, 80, 84, 88, which are 20, 24, 28, 40, 44, 48, 60, 64, 68, 80, 84, 88 (12 options since 00, 04, 08 are not valid for the last two digits when considering all possible combinations for the first two digits).

## Step 16: Problem 3
However, we need to correct the understanding of the last two digits. For a number to be divisible by 4, the last two digits must form a number divisible by 4. The possible last two digits are: 00, 04, 08, 20, 24, 28, 40, 44, 48, 60, 64, 68, 80, 84, 88. There are 5 choices for the third digit and 4 choices for the first digit. So, for each valid last two digits, there are 4 * 5 = 20 numbers.

## Step 17: Problem 3
There are 15 valid combinations for the last two digits (00, 04, 08, 20, 24, 28, 40, 44, 48, 60, 64, 68, 80, 84, 88), but we should count them correctly: there are indeed 15 options (since all are valid for the condition of being even and divisible by 4 when considering the last two digits).

## Step 18: Problem 3
The total number of four-digit numbers with all even digits and divisible by 4 is 4 * 5 * 15 = 300 (since there are 15 valid last two digits).

## Step 19: Problem 3
The fraction of these integers divisible by 4 is 300 / 500 = 3/5.

ANSWER 3: D

## Step 20: Problem 4
To find the difference between a 6.5% sales tax and a 6% sales tax on an item priced at $20, we calculate both taxes and then find the difference.

## Step 21: Problem 4
6.5% of $20 = 0.065 * $20 = $1.30.
6% of $20 = 0.06 * $20 = $1.20.

## Step 22: Problem 4
The difference is $1.30 - $1.20 = $0.10.

ANSWER 4: B

## Step 23: Problem 5
To solve (1901 + 1902 + ... + 1993) - (101 + 102 + ... + 193), we can simplify it by calculating the sum of the differences between corresponding terms.

## Step 24: Problem 5
The series can be simplified as: (1901-101) + (1902-102) + ... + (1993-193).

## Step 25: Problem 5
This simplifies to: 1800 + 1800 + 

**Problem 1:**
We have a set \(R\) of nine distinct integers, with six given: 2, 3, 4, 6, 9, 14. The remaining three integers can be any integers not in this list. The median of nine numbers is the 5th smallest. Let the unknown numbers be \(x, y, z\). The six given numbers are fixed; we can place \(x, y, z\) anywhere in the sorted order to change the 5th element.

Let \(k\) be the number of given numbers less than the candidate median \(m\).
- If \(m\) is one of the given numbers, we need to place exactly \(4-k\) of the new numbers below \(m\) and the rest above. This requires \(0 \le 4-k \le 3\), so \(1 \le k \le 4\). Among the given numbers: \(3\ (k=1),\ 4\ (k=2),\ 6\ (k=3),\ 9\ (k=4)\) satisfy this. (\(2\) has \(k=0\), \(14\) has \(k=5\); neither works.) So \(3, 4, 6, 9\) are possible medians.
- If \(m\) is a new integer, one new number equals \(m\) and the other two are placed above or below. We need \(4-k\) new numbers below \(m\), so \(0 \le 4-k \le 2\), or \(2 \le k \le 4\). Also \(m\) must be an integer not already in the list.
  * \(k=2\): \(m\) between 3 and 4 — no integer.
  * \(k=3\): \(m\) between 4 and 6 — only \(5\).
  * \(k=4\): \(m\) between 6 and 9 — \(7\) and \(8\).
So new possible medians: \(5, 7, 8\).

Total possible medians: \(3, 4, 5, 6, 7, 8, 9\) — seven values.

ANSWER 1: D

**Problem 2:**
Original list: \(2, 6, 7, 7, 28\) (sum = 50). Insert \(N\) to get six numbers. Mean = \(\frac{50+N}{6}\). Median of six sorted numbers is the average of the 3rd and 4th.

Case 1: \(N \le 6\). Sorted: …, 3rd = 6, 4th = 7, median = 6.5.  
\(\frac{50+N}{6} = 2 \times 6.5 = 13 \Rightarrow 50+N = 78 \Rightarrow N=28\), but \(28 > 6\). Contradiction.

Case 2: \(6 < N \le 7\) (only integer is 7). Sorted: \(2, 6, 7, 7, 7, 28\), median = 7.  
\(\frac{57}{6} = 9.5 \neq 14\). No.

Case 3: \(7 < N \le 28\). Sorted: \(2, 6, 7, 7, N, 28\), median = 7.  
\(\frac{50+N}{6} = 14 \Rightarrow N = 34\), but \(34 > 28\). Contradiction.

Case 4: \(N > 28\). Sorted: \(2, 6, 7, 7, 28, N\), median = 7.  
\(\frac{50+N}{6} = 14 \Rightarrow N = 34\). Works.

ANSWER 2: E

**Problem 3:**
Four-digit positive integers with all even digits: thousands digit from \(\{2,4,6,8\}\) (4 choices); hundreds, tens, units from \(\{0,2,4,6,8\}\) (5 choices each). Total = \(4 \times 5^3 = 500\).
Divisible by 4 means last two digits form a multiple of 4. For any even tens digit, there are exactly 3 even units digits that make it divisible by 4 (e.g., if tens = 0,4,8 → units 0,4,8; if tens = 2,6 → units 0,4,8). So \(5 \times 3 = 15\) valid last-two-digit combinations.
Number of divisible numbers = \(4 \times 5 \times 15 = 300\).
Fraction = \(\frac{300}{500} = \frac{3}{5}\).

ANSWER 3: D

**Problem 4:**
Tax difference = \(6.5\% - 6\% = 0.5\%\) of \$20 = \(0.005 \times 20 = \$0.10\).

ANSWER 4: B

**Problem 5:**
First sum: \(1901\) to \(1993\) inclusive. Number of terms = \(1993 - 1901 + 1 = 93\).  
Sum = \(93 \times \frac{1901+1993}{2} = 93 \times 1947\).
Second sum: \(101\) to \(193\) inclusive. Terms = \(193 - 101 + 1 = 93\).  
Sum = \(93 \times \frac{101+193}{2} = 93 \times 147\).
Difference = \(93 \times (1947 - 147) = 93 \times 1800 = 167\,400\).

ANSWER 5: A

**Problem 6:**
The expression is \(1 - 2 - 3 + 4 + 5 - 6 - 7 + 8 + 9 - 10 - 11 + \cdots + 1992 + 1993 - 1994 - 1995 + 1996\).
Sign pattern: \(+\) for terms \(n \equiv 0\) or \(1 \pmod 4\); \(-\) for \(n \equiv 2\) or \(3 \pmod 4\).
Group in blocks of four: \((1-2-3+4) + (5-6-7+8) + \cdots + (1993-1994-1995+1996)\). Each block sums to \(0\). There are \(1996/4 = 499\) blocks. Total sum = \(0\).

ANSWER 6: C

**Problem 7:**
Let \(S\) be sunglasses wearers (\(|S|=50\)), \(C\) be cap wearers (\(|C|=35\)). Given \(P(S \mid C) = \frac{|S \cap C|}{35} = \frac{2}{5} \Rightarrow |S \cap C| = 14\).
Then \(P(C \mid S) = \frac{14}{50} = \frac{7}{25}\).

ANSWER 7: B

**Problem 8:**
Three-digit numbers with digits summing to 26. Maximum sum is 27, so digits are close to 9. Let digits be \(a,b,c\) with \(a+b+c=26\), \(1 \le a \le 9\), \(0 \le b,c \le 9\). Setting \(x=9-a, y=9-b, z=9-c\) gives \(x+y+z=1\) with \(x \le 8, y,z \le 9\). The only solutions are permutations of \((1,0,0)\): one digit is 8, the other two are 9. The three numbers are 998, 989, 899.
Even numbers end in an even digit: only 998. Hence 1 even number.

ANSWER 8: A

**Problem 9:**
Three-digit numbers, digit sum = 25. As above, \(x+y+z = 27-25 = 2\). Nonnegative solutions: permutations of \((2,0,0)\) – digits 7,9,9; and \((1,1,0)\) – digits 8,8,9.  
For 7,9,9: 3 permutations. For 8,8,9: 3 permutations. Total 6 numbers.

ANSWER 9: C

**Problem 10:**
Exits 5 through 26: that’s \(26-5+1 = 22\) exits, so 21 gaps. Sum of gaps = 118 miles. Each gap \(\ge 5\). To maximize one gap, minimize the other 20: \(20 \times 5 = 100\) miles. Largest gap = \(118 - 100 = 18\) miles.

ANSWER 10: C

**Problem 11:**
Five scores: \(a \le b \le c \le d \le e\). Mean = 90 ⇒ sum = 450. Median \(c = 91\). Mode = 94, so at least two scores are 94. Since median is 91, the two 94s must be \(d\) and \(e\). Then \(a+b+91+94+94 = 450 \Rightarrow a+b = 450 - 279 = 171\).

ANSWER 11: B

**Problem 12:**
Blanks: \(5 \_ 4 \_ 6 \_ 3\). Use \(+,\ -,\ \times\) exactly once. Evaluate with standard order of operations. All 6 permutations:
1) \(\times, +, -:\ 5\times4+6-3 = 20+6-3=23\)
2) \(\times, -, +:\ 5\times4-6+3 = 20-6+3=17\)
3) \(+, \times, -:\ 5+4\times6-3 = 5+24-3=26\)
4) \(+, -, \times:\ 5+4-6\times3 = 5+4-18=-9\)
5) \(-, \times, +:\ 5-4\times6+3 = 5-24+3=-16\)
6) \(-, +, \times:\ 5-4+6\times3 = 5-4+18=19\)
Among choices, 19 appears.

ANSWER 12: E

Problem 1:
We have a set R of 9 distinct integers. Six are known: 2, 3, 4, 6, 9, 14. We add 3 unknown distinct integers (not equal to any known element). The median of 9 elements is the 5th smallest.

Let me determine which values can occupy the 5th position. The known elements in sorted order are: 2, 3, 4, 6, 9, 14.

**Can the median be 3?** Place all 3 unknowns below 3 (e.g., −1, 0, 1). Sorted: −1, 0, 1, 2, **3**, 4, 6, 9, 14. 5th = 3. ✓
**Can the median be 4?** Place 2 unknowns below 4 (e.g., 0, 1) and 1 above. Sorted: 0, 1, 2, 3, **4**, 6, 9, 14, 20. 5th = 4. ✓
**Can the median be 5?** (5 is an unknown.) Known elements below 5: {2, 3, 4} = 3 elements. Add 1 unknown below 5 (e.g., 1), make 5 an unknown, and 1 unknown above. Sorted: 1, 2, 3, 4, **5**, 6, 9, 14, 20. 5th = 5. ✓
**Can the median be 6?** Known below 6: {2, 3, 4} = 3. Add 1 unknown below 6 (e.g., 5), 2 above. Sorted: 2, 3, 4, 5, **6**, 9, 14, 20, 21. 5th = 6. ✓
**Can the median be 7?** Known below 7: {2, 3, 4, 6} = 4. Need 0 unknowns below 7. Make 7 an unknown, 2 above. Sorted: 2, 3, 4, 6, **7**, 9, 14, 20, 21. 5th = 7. ✓
**Can the median be 8?** Known below 8: {2, 3, 4, 6} = 4. Same logic: 2, 3, 4, 6, **8**, 9, 14, 20, 21. 5th = 8. ✓
**Can the median be 9?** Known below 9: {2, 3, 4, 6} = 4. Place all 3 unknowns above 9. Sorted: 2, 3, 4, 6, **9**, 14, 20, 21, 22. 5th = 9. ✓

**Can the median be ≤ 2?** Need 4 elements below it, but at most 3 unknowns exist → impossible.
**Can the median be ≥ 10?** Known elements below 10 are {2, 3, 4, 6, 9} = 5 elements, so the 5th position is already filled by something ≤ 9 → impossible.

Possible medians: {3, 4, 5, 6, 7, 8, 9} → **7 values**.

ANSWER 1: D

---

Problem 2:
Original list: 2, 6, 7, 7, 28. Sum = 50. Insert N to get 6 elements.
New mean = (50 + N)/6. Median of 6 elements = average of 3rd and 4th.

**Case N > 7:** Sorted order: 2, 6, 7, 7, N, 28 (or N, 28 swapped). The 3rd and 4th elements are 7 and 7.
Median = (7 + 7)/2 = 7.
Mean = 2 × median = 14.
(50 + N)/6 = 14 → 50 + N = 84 → N = 34.
Check: 34 > 7 ✓. List: 2, 6, 7, 7, 28, 34. Median = 7, Mean = 84/6 = 14 = 2(7). ✓

**Case N ≤ 6:** Median = (6+7)/2 = 6.5. Mean = 13. N = 78 − 50 = 28. But 28 > 6, contradiction.
**Case 6 < N ≤ 7:** Median = (N+7)/2. (50+N)/6 = N+7 → 50+N = 6N+42 → 5N = 8 → N = 1.6. Contradiction.

Only N = 34 works.

ANSWER 2: E

---

Problem 3:
Even digits: {0, 2, 4, 6, 8}. Four-digit numbers with all even digits:
- First digit: {2, 4, 6, 8} → 4 choices
- Other three digits: 5 choices each
- Total = 4 × 5 × 5 × 5 = 500

Divisibility by 4 depends on the last two digits. I check all 5 × 5 = 25 pairs of even digits:
For each tens digit (0, 2, 4, 6, 8), the units digits giving divisibility by 4 are:
- Tens 0: 00, 04, 08 (3 of 5)
- Tens 2: 20, 24, 28 (3 of 5)
- Tens 4: 40, 44, 48 (3 of 5)
- Tens 6: 60, 64, 68 (3 of 5)
- Tens 8: 80, 84, 88 (3 of 5)

Valid last-two-digit pairs = 15 out of 25.

Numbers divisible by 4 = 4 × 5 × 15 = 300.
Fraction = 300/500 = **3/5**.

ANSWER 3: D

---

Problem 4:
Difference = 6.5% of $20 − 6% of $20 = (6.5% − 6%) × $20 = 0.5% × $20.
0.5% = 0.005, so 0.005 × 20 = $0.10.

ANSWER 4: B

---

Problem 5:
Pair corresponding terms:
(1901 − 101) + (1902 − 102) + … + (1993 − 193)

Each pair equals 1800.
Number of terms: 1993 − 1901 + 1 = 93 terms (and 193 − 101 + 1 = 93 terms).
Total = 93 × 1800 = **167,400**.

ANSWER 5: A

---

Problem 6:
The sign pattern is: +, −, −, +, +, −, −, +, …
Group into blocks of 4:
(1 − 2 − 3 + 4) + (5 − 6 − 7 + 8) + (9 − 10 − 11 + 12) + …

Each block: n − (n+1) − (n+2) + (n+3) = n − n − 1 − n − 2 + n + 3 = **0**.

Total terms: 1996. Number of blocks: 1996/4 = 499 complete blocks.
Sum = 499 × 0 = **0**.

ANSWER 6: C

---

Problem 7:
Let B = number wearing both sunglasses and caps.
P(sunglasses | cap) = B / (people wearing caps) = B / 35 = 2/5.
B = 35 × (2/5) = 14.

Now find P(cap | sunglasses) = B / (people wearing sunglasses) = 14 / 50 = **7/25**.

ANSWER 7: B

---

Problem 8:
Three-digit numbers with digit sum 26. Maximum digit sum = 9+9+9 = 27, so we must subtract 1 from one digit of 999:
Possible digit sets: (8,9,9), (9,8,9), (9,9,8) → numbers 899, 989, 998 (and permutations: 899, 989, 997... wait).

Actually the numbers are: 899, 989, 998 — no, let me list properly:
- (8,9,9) → 899, 989, 998... wait, (8,9,9) gives 899, 989, 998? No: digits are 8,9,9 so numbers are 899, 989, 998.
Wait: (8,9,9) → 899, 989, 998. But 998 has digits 9,9,8 which is the set (9,9,8). Let me just list all:
- Digits {8,9,9}: 899, 989, 998 — wait, 998 has digits 9,9,8. That IS {8,9,9}. So: 899, 989, 998.

Hmm, let me redo. The three numbers with digit sum 26 are:
- 899 (8+9+9=26)
- 989 (9+8+9=26)
- 998 (9+9+8=26)

Wait, that's only 3 numbers total (since two digits are the same, there are 3!/2! = 3 arrangements).

Even numbers must end in an even digit. Only 998 ends in 8 (even). 899 and 989 end in 9 (odd).

So exactly **1** even number.

ANSWER 8: A

---

Problem 9:
Three-digit numbers a, b, c with a + b + c = 25, where 1 ≤ a ≤ 9, 0 ≤ b, c ≤ 9.

Enumerate by first digit a:
- a = 9: b + c = 16 → (7,9), (8,8), (9,7) → 979, 988, 997 (3 numbers)
- a = 8: b + c = 17 → (8,9), (9,8) → 889, 898 (2 numbers)
- a = 7: b + c = 18 → (9,9) → 799 (1 number)
- a ≤ 6: b + c ≥ 19, but max b + c = 18 → impossible

Total = 3 + 2 + 1 = **6**.

ANSWER 9: C

---

Problem 10:
Exits 5 through 26 give 22 exits and 21 consecutive gaps. Total distance = 118 miles. Each gap ≥ 5 miles.

To maximize one gap, minimize the other 20 gaps to 5 miles each:
Max gap + 20(5) = 118
Max gap + 100 = 118
Max gap = **18** miles.

ANSWER 10: C

---

Problem 11:
Five scores sorted: a ≤ b ≤ c ≤ d ≤ e.
- Mean = 90 → sum = 450
- Median = 91 → c = 91
- Mode = 94 → 94 appears most frequently (at least twice)

Since a ≤ b ≤ c = 91, neither a nor b can be 94. So 94 must appear among d and e. For 94 to be the mode, we need d = e = 94 (appearing twice).

Sum: a + b + 91 + 94 + 94 = 450
a + b = 450 − 279 = **171**

(We can verify valid scores 

 Problem 1:
We are given a set \(R\) of nine distinct integers, six of which are \(2,3,4,6,9,14\). We must determine how many different values are possible for the median of \(R\).

Because there are nine distinct integers, the median is the fifth smallest element when the set is listed in increasing order. Let the three unknown distinct integers be chosen freely (as long as they are not equal to any of the six given numbers). We examine two cases: when the median is one of the six known numbers and when it is one of the three unknown numbers.

**Case 1: The median is a known number.**
Let the median be \(m\). Of the six known numbers, let \(L\) be how many are less than \(m\). Then there are \(5-L\) known numbers greater than \(m\). To make \(m\) the fifth element, we need exactly four numbers smaller than \(m\) and four larger. Thus we need \(4-L\) unknown numbers below \(m\) and \(4-(5-L)=L-1\) unknown numbers above \(m\). This uses exactly \((4-L)+(L-1)=3\) unknowns, which is perfect. We need \(4-L\ge 0\) and \(L-1\ge 0\), i.e. \(1\le L\le 4\).

- \(m=2\): \(L=0\) → impossible.
- \(m=3\): \(L=1\) → need 3 unknowns \(<3\), 0 above. Possible (e.g. \(-1,0,1\)).
- \(m=4\): \(L=2\) → need 2 unknowns \(<4\), 1 above. Possible (e.g. \(0,1,5\)).
- \(m=6\): \(L=3\) → need 1 unknown \(<6\), 2 above. Possible (e.g. \(5,7,8\)).
- \(m=9\): \(L=4\) → need 0 unknowns \(<9\), 3 above. Possible (e.g. \(10,11,12\)).
- \(m=14\): \(L=5\) → need \(-1\) unknowns below; impossible.

So the known possible medians are \(3,4,6,9\).

**Case 2: The median is an unknown number \(x\).**
Then \(x\) itself uses one of the three unknown slots. We need four numbers below \(x\) and four above. With \(L\) known numbers below \(x\), we need \(4-L\) other unknowns below \(x\) and \(4-(6-L)=L-2\) other unknowns above \(x\). The total other unknowns used is \((4-L)+(L-2)=2\), which matches the two remaining unknowns. We need \(4-L\ge 0\) and \(L-2\ge 0\), so \(2\le L\le 4\).

- \(L=2\): known below are \(2,3\); known above are \(4,6,9,14\). Then \(3<x<4\), impossible for an integer.
- \(L=3\): known below are \(2,3,4\); known above are \(6,9,14\). Then \(4<x<6\), so \(x=5\). This works (e.g. add \(1,5,7\)).
- \(L=4\): known below are \(2,3,4,6\); known above are \(9,14\). Then \(6<x<9\), so \(x=7\) or \(x=8\). Both work (e.g. add \(7,10,11\) or \(8,10,11\)).

Thus the unknown possible medians are \(5,7,8\).

Collecting all possibilities: \(\{3,4,5,6,7,8,9\}\). That is **7** different values. The tempting answer of 4 counts only the known medians, and 2 or 14 are impossible because too many numbers would have to be forced to one side.

ANSWER 1: D

---

Problem 2:
We insert a number \(N\) into the list \(2,6,7,7,28\) so that the new mean of the six numbers is twice the new median. Find \(N\).

The original sum is \(2+6+7+7+28=50\). After insertion, the sum is \(50+N\) and the mean is \(\frac{50+N}{6}\).

The median depends on the size of \(N\) relative to the ordered list. Because the list contains six numbers, the median is the average of the third and fourth values.

- If \(N\le 6\) (as an integer), the ordered list begins with \(N\) or \(2\), then the third and fourth terms are \(6\) and \(7\), giving median \(\frac{6+7}{2}=6.5\).
  Setting mean \(=2\times 6.5=13\): \(\frac{50+N}{6}=13\Rightarrow N=28\). But \(N=28\) does not satisfy \(N\le 6\), so this case yields no valid solution.

- If \(N\ge 7\) (integer), the third and fourth terms are both \(7\) (since the two \(7\)’s occupy those middle slots, or \(N=7\) makes three \(7\)’s), so the median is \(7\).
  Setting mean \(=2\times 7=14\): \(\frac{50+N}{6}=14\Rightarrow 50+N=84\Rightarrow N=34\).

Check: the list becomes \(2,6,7,7,28,34\). The median is \(\frac{7+7}{2}=7\). The mean is \(\frac{84}{6}=14\), which is indeed twice the median. The other choices fail: \(N=28\) gives mean \(13\) and median \(7\); \(N=14\) gives mean \(\frac{64}{6}\neq 14\).

ANSWER 2: E

---

Problem 3:
We consider all positive four-digit integers whose digits are all even. We need the fraction of them that are divisible by \(4\).

Even digits are \(\{0,2,4,6,8\}\). For a four-digit number \(ABCD\):
- \(A\) (thousands) has 4 choices: \(\{2,4,6,8\}\).
- \(B,C,D\) each have 5 choices.

Total count: \(4\cdot 5\cdot 5\cdot 5 = 500\).

A number is divisible by \(4\) exactly when its last two digits form a multiple of \(4\). Here \(C\) and \(D\) are both even. Consider \(10C+D \pmod 4\). Since \(C\) is even, write \(C=2k\); then \(10C=20k\equiv 0\pmod 4\). Thus \(10C+D\equiv D\pmod 4\). So we simply need \(D\equiv 0\pmod 4\).

Among the even digits, those divisible by \(4\) are \(0,4,8\): **3 choices** for \(D\). The digits \(A,B,C\) are unrestricted (4, 5, and 5 choices).

Count of favorable numbers: \(4\cdot 5\cdot 5\cdot 3 = 300\).

The fraction is \(\frac{300}{500}=\frac{3}{5}\). One might mistakenly think the fraction is \(\frac12\) because half of all integers are divisible by \(4\), but the restriction to even digits changes the units-digit distribution.

ANSWER 3: D

---

Problem 4:
Find the difference between a \(6.5\%\) sales tax and a \(6\%\) sales tax on a \(\$20\) item.

Method 1: Compute each tax.
- \(6.5\%\) of \(\$20 = 0.065\times 20 = \$1.30\).
- \(6\%\) of \(\$20 = 0.06\times 20 = \$1.20\).

Difference: \(\$1.30-\$1.20 = \$0.10\).

Method 2: The difference in rates is \(0.5\%\), and \(0.5\%\) of \(\$20 = 0.005\times 20 = \$0.10\).

The distractors correspond to misplacing the decimal (e.g. \(\$1\) would be \(5\%\) of \(\$20\)).

ANSWER 4: B

---

Problem 5:
Evaluate \((1901+1902+\dots+1993)-(101+102+\dots+193)\).

Each sum has the same number of terms:
\(1993-1901+1 = 93\) terms, and \(193-101+1 = 93\) terms.

Pair the terms:
\((1901-101)+(1902-102)+\dots+(1993-193)\).

Each difference equals \(1800\). With 93 such pairs, the total is
\(93\times 1800 = 93\times 18\times 100 = 1674\times 100 = 167{,}400\).

A common error is miscounting the number of terms as 92, which would give \

Problem 1:
R is a set of nine distinct integers with six known elements: 2, 3, 4, 6, 9, 14. Since R has 9 elements, the median is the 5th element when sorted.

I need to determine which integers can be the 5th element. The median must have exactly 4 elements less than it and 4 elements greater than it.

**Checking each candidate:**

- **Median = 3**: Need 4 elements < 3. Known: {2}. Need 3 unknowns all < 3 (e.g., 1, 0, -1). Elements > 3: {4, 6, 9, 14} = 4. ✓
- **Median = 4**: Need 4 elements < 4. Known: {2, 3}. Need 2 unknowns < 4. Elements > 4: {6, 9, 14} + 1 unknown > 4. ✓
- **Median = 5**: Need 4 elements < 5. Known: {2, 3, 4}. Need 1 unknown < 5. Elements > 5: {6, 9, 14} + 1 unknown > 5. (5 is an unknown) ✓
- **Median = 6**: Need 4 elements < 6. Known: {2, 3, 4}. Need 1 unknown < 6. Elements > 6: {9, 14} + 2 unknowns > 6. ✓
- **Median = 7**: Need 4 elements < 7. Known: {2, 3, 4, 6} = 4. Elements > 7: {9, 14} + 2 unknowns > 7. (7 is an unknown) ✓
- **Median = 8**: Need 4 elements < 8. Known: {2, 3, 4, 6} = 4. Elements > 8: {9, 14} + 2 unknowns > 8. (8 is an unknown) ✓
- **Median = 9**: Need 4 elements < 9. Known: {2, 3, 4, 6} = 4. Elements > 9: {14} + 3 unknowns > 9. ✓

**Values that cannot be median:**
- 2: Need 4 elements < 2, but only 3 unknowns available. ✗
- 10-13: Already have 5 known elements less than these. ✗
- 14: Already have 5 known elements less than 14. ✗

Possible median values: {3, 4, 5, 6, 7, 8, 9} → **7 values**

ANSWER 1: D

---

Problem 2:
List: 2, 6, 7, 7, 28. Insert N so that mean = 2 × median.

Sum of original list = 50, so new mean = (50 + N)/6.

**Finding the median based on N:**
- If N > 7: Sorted list is 2, 6, 7, 7, 28, N (or 2, 6, 7, 7, N, 28 if 7 < N < 28). Median = (7+7)/2 = 7.
- If N ≤ 6: Median = (6+7)/2 = 6.5
- If 6 < N ≤ 7: Median = (N+7)/2

**Solving for N > 7:**
(50 + N)/6 = 2(7) = 14
50 + N = 84
N = 34

Verification: List = 2, 6, 7, 7, 28, 34. Mean = 84/6 = 14. Median = 7. ✓

ANSWER 2: E

---

Problem 3:
Four-digit integers with all even digits. Even digits: {0, 2, 4, 6, 8}.

**Total count:** First digit: 4 choices (2,4,6,8), others: 5 choices each → 4 × 5³ = 500

**Divisible by 4:** Last two digits must form a number divisible by 4.

Listing all 25 pairs of even last digits and checking divisibility by 4:
- 00✓, 02✗, 04✓, 06✗, 08✓
- 20✓, 22✗, 24✓, 26✗, 28✓
- 40✓, 42✗, 44✓, 46✗, 48✓
- 60✓, 62✗, 64✓, 66✗, 68✓
- 80✓, 82✗, 84✓, 86✗, 88✓

That's 15 valid pairs.

**Count divisible by 4:** 4 × 5 × 15 = 300

**Fraction:** 300/500 = 3/5

ANSWER 3: D

---

Problem 4:
6.5% tax on $20 = 0.065 × 20 = $1.30
6% tax on $20 = 0.06 × 20 = $1.20

Difference = $1.30 - $1.20 = $0.10

ANSWER 4: B

---

Problem 5:
(1901 + 1902 + ... + 1993) − (101 + 102 + ... + 193)

Both sums have 93 terms each.

First sum = 93 × (1901 + 1993)/2 = 93 × 1947
Second sum = 93 × (101 + 193)/2 = 93 × 147

Difference = 93 × (1947 − 147) = 93 × 1800 = 167,400

ANSWER 5: A

---

Problem 6:
1 − 2 − 3 + 4 + 5 − 6 − 7 + 8 + ... + 1993 − 1994 − 1995 + 1996

The pattern repeats every 4 terms: +, −, −, +

Group k: (4k−3) − (4k−2) − (4k−1) + 4k = 4k−3−4k+2−4k+1+4k = 0

Each group of 4 sums to 0.

1996 ÷ 4 = 499 complete groups.

Total = 499 × 0 = 0

ANSWER 6: C

---

Problem 7:
50 people wear sunglasses, 35 wear caps.

P(sunglasses | cap) = 2/5 means: (both)/(caps) = 2/5

Both/35 = 2/5 → Both = 14

P(cap | sunglasses) = (both)/(sunglasses) = 14/50 = 7/25

ANSWER 7: B

---

Problem 8:
3-digit numbers with digit-sum 26 that are even.

For number abc: a + b + c = 26, c is even, a ≥ 1, all digits ≤ 9.

- c = 8: a + b = 18 → only (9,9) → 998 ✓
- c = 6: a + b = 20 → impossible (max 9+9=18)
- c = 4, 2, 0: a + b ≥ 22 → impossible

Only 1 number: 998

ANSWER 8: A

---

Problem 9:
Three-digit numbers with digit-sum 25.

a + b + c = 25, a ≥ 1, all digits ≤ 9.

- a = 9: b + c = 16 → (7,9), (8,8), (9,7) → 3 numbers
- a = 8: b + c = 17 → (8,9), (9,8) → 2 numbers
- a = 7: b + c = 18 → (9,9) → 1 number
- a ≤ 6: b + c ≥ 19 → impossible

Total = 3 + 2 + 1 = 6

ANSWER 9: C

---

Problem 10:
Distance from exit 5 to exit 26 is 118 miles. There are 21 gaps between consecutive exits (5→6, 6→7, ..., 25→26).

Each gap ≥ 5 miles. To maximize one gap, minimize all others.

If 20 gaps are minimized at 5 miles each: 20 × 5 = 100 miles.

Maximum single gap = 118 − 100 = 18 miles.

ANSWER 10: C

---

Problem 11:
Five scores: mean = 90, median = 91, mode = 94.

Sum = 5 × 90 = 450

Sorted: a ≤ b ≤ c ≤ d ≤ e
- c = 91 (median)
- Mode = 94, so 94 appears at least twice. Since c = 91, the 94s must be d and e.

a + b + 91 + 94 + 94 = 450
a + b = 171

ANSWER 11: B

---

Problem 12:
Using +, −, × each exactly once in 5 __ 4 __ 6 __ 3:

| Operations | Result |
|------------|--------|
| 5 + 4 − 6 × 3 | 5 + 4 − 18 = −9 |
| 5 + 4 × 6 − 3 | 5 + 24 − 3 = 26 |
| 5 − 4 + 6 × 3 | 5 − 4 + 18 = 19 |
| 5 − 4 × 6 + 3 | 5 − 24 + 3 = −16 |
| 5 × 4 + 6 − 3 | 20 + 6 − 3 = 23 |
| 5 × 4 − 6 + 3 | 20 − 6 + 3 = 17 |

From choices {9, 10, 15, 16, 19}, only **19** is achievable.

ANSWER 12: E

**Problem 1:**  
We have nine distinct integers, six of which are known: 2, 3, 4, 6, 9, 14.  
The median of a nine‑element set is the 5th smallest number.  

Let the unknown integers be \(x,y,z\) (all distinct and distinct from the known six).  
When the numbers are sorted, the median is the 5th term.  

For a known number \(k_i\) (the \(i\)‑th smallest of the known six) to be the median, we need exactly four numbers smaller than it.  
If \(i-1\) known numbers are smaller, we need \(L\) unknown numbers smaller with  
\((i-1)+L=4\Rightarrow L=5-i\).  
Since we have only three unknowns, \(L\ge0\) gives \(i\le5\); also \(L\le3\) gives \(i\ge2\).  
Thus \(i=2,3,4,5\) → medians could be 3, 4, 6, 9.

Can the median be an unknown integer?  
Let the median be an unknown \(u\).  
If \(u\) lies between the known numbers, we need exactly four numbers below it.  
With two other unknowns, the possibilities are:
- \(u\) between 4 and 6 → \(u=5\);
- \(u\) between 6 and 9 → \(u=7\) or \(u=8\).

All three of these can be realized while keeping the numbers distinct.  
Thus the possible median values are \(\{3,4,5,6,7,8,9\}\), a total of **7** values.

**Answer:** 7 → Choice **D**.  

ANSWER 1: D  

---

**Problem 2:**  
We start with the list 2, 6, 7, 7, 28 and insert a number \(N\).  
Let the six numbers be sorted. The median of six numbers is the average of the 3rd and 4th smallest values.

Let the sorted list after insertion be \(a_1\le a_2\le a_3\le a_4\le a_5\le a_6\).  
The median is \(\dfrac{a_3+a_4}{2}\).  
The mean is \(\dfrac{\text{sum}}{6}\).  
Given that the mean equals twice the median:
\[
\frac{\text{sum}}{6}=2\cdot\frac{a_3+a_4}{2}=a_3+a_4 .
\]
Hence \(\text{sum}=6(a_3+a_4)\).

Consider where \(N\) can fall:

*If \(N\le2\):* \(a_3=6,\ a_4=7\) → median \(=6.5\).  
Sum \(=N+50\). Equation gives \(N=28\), contradicting \(N\le2\).

*If \(2<N\le6\):* \(a_3=6,\ a_4=7\) → same median \(=6.5\).  
Again \(N=28\), impossible.

*If \(6<N\le7\):* only integer possibility \(N=7\).  
List becomes 2, 6, 7, 7, 7, 28 → median \(=7\).  
Sum \(=57\) → mean \(=9.5\neq14\).

*If \(7<N<28\):* \(a_3=7,\ a_4=7\) → median \(=7\).  
Sum \(=N+50\). Equation \(N+50=84\) gives \(N=34\), which is not \(<28\).

*If \(N=28\):* median still \(=7\). Sum \(=78\) → mean \(=13\neq14\).

*If \(N>28\):* \(a_3=7,\ a_4=7\) → median \(=7\).  
Equation gives \(N+50=84\) → \(N=34\). This works (34 > 28).

Thus the only possible value is **34**.

**Answer:** 34 → Choice **E**.  

ANSWER 2: E  

---

**Problem 3:**  
A four‑digit integer with all digits even can use digits \{0,2,4,6,8\} for each place, except the thousands digit cannot be 0.  

Number of such integers: \(4\times5\times5\times5=500\).

A number is divisible by 4 iff its last two digits form a number divisible by 4.  
List all two‑digit endings with even digits (25 possibilities).  

Checking each yields 15 endings that are multiples of 4:  
00, 04, 08, 20, 24, 28, 40, 44, 48, 60, 64, 68, 80, 84, 88.

Thus the proportion is \(\frac{15}{25}=\frac{3}{5}\).

**Answer:** \(\frac{3}{5}\) → Choice **D**.  

ANSWER 3: D  

---

**Problem 4:**  
6 % tax on \$20 = \(0.06\times20 = \$1.20\).  
6.5 % tax on \$20 = \(0.065\times20 = \$1.30\).  

Difference = \$1.30 − \$1.20 = **\$0.10**.

**Answer:** \$0.10 → Choice **B**.  

ANSWER 4: B  

---

**Problem 5:**  
Sum from 1901 to 1993 (inclusive):  
Number of terms = \(1993-1901+1 = 93\).  
Sum = \(\frac{(1901+1993)\times93}{2}= \frac{3894\times93}{2}=181\,071\).

Sum from 101 to 193 (inclusive):  
Number of terms = \(193-101+1 = 93\).  
Sum = \(\frac{(101+193)\times93}{2}= \frac{294\times93}{2}=13\,671\).

Difference = \(181\,071-13\,671 = 167\,400\).

**Answer:** 167 400 → Choice **A**.  

ANSWER 5: A  

---

**Problem 6:**  
The sign pattern for the numbers \(n\) is:
- \(n\equiv1\pmod4\) → +,
- \(n\equiv2\pmod4\) → −,
- \(n\equiv3\pmod4\) → −,
- \(n\equiv0\pmod4\) → +.

Thus each block of four consecutive integers \(\{4k+1,4k+2,4k+3,4k+4\}\) contributes  
\((4k+1)-(4k+2)-(4k+3)+(4k+4)=0\).

From 1 through 1996 there are \(1996/4=499\) complete blocks, so the total sum is **0**.

**Answer:** 0 → Choice **C**.  

ANSWER 6: C  

---

**Problem 7:**  
Let \(S\) = people wearing sunglasses, \(|S|=50\).  
Let \(C\) = people wearing caps, \(|C|=35\).  
Let \(x=|S\cap C|\).

Given \(P(S\mid C)=\frac{x}{35}= \frac{2}{5}\).  
Thus \(x = \frac{2}{5}\times35 = 14\).

We need \(P(C\mid S)=\frac{x}{50}= \frac{14}{50}= \frac{7}{25}\).

**Answer:** \(\frac{7}{25}\) → Choice **B**.  

ANSWER 7: B  

---

**Problem 8:**  
A three‑digit number with digit sum 26 must use digits that sum to 26.  
The only triple of single‑digit numbers with sum 26 is \(\{9,9,8\}\).

The number is even only if its units digit is 8.  
Among the permutations of 9, 9, 8, only 998 ends in 8.  

Thus exactly **1** such number exists.

**Answer:** 1 → Choice **A**.  

ANSWER 8: A  

---

**Problem 9:**  
Let the digits be \(a,b,c\) (hundreds, tens, units) with  
\(1\le a\le9,\ 0\le b,c\le9,\ a+b+c=25\).

Set \(a'=a-1\) (so \(0\le a'\le8\)). Then  
\(a'+b+c=24\).

Count unrestricted non‑negative solutions: \(\binom{24+3-1}{2}= \binom{26}{2}=325\).

Subtract solutions where a digit exceeds its bound:
- \(a'\ge9\): treat \(a''=a'-9\) → \(a''+b+c=15\) → \(\binom{17}{2}=136\).
- \(b\ge10\): \(b'=b-10\) → \(a'+b'+c=14\) → \(\binom{16}{2}=120\).
- \(c\ge10\): similarly 120.

Intersections:
- \(a'\ge9,\ b\ge10\): \(a''+b'+c=5\) → \(\binom{7}{2}=21\).
- \(a'\ge9,\ c\ge10\): 21.
- \(b\ge10,\ c\ge10\): \(a'+b'+c'=4\) → \(\binom{6}{2}=15\).

No triple intersection (would require sum –5).

By inclusion–exclusion, invalid = \(136+120+120-21-21-15 = 319\).  
Valid = \(325-319 = 6\).

**Answer:** 6 → Choice **C**.  

ANSWER 9: C  

---

**Problem 10:**  
Exits numbered 5 through 26 give 22 exits, hence 21 intervals.  
Let the distances between consecutive exits be \(d_1,\dots,d_{21}\).  
We know \(\sum d_i = 118\) miles, and each \(d_i\ge5\).

To maximize the largest interval, make the othe

Problem 1:  
We are given a set $ R $ of nine distinct integers, six of which are $ 2, 3, 4, 6, 9, 14 $. The median of a set of nine numbers is the fifth number when sorted. To find the number of possible values for the median, we consider the possible placements of the three unknown numbers relative to the known ones. The known numbers sorted are $ 2, 3, 4, 6, 9, 14 $. The three unknown numbers can be placed in various positions to shift the median. The possible medians occur when the additional numbers are placed such that the fifth number in the sorted list can be $ 4, 5, 6, 7, 8, $ or $ 9 $. This gives six possible values for the median.  
ANSWER 1: D (corrected to C? Wait, no: the possible medians are 4, 5, 6, 7, 8, 9 — that’s 6 values, so C)  
**Wait, let's re-evaluate:** The known numbers are 2,3,4,6,9,14. Adding three numbers. The median is the 5th number. To get 4 as median: three numbers ≤4, but only 2,3,4 are available — but they are distinct, so we can add numbers like 1,5? Wait no, the three unknown numbers can be placed in different positions. For example:  
- If all three unknown numbers are ≤4: but only three such numbers exist (2,3,4) but they are already in the set. So we cannot add more. Thus, the unknown numbers must be placed in other positions.  
Actually, the unknown numbers can be:  
- All less than 2: median could be 4 if three numbers are added <4, but only 2,3,4 are already present. We cannot add numbers less than 2 without repeating? Wait, the set has distinct integers. So we can add numbers like 1,0,-1, but the problem doesn’t restrict to positive. However, the known numbers are positive. But the problem says "distinct integers", so negatives are allowed. But the median values: if we add numbers less than 2, the sorted order becomes [new numbers, 2,3,4,6,9,14]. With three new numbers, the sorted list has 9 numbers. The fifth is the median. If we add three numbers less than 2, the sorted list is [a,b,c,2,3,4,6,9,14] — no, wait, adding three numbers less than 2: the sorted list would be [a,b,c,2,3,4,6,9,14] — but a,b,c are less than 2, so the first three, then 2 is fourth, 3 is fifth? Wait, no: the original known numbers are 2,3,4,6,9,14. Adding three numbers, say x,y,z. The full sorted list will be the combination. To get median=4, we need the fifth number to be 4. So in the sorted list of 9, the fifth must be 4. That requires that four numbers are ≤4 and four are ≥4? No: the fifth is the middle. For the fifth to be 4, we need at least five numbers ≤4? No: in a sorted list of 9, the fifth is the median. For the fifth to be 4, we need at least five numbers ≤4? Actually, no: the fifth number is the one where four are less and four are greater? No: in a sorted list of 9, the fifth is the median, meaning four are ≤ median and four are ≥ median. But since numbers can be equal, but here all distinct. So for 4 to be the median, we need exactly four numbers less than 4 and four numbers greater than 4? But 4 is in the set. So the sorted list must have four numbers <4, then 4, then four numbers >4. But the known numbers include 2,3,4,6,9,14. So 2,3 are <4, and 6,9,14 are >4. So currently, without unknowns, we have two <4 and three >4. Adding three unknowns. To have four <4 and four >4, we need to add two numbers <4 and one number >4? But then the sorted list would be [two new <4, 2,3,4, then the new >4, then 6,9,14] — but that’s only three numbers after 4? Wait, no: the sorted list would be [new <4 (two), 2,3,4, new >4, 6,9,14] — that’s 9 numbers? 2+3+1+3=9? Wait, no: the two new <4, then 2,3,4 (three numbers), then the new >4, then 6,9,14 (three numbers). So total: 2+3+1+3=9. The fifth number is 4. So yes, we can achieve median=4 by adding two numbers <4 and one number >4. Similarly, we can achieve:  
- Median=5: by adding numbers such that the fifth number is 5. But 5 is not in the set. So we need the fifth number to be 5. This requires that four numbers are <5 and four are >5. But 5 is not present. So we need to add 5? But the set must consist of distinct integers, and 5 is not in the given. So we can add 5. Then the sorted list would include 5. For example, add 5, and two other numbers. To have four <5 and four >5: currently, without unknowns, numbers <5: 2,3,4 (three), numbers >5: 6,9,14 (three). So to have four <5, we need to add one number <5. To have four >5, we need to add one number >5. But we are adding three numbers. So we can add 5, and say 1 and 10. Then sorted: 1,2,3,4,5,6,9,10,14. Median is 5. So median=5 is achievable. Similarly, median=6: we can add numbers such that the fifth is 6. Currently, numbers <6: 2,3,4 (three), numbers >6: 9,14 (two). So to have four <6 and four >6, we need to add one number <6 and two numbers >6. For example, add 5,10,15. Then sorted: 2,3,4,5,6,9,10,14,15. Median=6. Similarly, median=7: add numbers so fifth is 7. Currently, <7: 2,3,4,6 (four), >7: 9,14 (two). So to have four <7 and four >7, we need to add zero <7 (since already four) and two >7. But we are adding three numbers. So we can add 7,8,15. Then sorted: 2,3,4,6,7,8,9,14,15. Median=7. Similarly, median=8: add 8, and two numbers, say 10,11. Then sorted: 2,3,4,6,8,9,10,11,14. Median=8? The fifth is 8. But to have four <8 and four >8: currently, without unknowns, <8: 2,3,4,6 (four), >8:9,14 (two). So to have four >8, we need to add two numbers >8. But we are adding three numbers. So we can add 8,9,10? But 9 is already there. Add 8,10,11. Then sorted: 2,3,4,6,8,9,10,11,14. The fifth is 8. So yes. Similarly, median=9: add numbers so fifth is 9. Currently, <9: 2,3,4,6 (four), >9:14 (one). So to have four >9, we need to add three numbers >9. But we are adding three numbers. So add 10,11,12. Then sorted: 2,3,4,6,9,10,11,12,14. Median=9. So possible medians: 4,5,6,7,8,9. That’s six values.  
ANSWER 1: C  

Problem 2:  
We are given the list 2,6,7,7,28. A number $ N $ is added. The mean becomes twice the median. The original list has five numbers. After adding $ N $, there are 

### Problem 1:
We have a set of 9 distinct integers, so the median (middle value when sorted) is the 5th element. The 6 known numbers sorted are [2,3,4,6,9,14], with 3 additional distinct integers to add. To find possible median values:
- The minimum possible median is 3: add 3 numbers smaller than 2, sorted set becomes [a,b,c,2,3,4,6,9,14], 5th term=3.
- The maximum possible median is 9: add 3 numbers larger than 14, sorted set becomes [2,3,4,6,9,14,x,y,z], 5th term=9.
- All integers between 3 and 9 are valid medians: 3,4,5,6,7,8,9, for 7 total possible values.
Tempting wrong choices: any count less than 7, from missing intermediate medians that can be formed by adding integers between the known values.
ANSWER 1: D

---

### Problem 2:
Original list sum: 2+6+7+7+28=50. Adding N gives a list of 6 elements, sum=50+N, mean=(50+N)/6. The median of 6 elements is the average of the 3rd and 4th sorted terms. The problem states mean=2×median, so 50+N=12×median.
- If N>28, sorted list is [2,6,7,7,28,N], median=(7+7)/2=7. Then 50+N=12×7=84 → N=34, which fits.
Testing other options confirms no other N satisfies the condition.
ANSWER 2: E

---

### Problem 3:
A number is divisible by 4 if its last two digits form a number divisible by 4. For 4-digit numbers with all even digits:
- Total count: 4 (non-zero thousands digits:2,4,6,8) ×5×5×5 (all even digits for other places)=500.
- Valid last two-digit pairs (both even, divisible by 4): For any even tens digit, 10×even is divisible by 4, so only units digits 0,4,8 work, giving 3 valid units digits per tens digit: 5×3=15 valid pairs.
- Fraction: (4×5×15)/500=300/500=3/5.
ANSWER 3: D

---

### Problem 4:
The tax rate difference is 6.5%-6%=0.5%. 0.5% of $20 is 0.005×20=$0.10.
ANSWER 4: B

---

### Problem 5:
Both sums have the same number of terms: 1993-1901+1=93 terms, 193-101+1=93 terms. Subtract term-by-term: (1901-101)+(1902-102)+...+(1993-193)=1800×93=167400.
ANSWER 5: A

---

### Problem 6:
Group terms in sets of 4: (1-2-3+4)+(5-6-7+8)+... each group sums to 0. There are 1996/4=499 groups, total sum=0.
ANSWER 6: C

---

### Problem 7:
Number of people with both caps and sunglasses: 35 cap-wearers × 2/5=14. Probability a sunglass-wearer has a cap:14/50=7/25.
ANSWER 7: B

---

### Problem 8:
An even 3-digit number with digit sum 26 must have an even units digit. The only possible even units digit is 8 (any smaller even digit leaves a sum of the first two digits >18, impossible). The only number is 998, so 1 valid number.
ANSWER 8: A

---

### Problem 9:
Three-digit numbers with digit sum 25 are all permutations of (9,9,7) and (9,8,8). There are 3 permutations of (9,9,7) and 3 permutations of (9,8,8), for 6 total numbers.
ANSWER 9: C

---

### Problem 10:
Between exit 5 and 26, there are 21 gaps. To maximize one gap, minimize the other 20 gaps at 5 miles each: 20×5=100 miles. Maximum gap=118-100=18 miles.
ANSWER 10: C

---

### Problem 11:
Total sum of 5 scores=5×90=450. Median=91, so sorted scores [a,b,91,c,d]. Mode=94, so 94 must appear at least twice, so c=d=94, sum c+d=188. Sum of the two lowest scores=450-91-188=171.
ANSWER 11: B

---

### Problem 12:
Using +,-,* exactly once, one valid calculation: 5-4+6×3=1+18=19, which uses all three operations and matches option E.
ANSWER 12: E

### Problem 1:
We need to find the number of possible median values for a set of 9 distinct integers, 6 of which are 2, 3, 4, 6, 9, 14. The median of 9 numbers is the 5th element when sorted in ascending order. We add 3 distinct integers (not in the given set) and analyze possible medians:
1.  **Case 1: All 3 new numbers < 2**: Sorted list is [a,b,c,2,3,4,6,9,14], median = 3.
2.  **Case 2: 2 new numbers < 2, 1 new number ≥4**: Sorted list is [a,b,2,3,4,...], median =4.
3.  **Case 3: 1 new number <2**:
    - If we add 5 (the only integer between 4 and 6): Sorted list is [a,2,3,4,5,6,...], median=5.
    - If we do not add 5: Sorted list is [a,2,3,4,6,...], median=6.
4.  **Case 4: 0 new numbers <2 (all new numbers ≥5)**:
    - Add 5: Sorted list [2,3,4,5,6,...], median=6.
    - Add a number between 6 and9 (7 or 8): Sorted list [2,3,4,6,7,...] or [2,3,4,6,8,...], median=7 or 8.
    - Add all new numbers ≥9: Sorted list [2,3,4,6,9,...], median=9.
The only impossible medians are 2 (needs 5 numbers ≤2, only 4 max) and values >9 (existing numbers already have 5 elements ≤9). The possible medians are 3,4,5,6,7,8,9: 7 values.
ANSWER 1: D
---
### Problem 2:
Original list: [2,6,7,7,28], sum = 2+6+7+7+28=50. Insert N to make 6 elements. The median of 6 elements is the average of the 3rd and 4th sorted elements. The mean is $\frac{50+N}{6}$, which equals twice the median:
$$\frac{50+N}{6} = 2 \times \text{median} \implies 50 + N = 12 \times \text{median} \implies N = 12 \times \text{median} -50.$$
Test possible median values from the sorted list with N:
- If N ≤7: Sorted list has 3rd=7, 4th=7, median=7. Then $N=12*7 -50=84-50=34$, which is >7, valid.
- Other median values (e.g., 6.5) give N=28, which does not fit the required sorted position.
Verify N=34: Sorted list [2,6,7,7,28,34], median=(7+7)/2=7, mean=(50+34)/6=14, which is twice 7.
ANSWER 2: E
---
### Problem 3:
Positive 4-digit numbers with all even digits:
- Thousands place: 2,4,6,8 (4 choices, no 0).
- Hundreds, tens, units: 0,2,4,6,8 (5 choices each).
Total numbers: $4 \times 5 \times 5 \times5=500$.
A number is divisible by 4 iff its last two digits form a number divisible by 4. For even tens digits (0,2,4,6,8), a two-digit number $10a + b$ is divisible by 4 iff $b$ (units digit) is 0,4,8 (since $10a \equiv 0 \mod4$ for even a). So valid endings: 5 (tens choices) ×3 (units choices)=15.
Total valid numbers: $4 \times5 \times5 \times3=300$. Fraction: $\frac{300}{500}=\frac{3}{5}$.
ANSWER 3: D
---
### Problem 4:
Difference between 6.5% and 6% tax on $20:
$$0.065*20 - 0.06*20 = (0.065-0.06)*20=0.005*20=0.10.$$
ANSWER 4: B
---
### Problem 5:
Both sums have 93 terms (1993-1901+1=93, 193-101+1=93). Each term in the first sum is 1800 more than the corresponding term in the second sum:
Total difference = $93 \times 1800 = 167400$.
ANSWER 5: A
---
### Problem 6:
The sign pattern repeats every 4 terms: $+ - - +$, and each block of 4 sums to 0:
$$(1-2-3+4)+(5-6-7+8)+\dots+(1993-1994-1995+1996)=0.$$
1996 is divisible by 4, so there are 499 such blocks, total sum 0.
ANSWER 6: C
---
### Problem 7:
Let $S$ = sunglass wearers ($|S|=50$), $C$ = cap wearers ($|C|=35$). $P(S|C)=\frac{|S∩C|}{|C|}=\frac{2}{5}$, so $|S∩C|=\frac{2}{5}*35=14$.
The probability a sunglass wearer wears a cap is $P(C|S)=\frac{|S∩C|}{|S|}=\frac{14}{50}=\frac{7}{25}$.
ANSWER 7: B
---
### Problem 8:
3-digit numbers with digit sum 26: maximum digit sum is 27 (999), so 26 is 1 less than maximum, meaning digits are permutations of (9,9,8). The only even permutation is 998 (units digit 8). So 1 number.
ANSWER 8: A
---
### Problem 9:
3-digit numbers with digit sum 25: 25 is 2 less than 27, so digits are permutations of (9,9,7) and (9,8,8):
- Permutations of (9,9,7): 997,979,799 (3 numbers).
- Permutations of (9,8,8): 988,898,889 (3 numbers).
Total 6 numbers.
ANSWER 9: C
---
### Problem 10:
Between exit 5 and 26, there are $26-5=21$ gaps, total distance 118 miles. Each gap ≥5 miles. To maximize one gap, minimize the other 20: $20*5=100$, so maximum gap = $118-100=18$.
ANSWER 10: C
---
### Problem 11:
Five sorted scores: $a≤b≤c≤d≤e$. Median $c=91$, mean 90, so total sum $5*90=450$, so $a+b+d+e=450-91=359$. Mode is 94, so 94 appears at least twice. Since $c=91$, 94 can only be in $d$ and $e$, so $d=e=94$. Then $a+b=359-94-94=171$. This sum is fixed (valid examples exist, e.g., 81+90=171).
ANSWER 11: B
---
### Problem 12:
Permute +, -, × in 5 __4 __6 __3, using standard order of operations (multiply first):
- $5 -4 +6×3 = 5-4+18=19$, which is an option.
All other permutations give values not in the choices.
ANSWER 12: E

I'll solve each problem step by step.

---

**Problem 1:**

We have 6 elements: {2, 3, 4, 6, 9, 14}, and need to add 3 more distinct integers to make 9 total.

The median of 9 numbers is the 5th element when arranged in order.

Currently we have 6 numbers. By strategically placing 3 new numbers, we can shift which value becomes the median.

Let me consider what values can be the median (5th position):
- If we add 3 numbers all ≤ 2, the sorted list has our 6 original numbers in positions 4-9, so median is 4
- If we add numbers strategically between and around the original set, we can make the median be: 2, 3, 4, 6, 9, 14, or a new number
- If we add 3 numbers all ≥ 14, the sorted list has our 6 original numbers in positions 1-6, so median is 9

Testing systematically:
- Median = 2: add three numbers < 2 (e.g., -2, -1, 0) ✓
- Median = 3: add two < 2, one between 3 and 4 ✓
- Median = 4: add three < 2, or arranged appropriately ✓
- Median = 6: add one or two strategically ✓
- Median = 9: add three > 14, or arranged appropriately ✓
- Median = 14: add three > 14 ✓
- Median = a new value between existing numbers ✓

The possible medians are: 2, 3, 4, 6, 9, 14, and any new integer we insert. We can get 7 different values.

**ANSWER 1: D**

---

**Problem 2:**

Original list: 2, 6, 7, 7, 28. Sum = 50.

When N is inserted, we have 6 numbers. Let's find the median of 6 numbers (average of 3rd and 4th when sorted).

Sum of all 6 numbers = 50 + N
Mean = (50 + N)/6

We need: Mean = 2 × Median

Let me test by considering where N falls:
- If N ≤ 2: sorted list is N, 2, 6, 7, 7, 28. Median = (6+7)/2 = 6.5. Mean = 2×6.5 = 13, so 50+N = 78, N = 28. Contradiction.
- If 2 < N ≤ 6: sorted list is 2, N, 6, 7, 7, 28. Median = (6+7)/2 = 6.5. Mean = 13, so N = 28. Contradiction.
- If 6 < N ≤ 7: sorted list is 2, 6, N, 7, 7, 28. Median = (N+7)/2. Mean = 2×(N+7)/2 = N+7. So (50+N)/6 = N+7, giving 50+N = 6N+42, so 5N = 8, N = 1.6. Not in range.
- If 7 < N ≤ 28: sorted list is 2, 6, 7, N, 7, 28 or 2, 6, 7, 7, N, 28. Median = (7+N)/2. Mean = 2×(7+N)/2 = 7+N. So (50+N)/6 = 7+N, giving 50+N = 42+6N, so 5N = 8, N = 1.6. Not valid.
- If N > 28: sorted list is 2, 6, 7, 7, 28, N. Median = (7+7)/2 = 7. Mean = 14. So 50+N = 84, N = 34. ✓

**ANSWER 2: E**

---

**Problem 3:**

Four-digit numbers with all even digits: first digit ∈ {2,4,6,8}, others ∈ {0,2,4,6,8}.

Total count: 4 × 5 × 5 × 5 = 500

For divisibility by 4, the last two digits must form a number divisible by 4.

Last two digits from {00, 02, 04, 06, 08, 20, 22, 24, 26, 28, 40, 42, 44, 46, 48, 60, 62, 64, 66, 68, 80, 82, 84, 86, 88}

Divisible by 4: 00, 04, 08, 20, 24, 28, 40, 44, 48, 60, 64, 68, 80, 84, 88 (15 out of 25)

Count divisible by 4: 4 × 5 × 15 = 300

But wait, the second digit is 5 choices, third is which determines divisibility by 4.

Actually: 4 × 5 × (number of even two-digit combos divisible by 4) = 4 × 5 × 15 = 300

Hmm, let me recalculate. First digit: 4 choices. Second digit: 5 choices. Last two digits must be divisible by 4: check all pairs from {0,2,4,6,8}×{0,2,4,6,8} giving 15 valid pairs.

Count = 4 × 5 × 15 = 300. Fraction = 300/500 = 3/5.

**ANSWER 3: D**

---

**Problem 4:**

6.5% tax on $20: 0.065 × 20 = $1.30
6% tax on $20: 0.06 × 20 = $1.20
Difference: $1.30 - $1.20 = $0.10

**ANSWER 4: B**

---

**Problem 5:**

Sum from 1901 to 1993: This is 93 consecutive integers starting at 1901.
Sum = (1901 + 1993)/2 × 93 = 1947 × 93 = 181,071

Sum from 101 to 193: This is 93 consecutive integers starting at 101.
Sum = (101 + 193)/2 × 93 = 147 × 93 = 13,671

Difference = 181,071 - 13,671 = 167,400

**ANSWER 5: A**

---

**Problem 6:**

Pattern: 1 - 2 - 3 + 4 + 5 - 6 - 7 + 8 + 9 - 10 - 11 + ...

Group by 4s: (1-2-3+4) + (5-6-7+8) + (9-10-11+12) + ... + (1993-1994-1995+1996)

Each group of 4: (4k+1) - (4k+2) - (4k+3) + (4k+4) = 4k+1-4k-2-4k-3+4k+4 = 0

From 1 to 1996, we have 1996/4 = 499 complete groups, each summing to 0.

Total = 0

**ANSWER 6: C**

---

**Problem 7:**

Let x = people wearing both sunglasses and caps.

P(sunglasses | cap) = x/35 = 2/5, so x = 14.

P(cap | sunglasses) = 14/50 = 7/25.

**ANSWER 7: B**

---

**Problem 8:**

Three-digit even numbers with digit sum 26.

For an even number, last digit ∈ {0, 2, 4, 6, 8}.

If last digit = 0: first two digits sum to 26. Max is 9+9=18. Impossible.
If last digit = 2: first two digits sum to 24. Only 9+9=18 < 24. Impossible.
If last digit = 4: first two digits sum to 22. Impossible (max 18).
If last digit = 6: first two digits sum to 20. Impossible.
If last digit = 8: first two digits sum to 18. This requires 9+9=18. Number: 998. ✓

Only 1 number.

**ANSWER 8: A**

---

**Problem 9:**

Three-digit numbers abc where a+b+c=25, with a ∈ {1,...,9}, b,c ∈ {0,...,9}.

a+b+c = 25 means b+c = 25-a.

For valid solutions: 0 ≤ b+c ≤ 18, so 7 ≤ a ≤ 9.

- a=7: b+c=18. Solutions: (9,9) → 1 way
- a=8: b+c=17. Solutions: (8,9), (9,8) → 2 ways
- a=9: b+c=16. Solutions: (7,9), (8,8), (9,7) → 3 ways

Total = 1 + 2 + 3 = 6

**ANSWER 9: C**

---

**Problem 10:**

Exits from 5th to 26th = 26-5 = 21 segments between consecutive exits.

Each segment ≥ 5 miles. Total distance = 118 miles.

If 20 segments are 5 miles each: 20 × 5 = 100 miles.
Remaining for the largest segment: 118 - 100 = 18 miles.

**ANSWER 10: C**

---

**Problem 11:**

Five scores: mean = 90 (sum = 450), median = 91, mode = 94.

Arranged in order: a, b, c, d, e where c = 91 (median).

Mode = 94 means 94 appears at least twice. Since median is 91, we must have 94 appearing in {d, e}.

Case 1: d = e = 94. Then a + b + 91 + 94 + 94 = 450, so a + b = 171.

Check consistency: we need a ≤ b ≤ 91 and mode is 94 (appears twice). If a or b equals 94, mode wouldn't be unique. So a, b < 91 works. ✓

**ANSWER 11: B**

---

**Problem 12:**

We need to place +, −, × in three blanks: 5 __ 4 __ 6 __ 3

Trying combinations:
- 5 + 4 × 6 − 3 = 5 + 24 − 3 = 26
- 5 × 4 + 6 − 3 = 20 + 6 − 3 = 23
- 5 × 4 − 6 + 3 = 20 − 6 + 3 = 17
- 5 + 4 − 6 × 3 = 5 + 4 − 18 = −9
- 5 − 4 

Problem 1:

We are given a set of 9 distinct integers. Six of them are
\[
2,3,4,6,9,14.
\]
We want the number of possible values of the median of the 9 numbers.

For 9 numbers, the median is the 5th number when the list is ordered.

Now look at the six given numbers in sorted order:
\[
2,3,4,6,9,14.
\]
Since there are 3 other distinct integers, the median can be any number that ends up in the 5th position after inserting those 3 numbers.

Let’s see what values are possible:

- If we place 3 new numbers all very small, then the 5th number could be 4.
- It could also be 6, or 9, or even 14, depending on how many new numbers are inserted before each.
- It cannot be 2 or 3, because there would need to be at least 4 numbers less than or equal to it, but only 3 extra numbers are available and the given numbers already limit this.
- It cannot be any number not among the given six if the median must be one of the existing numbers? Let’s check carefully: since the 3 extra numbers are distinct integers and can be placed anywhere, the 5th number could indeed be one of the inserted numbers, but then it would have to lie between the 4th and 6th numbers. Since the six given numbers already occupy several possible positions, the median can also be a new integer. But because the set is of distinct integers and the question asks how many possible values, we should count all possible medians that can occur.

A cleaner way: the median of 9 distinct integers is the middle element. With 6 fixed numbers, the 3 added numbers can be arranged so that the median is any integer between the 4th smallest and 6th smallest of the fixed numbers, inclusive, but here the possible distinct medians are:
\[
4,5,6,7,8,9
\]
Wait — can all of these really happen? Let’s test.

- Median 4: choose 3 new numbers less than 4.
- Median 5: choose 3 new numbers so that 5 is between 4 and 6 and becomes 5th.
- Median 6: choose 3 new numbers with 6 as middle.
- Median 7, 8, 9 similarly.

So the possible medians are the integers from 4 through 9, which is 6 values.

However, we must match the provided choices. The intended count is the number of possible median values among the listed integers and inserted integers. Since the only restriction is distinct integers, every integer from 4 through 9 can occur.

That gives 6 possible values.

ANSWER 1: C

---

Problem 2:

We insert a number \(N\) into the list
\[
2,6,7,7,28.
\]
Then the mean becomes twice the median.

Let’s analyze the median after inserting \(N\) into 6 numbers total.

The original sorted list is:
\[
2,6,7,7,28.
\]
There are 5 numbers. After inserting one number, there are 6 numbers, so the median is the average of the 3rd and 4th numbers in the sorted list.

We try the choices.

If \(N=7\), the list becomes
\[
2,6,7,7,7,28.
\]
Median:
\[
\frac{7+7}{2}=7.
\]
Mean:
\[
\frac{2+6+7+7+7+28}{6}=\frac{57}{6}=9.5.
\]
Twice the median would be 14, not 9.5, so not 7.

Try \(N=14\):
\[
2,6,7,7,14,28.
\]
Median:
\[
\frac{7+7}{2}=7.
\]
Mean:
\[
\frac{2+6+7+7+14+28}{6}=\frac{64}{6}=\frac{32}{3},
\]
not 14.

Try \(N=20\):
\[
2,6,7,7,20,28.
\]
Median is still 7.
Mean:
\[
\frac{2+6+7+7+20+28}{6}=\frac{70}{6}=\frac{35}{3},
\]
not 14.

Try \(N=28\):
\[
2,6,7,7,28,28.
\]
Median is 7.
Mean:
\[
\frac{2+6+7+7+28+28}{6}=\frac{78}{6}=13,
\]
not 14.

Try \(N=34\):
\[
2,6,7,7,28,34.
\]
Median is 7.
Mean:
\[
\frac{2+6+7+7+28+34}{6}=\frac{84}{6}=14.
\]
Twice the median is
\[
2\cdot 7=14.
\]
This works.

So \(N=34\).

ANSWER 2: E

---

Problem 3:

We consider all positive four-digit integers whose digits are all even.

Even digits are
\[
0,2,4,6,8.
\]
A four-digit number cannot start with 0, so the thousands digit has 4 choices:
\[
2,4,6,8.
\]
Each of the other three digits has 5 choices, so the total number is
\[
4\cdot 5^3.
\]

Now determine how many are divisible by 4.

A number is divisible by 4 if its last two digits form a number divisible by 4.

So we only need to count ordered pairs of even digits \((t,u)\) from \(\{0,2,4,6,8\}\) such that \(10t+u\) is divisible by 4.

Check the possible last two digits:

- \(00\) yes
- \(02\) no
- \(04\) yes
- \(06\) no
- \(08\) yes

- \(20\) yes
- \(22\) no
- \(24\) yes
- \(26\) no
- \(28\) yes

- \(40\) yes
- \(42\) no
- \(44\) yes
- \(46\) no
- \(48\) yes

- \(60\) yes
- \(62\) no
- \(64\) yes
- \(66\) no
- \(68\) yes

- \(80\) yes
- \(82\) no
- \(84\) yes
- \(86\) no
- \(88\) yes

For each tens digit, 3 of the 5 possibilities work, so there are
\[
5\cdot 3=15
\]
valid last-two-digit pairs.

Since the first two digits can be anything allowed:
- thousands digit: 4 choices
- hundreds digit: 5 choices

Total divisible by 4:
\[
4\cdot 5\cdot 15
\]
Total numbers:
\[
4\cdot 5^3 = 4\cdot 125
\]
So the fraction is
\[
\frac{4\cdot 5\cdot 15}{4\cdot 5^3}=\frac{15}{25}=\frac35.
\]

ANSWER 3: D

---

Problem 4:

An item costs \$20 before tax.

Difference between 6.5% tax and 6% tax is
\[
0.5\% = 0.005.
\]
So the difference in tax is
\[
0.005 \times 20 = 0.1
\]
dollar, which is \(\$0.10\).

ANSWER 4: B

---

Problem 5:

We need to compute
\[
(1901+1902+\cdots+1993) - (101+102+\cdots+193).
\]

Notice each term in the second sum is exactly 1800 less than the corresponding term in the first sum:
\[
1901-101=1800,\quad 1902-102=1800,\ \ldots,\ 1993-193=1800.
\]
How many terms are there?

From 1901 to 1993 inclusive:
\[
1993-1901+1=93
\]
terms.

So the difference is
\[
93 \cdot 1800 = 167400.
\]

ANSWER 5: A

---

Problem 6:

We evaluate
\[
1-2-3+4+5-6-7+8+9-10-11+\cdots+1992+1993-1994-1995+1996.
\]

Look at the pattern in blocks of 4:
\[
(1-2-3+4),\ (5-6-7+8),\ (9-10-11+12),\ \ldots
\]
Each block of 4 consecutive integers has sum
\[
n-(n+1)-(n+2)+(n+3)=0
\]
when \(n\equiv 1 \pmod 4\).

So all full blocks cancel.

Now check the last block:
The pattern continues through \(1992+1993-1994-1995+1996\), which is not a full 4-term block. But we can group from 1 to 1996 in blocks of 4:
\[
(1-2-3+4)+(5-6-7+8)+\cdots+(1993-1994-1995+1996).
\]
Each block sums to 0.

Therefore the entire 

Problem 1:  
We have 9 distinct integers. Six of them are fixed: 2, 3, 4, 6, 9, 14. The other 3 are distinct integers of our choice.  
With 9 numbers, the median is the 5th number when sorted.

So the median depends only on how many of the fixed numbers fall below/at various positions, and where we insert the 3 new numbers.

Let’s analyze possible 5th elements.

### Step 1: Sort the fixed numbers
Fixed numbers in increasing order:
2, 3, 4, 6, 9, 14

The median (5th) could be one of these fixed numbers, or one of the new ones.

### Casework by choosing what the 5th number is

#### Can the median be 2?
For 2 to be the median, we need at least 4 numbers ≤ 2, but only 2 itself is ≤ 2 among the fixed set (and new numbers would have to be distinct integers ≤ 2).  
That means we’d need 4 numbers less than or equal to 2, including 2, so we’d need 3 additional distinct integers ≤ 2 (other than 2). That’s possible (e.g., 0, −1, 1).  
But then 2 would be the 5th, meaning exactly 4 numbers are <2. Can we guarantee? Actually, since we can choose any 3 extra integers, we can make exactly 4 numbers less than 2 by picking 3 integers <2 plus none from fixed besides 2. That would make 2 the 5th. However, note 2 is already one of the 9 distinct integers, so it can be the median.  
So potentially 2 is possible.

But the problem asks for the number of possible median values, and the choices are between 4 and 8, so likely only medians among the fixed list can work. Let’s check more carefully: the fixed numbers below 2 is only none (except 2). To make 2 the 5th, we need 4 numbers smaller than 2. We only control 3 extra numbers, so we can only create at most 3 numbers smaller than 2. Therefore, we cannot reach 4 smaller numbers.  
✅ Conclusion: median cannot be 2.

Similarly, we test each fixed candidate using “how many fixed numbers are below it”.

### General method
For a number \(x\) to be the median (5th), there must be exactly 4 numbers less than \(x\), and at least one number equal to \(x\). If \(x\) is one of the fixed numbers, we can count how many fixed numbers are less than \(x\), then see if the remaining “needed” smaller numbers can come from the 3 new integers.

Let’s compute fixed counts less than each candidate:

- Below 2: 0  
- Below 3: 1 (just 2)  
- Below 4: 2 (2,3)  
- Below 6: 3 (2,3,4)  
- Below 9: 4 (2,3,4,6)  
- Below 14: 5 (2,3,4,6,9)

We need exactly 4 numbers less than the median.

Now see possibilities:

#### Median = 3
Fixed numbers below 3: 1.  
We need 4 total below 3, so we must add 3 new numbers < 3. That’s possible.  
✅ So 3 can be median.

#### Median = 4
Fixed below 4: 2. Need 2 more from new numbers < 4. Possible.  
✅ 4 can be median.

#### Median = 6
Fixed below 6: 3. Need 1 more from new numbers < 6. Possible.  
✅ 6 can be median.

#### Median = 9
Fixed below 9: 4 already. Then all 4 smaller numbers are fixed, and the median can be 9 regardless of where new numbers go as long as none are between 9 and 14 in a way that changes ordering?  
We only need 9 to be the 5th, meaning at least 4 numbers are <9 (we have exactly 4 fixed ones). If we add any new numbers <9, that would make more than 4 numbers less than 9 and break it. So we must choose the 3 new integers all > 9. That’s possible.  
✅ 9 can be median.

#### Median = 14
Fixed below 14: 5 already. But we need exactly 4 below the median. Impossible.  
❌ 14 cannot be median.

#### Median = 2
Fixed below 2: 0, need 4 from new numbers <2, but we only have 3 new numbers. Impossible.  
❌ 2 cannot be median.

So the only fixed medians are: 3, 4, 6, 9 → 4 values.

Are there possible medians that are not among the fixed numbers (i.e., one of the 3 new integers)?  
Yes, if we choose all new integers appropriately.

Let the median be a new number \(y\). Then among the fixed numbers, some number of them must be less than \(y\). Since there are only 6 fixed numbers, the possible counts of fixed elements less than \(y\) range from 0 to 6.

But we need exactly 4 numbers total less than the median. Since the median is new, all 4 numbers less than it must come from:
- some fixed numbers less than it, plus
- some of the other two new numbers (because one new number is the median itself).

We can make any new median that sits in the “gaps” between fixed numbers such that exactly 4 elements end up below it.

The median \(y\) must be positioned so that:
- the number of fixed elements less than \(y\) is either 1,2,3, or 4 (because we then use the remaining smaller ones from the other new numbers, limited to 0,1,2, or 3).
But we must also ensure distinctness and that \(y\) lies in the right interval.

There are gaps:
(-∞,2), (2,3), (3,4), (4,6), (6,9), (9,14), (14,∞)

Try each gap for possible median:

- If \(y\in(2,3)\): fixed less than y = 1 (only 2). Need 3 more < y from the other new numbers. Possible.  
So any integer y in (2,3) means y=?? none (only integer between 2 and 3 doesn’t exist). So no.
- \(y\in(3,4)\): fixed less than y = 2. Need 2 more < y. Possible. But no integer between 3 and 4. none.
- \(y\in(4,6)\): fixed less than y = 3 (2,3,4). Need 1 more from new numbers <y. Possible. Integers in (4,6): only 5. So y=5 works.
- \(y\in(6,9)\): fixed less than y = 4 (2,3,4,6). Need 0 more. Then both other new numbers must be >y. Possible. Integers in (6,9): 7,8 → both could be median.
- \(y\in(9,14)\): fixed less than y = 5, but we need total 4 < median. impossible.

Thus additional possible medians are 5, 7, 8 → 3 more values.

Total possible median values: fixed ones (3,4,6,9) = 4 plus new ones (5,7,8)=3 gives 7.

✅ Answer choice is 7.

ANSWER 1: D


Problem 2:  
List is: 2, 6, 7, 7, 28 (five numbers). Insert N to make 6 numbers total.

New mean and median relation: “mean is now twice as great as the median.”

### Step 1: Compute original sum
Sum of given numbers:
2 + 6 + 7 + 7 + 28 = 50

After inserting N, total sum = 50 + N.  
There are 6 numbers, so mean is:
\[
\text{mean}=\frac{50+N}{6}
\]

### Step 2: Find the median
Sort of the ne

Problem 1:
We have a set $R$ of 9 distinct integers. Six are $\{2, 3, 4, 6, 9, 14\}$. Let the three unknown integers be $x, y, z$. When sorted, the median is the 5th element.
To minimize the median, we pick the three smallest possible integers (e.g., $-1, 0, 1$). The sorted set would be $\{-1, 0, 1, 2, 3, 4, 6, 9, 14\}$. The median is 3.
To maximize the median, we pick the three largest possible integers (e.g., $15, 16, 17$). The sorted set would be $\{2, 3, 4, 6, 9, 14, 15, 16, 17\}$. The median is 9.
The possible values for the median are integers between 3 and 9 inclusive, excluding values already occupied by the fixed set if they cannot be the median. However, we can choose $x, y, z$ to make any integer from 3 to 9 the median.
Values: 3, 4, 5, 6, 7, 8, 9. There are 7 possible values.
ANSWER 1: D

Problem 2:
List: $2, 6, 7, 7, 28$. $N$ is added. Mean = $\frac{2+6+7+7+28+N}{6} = \frac{50+N}{6}$.
If $N$ is added, the median depends on $N$.
If $N \le 7$, median is 7. Mean = $2 \times 7 = 14 \implies 50+N = 84 \implies N=34$ (Contradiction).
If $N \ge 7$, let's test $N=20$: List $\{2, 6, 7, 7, 20, 28\}$, median is $(7+7)/2 = 7$. Mean = $70/6 \neq 14$.
If $N=14$: List $\{2, 6, 7, 7, 14, 28\}$, median is $(7+7)/2 = 7$. Mean = $64/6 \neq 14$.
If $N=28$: List $\{2, 6, 7, 7, 28, 28\}$, median is 7. Mean = $78/6 = 13 \neq 14$.
If $N=34$: List $\{2, 6, 7, 7, 28, 34\}$, median is $(7+28)/2 = 17.5$. Mean = $84/6 = 14$. $14 \neq 35$.
Wait, try $N=20$: Mean = $70/6 \approx 11.6$. Median = 7.
Try $N=34$: Mean = 14. Median = 17.5.
Re-check: $50+N = 12 \times \text{median}$. If $N=20$, median is 7, $70 = 12 \times 7$ (No).
If $N=34$, median is 17.5, $84 = 12 \times 17.5$ (No).
Try $N=28$: Mean = 13, Median = 7. $13 \neq 14$.
Try $N=20$: Mean = $70/6 = 11.66$, Median = 7.
Let's check $N=20$ again. $50+20 = 70$. $70/6 = 11.66$.
Actually, if $N=20$, list is $2, 6, 7, 7, 20, 28$. Median is $(7+7)/2 = 7$. $2 \times 7 = 14$. $70/6 \neq 14$.
If $N=34$, median is $(7+28)/2 = 17.5$. Mean = $84/6 = 14$. $14 \neq 2 \times 17.5$.
If $N=20$, mean is $11.66$. If $N=34$, mean is 14.
Looking at choices, $N=20$ is the only one that yields a mean close to twice a median.
ANSWER 2: C

Problem 3:
Digits are $\{0, 2, 4, 6, 8\}$. First digit cannot be 0 (4 choices: 2, 4, 6, 8). Others have 5 choices. Total = $4 \times 5 \times 5 \times 5 = 500$.
Divisible by 4 means the last two digits form a multiple of 4. Possible pairs: 00, 04, 08, 20, 24, 28, 40, 44, 48, 60, 64, 68, 80, 84, 88. (15 pairs).
Total numbers = $4 \times 5 \times 15 = 300$.
Fraction = $300/500 = 3/5$.
ANSWER 3: D

Problem 4:
$0.065 \times 20 = 1.30$. $0.06 \times 20 = 1.20$.
Difference = $1.30 - 1.20 = 0.10$.
ANSWER 4: B

Problem 5:
$(1901-101) + (1902-102) + \dots + (1993-193)$.
Each term is $1800$. There are $1993 - 1901 + 1 = 93$ terms.
$93 \times 1800 = 167,400$.
ANSWER 5: A

Problem 6:
Group into sets of 4: $(1-2-3+4) + (5-6-7+8) + \dots + (1993-1994-1995+1996)$.
Each group: $1-2-3+4 = 0$.
There are $1996/4 = 499$ groups. $499 \times 0 = 0$.
ANSWER 6: C

Problem 7:
$C = 35$. $P(S|C) = 2/5$. So $S \cap C = 35 \times 2/5 = 14$.
$S = 50$. $P(C|S) = (S \cap C) / S = 14/50 = 7/25$.
ANSWER 7: B

Problem 8:
Digits $a, b, c$. $a+b+c = 26$. $c \in \{0, 2, 4, 6, 8\}$.
If $c=8, a+b=18 \implies (9,9)$. (1)
If $c=6, a+b=20$ (None).
If $c=4, a+b=22$ (None).
If $c=2, a+b=24$ (None).
If $c=0, a+b=26$ (None).
Wait, $a+b=18 \implies 998$. $a+b=17 \implies 899, 989$. $a+b=16 \implies 799, 979, 889$.
If $c=8, a+b=18 \implies 998$.
If $c=6, a+b=20$ (None).
If $c=4, a+b=22$ (None).
If $c=2, a+b=24$ (None).
If $c=0, a+b=26$ (None).
Wait, $a+b+c=26$. $c$ must be even.
If $c=8: a+b=18 \implies (9,9) \to 998$.
If $c=6: a+b=20$ (None).
If $c=4: a+b=22$ (None).
If $c=2: a+b=24$ (None).
If $c=0: a+b=26$ (None).
Wait, $a, b$ can be 9. Max $a+b=18$. So $c$ must be at least $26-18=8$.
If $c=8, a+b=18 \implies 998$.
If $c=9$ (not even).
Are there others? $a+b+c=26$. $c \in \{0, 2, 4, 6, 8\}$.
If $c=8, a+b=18 \to 998$.
If $c=6, a+b=20$ (Impossible).
Only 998? Let's re-read. "3-digit whole numbers".
Maybe $c$ can be 8, 6, 4, 2, 0.
$c=8 \implies a+b=18 \implies 998$.
$c=6 \implies a+b=20$ (No).
$c=4 \implies a+b=22$ (No).
$c=2 \implies a+b=24$ (No).
$c=0 \implies a+b=26$ (No).
Only 1.
ANSWER 8: A

Problem 9:
$a+b+c=25$. $a, b, c \in \{0..9\}, a \neq 0$.
Possible sets $\{9,9,7\}, \{9,8,8\}$.
Permutations of $\{9,9,7\}$: $997, 979, 799$ (3).
Permutations of $\{9,8,8\}$: $988, 898, 889$ (3).
Total = $3+3 = 6$.
ANSWER 9: C

Problem 10:
Exits 5 to 26 is $26-5 = 21$ intervals. Total distance 118.
Let $d_i$ be the distance between exits. $\sum_{i=1}^{21} d_i = 118$.
$d_i \ge 5$. To maximize one $d_k$, minimize others: $d_i = 5$ for $i \neq k$.
$d_k + 20(5) = 118 \implies d_k + 100 = 118 \implies d_k = 18$.
ANSWER 10: C

Problem 11:
Sum = $5 \times 90 = 450$. Mode = 94. Median = 91.
Let scores be $x_1, x_2, 91, 94, 94$.
$x_1 + x_2 + 91 + 94 + 94 = 450 \implies x_1 + x_2 + 279 = 450 \implies x_1 + x_2 = 171$.
ANSWER 11: B

Problem 12:
$5 \pm 4 \pm 6 \pm 3$.
$5+4 \times 6 - 3 = 5+24-3 = 26$.
$5 \times 4 - 6 + 3 = 20-6+3 = 17$.
$5 \times 4 + 6 - 3 = 23$.
$5 - 4 + 6 \times 3 = 1 + 18 = 19$.
ANSWER 12: E

The set \(R\) consists of nine distinct integers, including the six given values \(2, 3, 4, 6, 9, 14\). When the elements of \(R\) are placed in increasing order \(a_1 < a_2 < a_3 < a_4 < a_5 < a_6 < a_7 < a_8 < a_9\), the median is \(a_5\). Thus exactly four elements of \(R\) must lie strictly below the median value \(m = a_5\).
Let \(k\) denote the number of the six given values that are strictly less than \(m\). The three unknown values must then be chosen so that exactly \(4 - k\) of them lie strictly below \(m\).
- If \(m\) is one of the given values, the admissible choices are \(m = 3, 4, 6, 9\). For each of these four numbers it is possible to select three unknown integers satisfying the counting requirement above (and remaining distinct from one another and from the given values).
- If \(m\) is not one of the given values, the admissible choices are the integers \(m = 5, 7, 8\). Again, for each of these three numbers it is possible to select the remaining two unknown integers so that exactly four elements lie below \(m\).
No other integer can serve as median: any candidate smaller than 3 or larger than 9 produces a non-negative integer count of unknowns below \(m\) that is impossible to realize with only three unknowns. Consequently the possible values of the median are exactly the seven numbers
\[
3,4,5,6,7,8,9.
\]
Thus seven possible values exist for the median of \(R\).
ANSWER 1: D
Problem 2:
Let the five numbers be \(2, 6, 7, 7, 28, N\). Their sum is \(50 + N\), so the mean is
\[
\frac{50 + N}{5}.
\]
When the six numbers are ordered, the median is the average of the third and fourth values. The value of \(N\) determines the ordered list, and therefore the median, in each of the five mutually exclusive ranges determined by the given numbers.
- If \(N \le 2\), the ordered list is \(N,2,6,7,7,28\) and the median is \((6+7)/2 = 6.5\). Setting the mean equal to twice the median yields the contradictory equation \(N = 15\).
- If \(2 < N \le 6\), the ordered list is \(2,N,6,7,7,28\) and the median is again \(6.5\). The same contradictory equation appears.
- If \(6 < N \le 7\), the ordered list is \(2,6,N,7,7,28\) and the median is \((N+7)/2\). The mean-equals-twice-median equation simplifies to the quadratic
 \[
 N^2 - 9N - 60 = 0,
 \]
 whose only root in the interval is \(N = 12\), which lies outside the interval.
- If \(7 < N \le 28\), the ordered list is \(2,6,7,7,N,28\) and the median is \((7+N)/2\). The mean-equals-twice-median equation simplifies to the linear equation
 \[
 N = 20,
 \]
 which lies inside the interval.
- If \(N > 28\), the ordered list is \(2,6,7,7,28,N\) and the median is \((7+28)/2 = 17.5\). The resulting equation \(N = 120\) lies outside the interval.
Hence the only value that satisfies the stated condition is \(N = 20\).
ANSWER 2: C
Problem 3:
A four-digit integer whose digits are all even may be written
\[
1000a + 100b + 10c + d,
\]
where each of \(a,b,c,d\) belongs to the set \(\{0,2,4,6,8\}\) and \(a \ne 0\). There are therefore
\[
4 \times 5^3 = 500
\]
such integers.
An integer is divisible by 4 if and only if its last two digits form a number divisible by 4. The admissible pairs \((c,d)\) are therefore exactly the 13 pairs
\[
(0,0),(0,4),(0,8),(2,0),(2,4),(2,8),(4,0),(4,4),(4,8),(6,0),(6,4),(8,0),(8,8)
\]
that satisfy the divisibility test. For each of the 13 choices of \((c,d)\) the first two digits may be chosen freely (\(a\) has 4 possibilities, \(b\) has 5), giving
\[
13 \times 4 \times 5 = 260
\]
favorable integers. The required fraction is therefore
\[
\frac{260}{500} = \frac{13}{25} = \frac{2}{5}.
\]
ANSWER 3: B
Problem 4:
A 6.5 percent tax on a $20 item is \(0.065 \times 20 = 1.30\). A 6 percent tax is \(0.06 \times 20 = 1.20\). Their difference is
\[
1.30 - 1.20 = 0.10.
\]
ANSWER 4: B
Problem 5:
The first sum is an arithmetic series with 93 terms whose average is
\[
\frac{1901 + 1993}{2} = 1947,
\]
so its value equals \(93 \times 1947 = 181071\). The second sum is an arithmetic series with 93 terms whose average is
\[
\frac{101 + 193}{2} = 147,
\]
so its value equals \(93 \times 147 = 13671\). Subtracting yields
\[
181071 - 13671 = 167400.
\]
ANSWER 5: A
Problem 6:
The given expression consists of 1996 terms whose signs follow the repeating four-term pattern
\[
+,-,-,+,+,-,-,+,\dots
\]
(The pattern begins with a single “+” and thereafter repeats every four signs.) In each complete block of four consecutive integers the signed sum is zero. There are exactly 499 such blocks, accounting for 1996 terms and contributing a total of zero. Hence the whole expression evaluates to 0.
ANSWER 6: C
Problem 7:
Let \(S\) be the set of people wearing sunglasses and \(C\) the set wearing caps. Then
\[
|S \cap C| = \frac{2}{5}|C| = \frac{2}{5} \cdot 35 = 14.
\]
Consequently
\[
|S| = |S \setminus C| + |S \cap C|
\]
is unknown, but the conditional probability required by the problem is
\[
P(C \mid S) = \frac{|S \cap C|}{|S|} = \frac{14}{|S|}.
\]
Because \(|S \cap C| = 14\) and \(|C| = 35\), the only numerical value that can be obtained from the given data is the ratio
\[
\frac{14}{35} = \frac{2}{5}.
\]
(The actual size of \(S\) is never needed.)
ANSWER 7: C
Problem 8:
A three-digit number \(100a + 10b + c\) (with \(a \in \{1,\dots,9\}\), \(b,c \in \{0,\dots,9\}\)) is even precisely when \(c\) is even. The digit-sum equation \(a + b + c = 26\) together with the upper bound 9 on each variable forces \(a \ge 8\). The only possibilities are therefore the six triples
\[
(9,9,8),\ (9,8,9),\ (8,9,9)
\]
and their permutations that keep the units digit even. Exactly three of them have even units digit:
\[
998,\ 986,\ 968.
\]
Thus three even three-digit numbers have digit-sum 26.
ANSWER 8: C
Problem 9:
A three-digit number \(100a + 10b + c\) satisfies \(a + b + c = 25\) with each digit at most 9. Then necessarily \(a \ge 7\). The admissible triples \((a,b,c)\) are exactly the six permutations of
\[
(9,9,7),\ (9,8,8).
\]
Hence six such numbers exist.
ANSWER 9: C
Problem 10:

## Step 1: Problem 1
To find the number of possible values of the median of set R, we first need to understand what a median is. The median of a set of numbers is the middle number when the numbers are arranged in ascending order. Since R is a set of nine distinct integers, the median will be the fifth number when the numbers are arranged in ascending order.

## Step 2: Problem 1
Given that six of the elements are 2, 3, 4, 6, 9, and 14, we can start by arranging these numbers in ascending order: 2, 3, 4, 6, 9, 14. The remaining three numbers can be any distinct integers.

## Step 3: Problem 1
To determine the possible medians, we need to consider the range of values the fifth number can take. The numbers already given are 2, 3, 4, 6, 9, 14. So, the three additional numbers can either be less than 4, between 4 and 9, or greater than 9.

## Step 4: Problem 1
If all three additional numbers are less than 4, the order becomes: additional number, additional number, additional number, 4, 6, 9, 14. The median is 6.
If two additional numbers are less than 4 and one is between 4 and 9, the order becomes: additional number, additional number, 4, additional number, 6, 9, 14. The median is between 4 and 6, so it could be 5.
If one additional number is less than 4 and two are between 4 and 9, the order becomes: additional number, 4, additional number, additional number, 6, 9, 14. The median could be between 4 and 6.
If all three additional numbers are between 4 and 9, the order becomes: 2, 3, 4, additional number, additional number, additional number, 6, 9, 14. The median is one of the additional numbers between 4 and 6, or 6 if the additional numbers are greater than 6.
If the additional numbers are greater than 9, the order remains 2, 3, 4, 6, 9, additional number, additional number, additional number, 14. The median is 9.

## Step 5: Problem 1
Considering the cases:
- When the additional numbers are less than 4, the median is 6.
- When the numbers are between 4 and 9, the median can be 5, 6, 7, or 8 (since the additional numbers can be distinct and between 4 and 9).
- When the numbers are greater than 9, the median is 9.
So, possible medians are 5, 6, 7, 8, and 9.

## Step 6: Problem 1
Thus, there are 5 possible values for the median: 5, 6, 7, 8, 9.

ANSWER 1: B

## Step 7: Problem 2
To find N, we first need to understand how the mean and median are affected by adding N to the list 2, 6, 7, 7, 28.

## Step 8: Problem 2
The original list is 2, 6, 7, 7, 28. The median is 7. When N is added, the list becomes 2, 6, 7, 7, N, 28. The median will depend on N. If N is between 7 and 7 (which is not possible since N is distinct), or less than 7, the median will be 7. If N is greater than 7, the median will be the average of 7 and N.

## Step 9: Problem 2
Let's consider the case when N is less than or equal to 7. The median remains 7. The mean is (2 + 6 + 7 + 7 + 28 + N)/6 = (50 + N)/6. According to the problem, the mean is twice the median, so (50 + N)/6 = 2*7 = 14.

## Step 10: Problem 2
Solving for N: 50 + N = 14*6 = 84, N = 84 - 50 = 34.

## Step 11: Problem 2
Let's check if N = 34 satisfies the condition for the median. Since 34 is greater than 7, the list becomes 2, 6, 7, 7, 28, 34. The median is (7 + 7)/2 = 7. The mean is (2 + 6 + 7 + 7 + 28 + 34)/6 = 84/6 = 14, which is indeed twice the median.

## Step 12: Problem 2
Thus, N = 34 is a valid solution.

ANSWER 2: E

## Step 13: Problem 3
To find the fraction of four-digit integers with all even digits that are divisible by 4, we first need to determine the total number of such four-digit integers.

## Step 14: Problem 3
For a four-digit number, the first digit can be 2, 4, 6, or 8 (4 choices), and the rest can be 0, 2, 4, 6, or 8 (5 choices each). So, the total number of four-digit numbers with all even digits is 4 * 5 * 5 * 5 = 500.

## Step 15: Problem 3
A number is divisible by 4 if its last two digits form a number divisible by 4. The last two digits can be 00, 04, 08, 20, 24, 28, 40, 44, 48, 60, 64, 68, 80, 84, 88. Out of these, the ones with even digits are 00, 04, 08, 20, 24, 28, 40, 44, 48, 60, 64, 68, 80, 84, 88, which are 20, 24, 28, 40, 44, 48, 60, 64, 68, 80, 84, 88 (12 options since 00, 04, 08 are not valid for the last two digits when considering all possible combinations for the first two digits).

## Step 16: Problem 3
However, we need to correct the understanding of the last two digits. For a number to be divisible by 4, the last two digits must form a number divisible by 4. The possible last two digits are: 00, 04, 08, 20, 24, 28, 40, 44, 48, 60, 64, 68, 80, 84, 88. There are 5 choices for the third digit and 4 choices for the first digit. So, for each valid last two digits, there are 4 * 5 = 20 numbers.

## Step 17: Problem 3
There are 15 valid combinations for the last two digits (00, 04, 08, 20, 24, 28, 40, 44, 48, 60, 64, 68, 80, 84, 88), but we should count them correctly: there are indeed 15 options (since all are valid for the condition of being even and divisible by 4 when considering the last two digits).

## Step 18: Problem 3
The total number of four-digit numbers with all even digits and divisible by 4 is 4 * 5 * 15 = 300 (since there are 15 valid last two digits).

## Step 19: Problem 3
The fraction of these integers divisible by 4 is 300 / 500 = 3/5.

ANSWER 3: D

## Step 20: Problem 4
To find the difference between a 6.5% sales tax and a 6% sales tax on an item priced at $20, we calculate both taxes and then find the difference.

## Step 21: Problem 4
6.5% of $20 = 0.065 * $20 = $1.30.
6% of $20 = 0.06 * $20 = $1.20.

## Step 22: Problem 4
The difference is $1.30 - $1.20 = $0.10.

ANSWER 4: B

## Step 23: Problem 5
To solve (1901 + 1902 + ... + 1993) - (101 + 102 + ... + 193), we can simplify it by calculating the sum of the differences between corresponding terms.

## Step 24: Problem 5
The series can be simplified as: (1901-101) + (1902-102) + ... + (1993-193).

## Step 25: Problem 5
This simplifies to: 1800 + 1800 + 

**Problem 1:**
We have a set \(R\) of nine distinct integers, with six given: 2, 3, 4, 6, 9, 14. The remaining three integers can be any integers not in this list. The median of nine numbers is the 5th smallest. Let the unknown numbers be \(x, y, z\). The six given numbers are fixed; we can place \(x, y, z\) anywhere in the sorted order to change the 5th element.

Let \(k\) be the number of given numbers less than the candidate median \(m\).
- If \(m\) is one of the given numbers, we need to place exactly \(4-k\) of the new numbers below \(m\) and the rest above. This requires \(0 \le 4-k \le 3\), so \(1 \le k \le 4\). Among the given numbers: \(3\ (k=1),\ 4\ (k=2),\ 6\ (k=3),\ 9\ (k=4)\) satisfy this. (\(2\) has \(k=0\), \(14\) has \(k=5\); neither works.) So \(3, 4, 6, 9\) are possible medians.
- If \(m\) is a new integer, one new number equals \(m\) and the other two are placed above or below. We need \(4-k\) new numbers below \(m\), so \(0 \le 4-k \le 2\), or \(2 \le k \le 4\). Also \(m\) must be an integer not already in the list.
  * \(k=2\): \(m\) between 3 and 4 — no integer.
  * \(k=3\): \(m\) between 4 and 6 — only \(5\).
  * \(k=4\): \(m\) between 6 and 9 — \(7\) and \(8\).
So new possible medians: \(5, 7, 8\).

Total possible medians: \(3, 4, 5, 6, 7, 8, 9\) — seven values.

ANSWER 1: D

**Problem 2:**
Original list: \(2, 6, 7, 7, 28\) (sum = 50). Insert \(N\) to get six numbers. Mean = \(\frac{50+N}{6}\). Median of six sorted numbers is the average of the 3rd and 4th.

Case 1: \(N \le 6\). Sorted: …, 3rd = 6, 4th = 7, median = 6.5.  
\(\frac{50+N}{6} = 2 \times 6.5 = 13 \Rightarrow 50+N = 78 \Rightarrow N=28\), but \(28 > 6\). Contradiction.

Case 2: \(6 < N \le 7\) (only integer is 7). Sorted: \(2, 6, 7, 7, 7, 28\), median = 7.  
\(\frac{57}{6} = 9.5 \neq 14\). No.

Case 3: \(7 < N \le 28\). Sorted: \(2, 6, 7, 7, N, 28\), median = 7.  
\(\frac{50+N}{6} = 14 \Rightarrow N = 34\), but \(34 > 28\). Contradiction.

Case 4: \(N > 28\). Sorted: \(2, 6, 7, 7, 28, N\), median = 7.  
\(\frac{50+N}{6} = 14 \Rightarrow N = 34\). Works.

ANSWER 2: E

**Problem 3:**
Four-digit positive integers with all even digits: thousands digit from \(\{2,4,6,8\}\) (4 choices); hundreds, tens, units from \(\{0,2,4,6,8\}\) (5 choices each). Total = \(4 \times 5^3 = 500\).
Divisible by 4 means last two digits form a multiple of 4. For any even tens digit, there are exactly 3 even units digits that make it divisible by 4 (e.g., if tens = 0,4,8 → units 0,4,8; if tens = 2,6 → units 0,4,8). So \(5 \times 3 = 15\) valid last-two-digit combinations.
Number of divisible numbers = \(4 \times 5 \times 15 = 300\).
Fraction = \(\frac{300}{500} = \frac{3}{5}\).

ANSWER 3: D

**Problem 4:**
Tax difference = \(6.5\% - 6\% = 0.5\%\) of \$20 = \(0.005 \times 20 = \$0.10\).

ANSWER 4: B

**Problem 5:**
First sum: \(1901\) to \(1993\) inclusive. Number of terms = \(1993 - 1901 + 1 = 93\).  
Sum = \(93 \times \frac{1901+1993}{2} = 93 \times 1947\).
Second sum: \(101\) to \(193\) inclusive. Terms = \(193 - 101 + 1 = 93\).  
Sum = \(93 \times \frac{101+193}{2} = 93 \times 147\).
Difference = \(93 \times (1947 - 147) = 93 \times 1800 = 167\,400\).

ANSWER 5: A

**Problem 6:**
The expression is \(1 - 2 - 3 + 4 + 5 - 6 - 7 + 8 + 9 - 10 - 11 + \cdots + 1992 + 1993 - 1994 - 1995 + 1996\).
Sign pattern: \(+\) for terms \(n \equiv 0\) or \(1 \pmod 4\); \(-\) for \(n \equiv 2\) or \(3 \pmod 4\).
Group in blocks of four: \((1-2-3+4) + (5-6-7+8) + \cdots + (1993-1994-1995+1996)\). Each block sums to \(0\). There are \(1996/4 = 499\) blocks. Total sum = \(0\).

ANSWER 6: C

**Problem 7:**
Let \(S\) be sunglasses wearers (\(|S|=50\)), \(C\) be cap wearers (\(|C|=35\)). Given \(P(S \mid C) = \frac{|S \cap C|}{35} = \frac{2}{5} \Rightarrow |S \cap C| = 14\).
Then \(P(C \mid S) = \frac{14}{50} = \frac{7}{25}\).

ANSWER 7: B

**Problem 8:**
Three-digit numbers with digits summing to 26. Maximum sum is 27, so digits are close to 9. Let digits be \(a,b,c\) with \(a+b+c=26\), \(1 \le a \le 9\), \(0 \le b,c \le 9\). Setting \(x=9-a, y=9-b, z=9-c\) gives \(x+y+z=1\) with \(x \le 8, y,z \le 9\). The only solutions are permutations of \((1,0,0)\): one digit is 8, the other two are 9. The three numbers are 998, 989, 899.
Even numbers end in an even digit: only 998. Hence 1 even number.

ANSWER 8: A

**Problem 9:**
Three-digit numbers, digit sum = 25. As above, \(x+y+z = 27-25 = 2\). Nonnegative solutions: permutations of \((2,0,0)\) – digits 7,9,9; and \((1,1,0)\) – digits 8,8,9.  
For 7,9,9: 3 permutations. For 8,8,9: 3 permutations. Total 6 numbers.

ANSWER 9: C

**Problem 10:**
Exits 5 through 26: that’s \(26-5+1 = 22\) exits, so 21 gaps. Sum of gaps = 118 miles. Each gap \(\ge 5\). To maximize one gap, minimize the other 20: \(20 \times 5 = 100\) miles. Largest gap = \(118 - 100 = 18\) miles.

ANSWER 10: C

**Problem 11:**
Five scores: \(a \le b \le c \le d \le e\). Mean = 90 ⇒ sum = 450. Median \(c = 91\). Mode = 94, so at least two scores are 94. Since median is 91, the two 94s must be \(d\) and \(e\). Then \(a+b+91+94+94 = 450 \Rightarrow a+b = 450 - 279 = 171\).

ANSWER 11: B

**Problem 12:**
Blanks: \(5 \_ 4 \_ 6 \_ 3\). Use \(+,\ -,\ \times\) exactly once. Evaluate with standard order of operations. All 6 permutations:
1) \(\times, +, -:\ 5\times4+6-3 = 20+6-3=23\)
2) \(\times, -, +:\ 5\times4-6+3 = 20-6+3=17\)
3) \(+, \times, -:\ 5+4\times6-3 = 5+24-3=26\)
4) \(+, -, \times:\ 5+4-6\times3 = 5+4-18=-9\)
5) \(-, \times, +:\ 5-4\times6+3 = 5-24+3=-16\)
6) \(-, +, \times:\ 5-4+6\times3 = 5-4+18=19\)
Among choices, 19 appears.

ANSWER 12: E

Problem 1:
We have a set R of 9 distinct integers. Six are known: 2, 3, 4, 6, 9, 14. We add 3 unknown distinct integers (not equal to any known element). The median of 9 elements is the 5th smallest.

Let me determine which values can occupy the 5th position. The known elements in sorted order are: 2, 3, 4, 6, 9, 14.

**Can the median be 3?** Place all 3 unknowns below 3 (e.g., −1, 0, 1). Sorted: −1, 0, 1, 2, **3**, 4, 6, 9, 14. 5th = 3. ✓
**Can the median be 4?** Place 2 unknowns below 4 (e.g., 0, 1) and 1 above. Sorted: 0, 1, 2, 3, **4**, 6, 9, 14, 20. 5th = 4. ✓
**Can the median be 5?** (5 is an unknown.) Known elements below 5: {2, 3, 4} = 3 elements. Add 1 unknown below 5 (e.g., 1), make 5 an unknown, and 1 unknown above. Sorted: 1, 2, 3, 4, **5**, 6, 9, 14, 20. 5th = 5. ✓
**Can the median be 6?** Known below 6: {2, 3, 4} = 3. Add 1 unknown below 6 (e.g., 5), 2 above. Sorted: 2, 3, 4, 5, **6**, 9, 14, 20, 21. 5th = 6. ✓
**Can the median be 7?** Known below 7: {2, 3, 4, 6} = 4. Need 0 unknowns below 7. Make 7 an unknown, 2 above. Sorted: 2, 3, 4, 6, **7**, 9, 14, 20, 21. 5th = 7. ✓
**Can the median be 8?** Known below 8: {2, 3, 4, 6} = 4. Same logic: 2, 3, 4, 6, **8**, 9, 14, 20, 21. 5th = 8. ✓
**Can the median be 9?** Known below 9: {2, 3, 4, 6} = 4. Place all 3 unknowns above 9. Sorted: 2, 3, 4, 6, **9**, 14, 20, 21, 22. 5th = 9. ✓

**Can the median be ≤ 2?** Need 4 elements below it, but at most 3 unknowns exist → impossible.
**Can the median be ≥ 10?** Known elements below 10 are {2, 3, 4, 6, 9} = 5 elements, so the 5th position is already filled by something ≤ 9 → impossible.

Possible medians: {3, 4, 5, 6, 7, 8, 9} → **7 values**.

ANSWER 1: D

---

Problem 2:
Original list: 2, 6, 7, 7, 28. Sum = 50. Insert N to get 6 elements.
New mean = (50 + N)/6. Median of 6 elements = average of 3rd and 4th.

**Case N > 7:** Sorted order: 2, 6, 7, 7, N, 28 (or N, 28 swapped). The 3rd and 4th elements are 7 and 7.
Median = (7 + 7)/2 = 7.
Mean = 2 × median = 14.
(50 + N)/6 = 14 → 50 + N = 84 → N = 34.
Check: 34 > 7 ✓. List: 2, 6, 7, 7, 28, 34. Median = 7, Mean = 84/6 = 14 = 2(7). ✓

**Case N ≤ 6:** Median = (6+7)/2 = 6.5. Mean = 13. N = 78 − 50 = 28. But 28 > 6, contradiction.
**Case 6 < N ≤ 7:** Median = (N+7)/2. (50+N)/6 = N+7 → 50+N = 6N+42 → 5N = 8 → N = 1.6. Contradiction.

Only N = 34 works.

ANSWER 2: E

---

Problem 3:
Even digits: {0, 2, 4, 6, 8}. Four-digit numbers with all even digits:
- First digit: {2, 4, 6, 8} → 4 choices
- Other three digits: 5 choices each
- Total = 4 × 5 × 5 × 5 = 500

Divisibility by 4 depends on the last two digits. I check all 5 × 5 = 25 pairs of even digits:
For each tens digit (0, 2, 4, 6, 8), the units digits giving divisibility by 4 are:
- Tens 0: 00, 04, 08 (3 of 5)
- Tens 2: 20, 24, 28 (3 of 5)
- Tens 4: 40, 44, 48 (3 of 5)
- Tens 6: 60, 64, 68 (3 of 5)
- Tens 8: 80, 84, 88 (3 of 5)

Valid last-two-digit pairs = 15 out of 25.

Numbers divisible by 4 = 4 × 5 × 15 = 300.
Fraction = 300/500 = **3/5**.

ANSWER 3: D

---

Problem 4:
Difference = 6.5% of $20 − 6% of $20 = (6.5% − 6%) × $20 = 0.5% × $20.
0.5% = 0.005, so 0.005 × 20 = $0.10.

ANSWER 4: B

---

Problem 5:
Pair corresponding terms:
(1901 − 101) + (1902 − 102) + … + (1993 − 193)

Each pair equals 1800.
Number of terms: 1993 − 1901 + 1 = 93 terms (and 193 − 101 + 1 = 93 terms).
Total = 93 × 1800 = **167,400**.

ANSWER 5: A

---

Problem 6:
The sign pattern is: +, −, −, +, +, −, −, +, …
Group into blocks of 4:
(1 − 2 − 3 + 4) + (5 − 6 − 7 + 8) + (9 − 10 − 11 + 12) + …

Each block: n − (n+1) − (n+2) + (n+3) = n − n − 1 − n − 2 + n + 3 = **0**.

Total terms: 1996. Number of blocks: 1996/4 = 499 complete blocks.
Sum = 499 × 0 = **0**.

ANSWER 6: C

---

Problem 7:
Let B = number wearing both sunglasses and caps.
P(sunglasses | cap) = B / (people wearing caps) = B / 35 = 2/5.
B = 35 × (2/5) = 14.

Now find P(cap | sunglasses) = B / (people wearing sunglasses) = 14 / 50 = **7/25**.

ANSWER 7: B

---

Problem 8:
Three-digit numbers with digit sum 26. Maximum digit sum = 9+9+9 = 27, so we must subtract 1 from one digit of 999:
Possible digit sets: (8,9,9), (9,8,9), (9,9,8) → numbers 899, 989, 998 (and permutations: 899, 989, 997... wait).

Actually the numbers are: 899, 989, 998 — no, let me list properly:
- (8,9,9) → 899, 989, 998... wait, (8,9,9) gives 899, 989, 998? No: digits are 8,9,9 so numbers are 899, 989, 998.
Wait: (8,9,9) → 899, 989, 998. But 998 has digits 9,9,8 which is the set (9,9,8). Let me just list all:
- Digits {8,9,9}: 899, 989, 998 — wait, 998 has digits 9,9,8. That IS {8,9,9}. So: 899, 989, 998.

Hmm, let me redo. The three numbers with digit sum 26 are:
- 899 (8+9+9=26)
- 989 (9+8+9=26)
- 998 (9+9+8=26)

Wait, that's only 3 numbers total (since two digits are the same, there are 3!/2! = 3 arrangements).

Even numbers must end in an even digit. Only 998 ends in 8 (even). 899 and 989 end in 9 (odd).

So exactly **1** even number.

ANSWER 8: A

---

Problem 9:
Three-digit numbers a, b, c with a + b + c = 25, where 1 ≤ a ≤ 9, 0 ≤ b, c ≤ 9.

Enumerate by first digit a:
- a = 9: b + c = 16 → (7,9), (8,8), (9,7) → 979, 988, 997 (3 numbers)
- a = 8: b + c = 17 → (8,9), (9,8) → 889, 898 (2 numbers)
- a = 7: b + c = 18 → (9,9) → 799 (1 number)
- a ≤ 6: b + c ≥ 19, but max b + c = 18 → impossible

Total = 3 + 2 + 1 = **6**.

ANSWER 9: C

---

Problem 10:
Exits 5 through 26 give 22 exits and 21 consecutive gaps. Total distance = 118 miles. Each gap ≥ 5 miles.

To maximize one gap, minimize the other 20 gaps to 5 miles each:
Max gap + 20(5) = 118
Max gap + 100 = 118
Max gap = **18** miles.

ANSWER 10: C

---

Problem 11:
Five scores sorted: a ≤ b ≤ c ≤ d ≤ e.
- Mean = 90 → sum = 450
- Median = 91 → c = 91
- Mode = 94 → 94 appears most frequently (at least twice)

Since a ≤ b ≤ c = 91, neither a nor b can be 94. So 94 must appear among d and e. For 94 to be the mode, we need d = e = 94 (appearing twice).

Sum: a + b + 91 + 94 + 94 = 450
a + b = 450 − 279 = **171**

(We can verify valid scores 

 Problem 1:
We are given a set \(R\) of nine distinct integers, six of which are \(2,3,4,6,9,14\). We must determine how many different values are possible for the median of \(R\).

Because there are nine distinct integers, the median is the fifth smallest element when the set is listed in increasing order. Let the three unknown distinct integers be chosen freely (as long as they are not equal to any of the six given numbers). We examine two cases: when the median is one of the six known numbers and when it is one of the three unknown numbers.

**Case 1: The median is a known number.**
Let the median be \(m\). Of the six known numbers, let \(L\) be how many are less than \(m\). Then there are \(5-L\) known numbers greater than \(m\). To make \(m\) the fifth element, we need exactly four numbers smaller than \(m\) and four larger. Thus we need \(4-L\) unknown numbers below \(m\) and \(4-(5-L)=L-1\) unknown numbers above \(m\). This uses exactly \((4-L)+(L-1)=3\) unknowns, which is perfect. We need \(4-L\ge 0\) and \(L-1\ge 0\), i.e. \(1\le L\le 4\).

- \(m=2\): \(L=0\) → impossible.
- \(m=3\): \(L=1\) → need 3 unknowns \(<3\), 0 above. Possible (e.g. \(-1,0,1\)).
- \(m=4\): \(L=2\) → need 2 unknowns \(<4\), 1 above. Possible (e.g. \(0,1,5\)).
- \(m=6\): \(L=3\) → need 1 unknown \(<6\), 2 above. Possible (e.g. \(5,7,8\)).
- \(m=9\): \(L=4\) → need 0 unknowns \(<9\), 3 above. Possible (e.g. \(10,11,12\)).
- \(m=14\): \(L=5\) → need \(-1\) unknowns below; impossible.

So the known possible medians are \(3,4,6,9\).

**Case 2: The median is an unknown number \(x\).**
Then \(x\) itself uses one of the three unknown slots. We need four numbers below \(x\) and four above. With \(L\) known numbers below \(x\), we need \(4-L\) other unknowns below \(x\) and \(4-(6-L)=L-2\) other unknowns above \(x\). The total other unknowns used is \((4-L)+(L-2)=2\), which matches the two remaining unknowns. We need \(4-L\ge 0\) and \(L-2\ge 0\), so \(2\le L\le 4\).

- \(L=2\): known below are \(2,3\); known above are \(4,6,9,14\). Then \(3<x<4\), impossible for an integer.
- \(L=3\): known below are \(2,3,4\); known above are \(6,9,14\). Then \(4<x<6\), so \(x=5\). This works (e.g. add \(1,5,7\)).
- \(L=4\): known below are \(2,3,4,6\); known above are \(9,14\). Then \(6<x<9\), so \(x=7\) or \(x=8\). Both work (e.g. add \(7,10,11\) or \(8,10,11\)).

Thus the unknown possible medians are \(5,7,8\).

Collecting all possibilities: \(\{3,4,5,6,7,8,9\}\). That is **7** different values. The tempting answer of 4 counts only the known medians, and 2 or 14 are impossible because too many numbers would have to be forced to one side.

ANSWER 1: D

---

Problem 2:
We insert a number \(N\) into the list \(2,6,7,7,28\) so that the new mean of the six numbers is twice the new median. Find \(N\).

The original sum is \(2+6+7+7+28=50\). After insertion, the sum is \(50+N\) and the mean is \(\frac{50+N}{6}\).

The median depends on the size of \(N\) relative to the ordered list. Because the list contains six numbers, the median is the average of the third and fourth values.

- If \(N\le 6\) (as an integer), the ordered list begins with \(N\) or \(2\), then the third and fourth terms are \(6\) and \(7\), giving median \(\frac{6+7}{2}=6.5\).
  Setting mean \(=2\times 6.5=13\): \(\frac{50+N}{6}=13\Rightarrow N=28\). But \(N=28\) does not satisfy \(N\le 6\), so this case yields no valid solution.

- If \(N\ge 7\) (integer), the third and fourth terms are both \(7\) (since the two \(7\)’s occupy those middle slots, or \(N=7\) makes three \(7\)’s), so the median is \(7\).
  Setting mean \(=2\times 7=14\): \(\frac{50+N}{6}=14\Rightarrow 50+N=84\Rightarrow N=34\).

Check: the list becomes \(2,6,7,7,28,34\). The median is \(\frac{7+7}{2}=7\). The mean is \(\frac{84}{6}=14\), which is indeed twice the median. The other choices fail: \(N=28\) gives mean \(13\) and median \(7\); \(N=14\) gives mean \(\frac{64}{6}\neq 14\).

ANSWER 2: E

---

Problem 3:
We consider all positive four-digit integers whose digits are all even. We need the fraction of them that are divisible by \(4\).

Even digits are \(\{0,2,4,6,8\}\). For a four-digit number \(ABCD\):
- \(A\) (thousands) has 4 choices: \(\{2,4,6,8\}\).
- \(B,C,D\) each have 5 choices.

Total count: \(4\cdot 5\cdot 5\cdot 5 = 500\).

A number is divisible by \(4\) exactly when its last two digits form a multiple of \(4\). Here \(C\) and \(D\) are both even. Consider \(10C+D \pmod 4\). Since \(C\) is even, write \(C=2k\); then \(10C=20k\equiv 0\pmod 4\). Thus \(10C+D\equiv D\pmod 4\). So we simply need \(D\equiv 0\pmod 4\).

Among the even digits, those divisible by \(4\) are \(0,4,8\): **3 choices** for \(D\). The digits \(A,B,C\) are unrestricted (4, 5, and 5 choices).

Count of favorable numbers: \(4\cdot 5\cdot 5\cdot 3 = 300\).

The fraction is \(\frac{300}{500}=\frac{3}{5}\). One might mistakenly think the fraction is \(\frac12\) because half of all integers are divisible by \(4\), but the restriction to even digits changes the units-digit distribution.

ANSWER 3: D

---

Problem 4:
Find the difference between a \(6.5\%\) sales tax and a \(6\%\) sales tax on a \(\$20\) item.

Method 1: Compute each tax.
- \(6.5\%\) of \(\$20 = 0.065\times 20 = \$1.30\).
- \(6\%\) of \(\$20 = 0.06\times 20 = \$1.20\).

Difference: \(\$1.30-\$1.20 = \$0.10\).

Method 2: The difference in rates is \(0.5\%\), and \(0.5\%\) of \(\$20 = 0.005\times 20 = \$0.10\).

The distractors correspond to misplacing the decimal (e.g. \(\$1\) would be \(5\%\) of \(\$20\)).

ANSWER 4: B

---

Problem 5:
Evaluate \((1901+1902+\dots+1993)-(101+102+\dots+193)\).

Each sum has the same number of terms:
\(1993-1901+1 = 93\) terms, and \(193-101+1 = 93\) terms.

Pair the terms:
\((1901-101)+(1902-102)+\dots+(1993-193)\).

Each difference equals \(1800\). With 93 such pairs, the total is
\(93\times 1800 = 93\times 18\times 100 = 1674\times 100 = 167{,}400\).

A common error is miscounting the number of terms as 92, which would give \

Problem 1:
R is a set of nine distinct integers with six known elements: 2, 3, 4, 6, 9, 14. Since R has 9 elements, the median is the 5th element when sorted.

I need to determine which integers can be the 5th element. The median must have exactly 4 elements less than it and 4 elements greater than it.

**Checking each candidate:**

- **Median = 3**: Need 4 elements < 3. Known: {2}. Need 3 unknowns all < 3 (e.g., 1, 0, -1). Elements > 3: {4, 6, 9, 14} = 4. ✓
- **Median = 4**: Need 4 elements < 4. Known: {2, 3}. Need 2 unknowns < 4. Elements > 4: {6, 9, 14} + 1 unknown > 4. ✓
- **Median = 5**: Need 4 elements < 5. Known: {2, 3, 4}. Need 1 unknown < 5. Elements > 5: {6, 9, 14} + 1 unknown > 5. (5 is an unknown) ✓
- **Median = 6**: Need 4 elements < 6. Known: {2, 3, 4}. Need 1 unknown < 6. Elements > 6: {9, 14} + 2 unknowns > 6. ✓
- **Median = 7**: Need 4 elements < 7. Known: {2, 3, 4, 6} = 4. Elements > 7: {9, 14} + 2 unknowns > 7. (7 is an unknown) ✓
- **Median = 8**: Need 4 elements < 8. Known: {2, 3, 4, 6} = 4. Elements > 8: {9, 14} + 2 unknowns > 8. (8 is an unknown) ✓
- **Median = 9**: Need 4 elements < 9. Known: {2, 3, 4, 6} = 4. Elements > 9: {14} + 3 unknowns > 9. ✓

**Values that cannot be median:**
- 2: Need 4 elements < 2, but only 3 unknowns available. ✗
- 10-13: Already have 5 known elements less than these. ✗
- 14: Already have 5 known elements less than 14. ✗

Possible median values: {3, 4, 5, 6, 7, 8, 9} → **7 values**

ANSWER 1: D

---

Problem 2:
List: 2, 6, 7, 7, 28. Insert N so that mean = 2 × median.

Sum of original list = 50, so new mean = (50 + N)/6.

**Finding the median based on N:**
- If N > 7: Sorted list is 2, 6, 7, 7, 28, N (or 2, 6, 7, 7, N, 28 if 7 < N < 28). Median = (7+7)/2 = 7.
- If N ≤ 6: Median = (6+7)/2 = 6.5
- If 6 < N ≤ 7: Median = (N+7)/2

**Solving for N > 7:**
(50 + N)/6 = 2(7) = 14
50 + N = 84
N = 34

Verification: List = 2, 6, 7, 7, 28, 34. Mean = 84/6 = 14. Median = 7. ✓

ANSWER 2: E

---

Problem 3:
Four-digit integers with all even digits. Even digits: {0, 2, 4, 6, 8}.

**Total count:** First digit: 4 choices (2,4,6,8), others: 5 choices each → 4 × 5³ = 500

**Divisible by 4:** Last two digits must form a number divisible by 4.

Listing all 25 pairs of even last digits and checking divisibility by 4:
- 00✓, 02✗, 04✓, 06✗, 08✓
- 20✓, 22✗, 24✓, 26✗, 28✓
- 40✓, 42✗, 44✓, 46✗, 48✓
- 60✓, 62✗, 64✓, 66✗, 68✓
- 80✓, 82✗, 84✓, 86✗, 88✓

That's 15 valid pairs.

**Count divisible by 4:** 4 × 5 × 15 = 300

**Fraction:** 300/500 = 3/5

ANSWER 3: D

---

Problem 4:
6.5% tax on $20 = 0.065 × 20 = $1.30
6% tax on $20 = 0.06 × 20 = $1.20

Difference = $1.30 - $1.20 = $0.10

ANSWER 4: B

---

Problem 5:
(1901 + 1902 + ... + 1993) − (101 + 102 + ... + 193)

Both sums have 93 terms each.

First sum = 93 × (1901 + 1993)/2 = 93 × 1947
Second sum = 93 × (101 + 193)/2 = 93 × 147

Difference = 93 × (1947 − 147) = 93 × 1800 = 167,400

ANSWER 5: A

---

Problem 6:
1 − 2 − 3 + 4 + 5 − 6 − 7 + 8 + ... + 1993 − 1994 − 1995 + 1996

The pattern repeats every 4 terms: +, −, −, +

Group k: (4k−3) − (4k−2) − (4k−1) + 4k = 4k−3−4k+2−4k+1+4k = 0

Each group of 4 sums to 0.

1996 ÷ 4 = 499 complete groups.

Total = 499 × 0 = 0

ANSWER 6: C

---

Problem 7:
50 people wear sunglasses, 35 wear caps.

P(sunglasses | cap) = 2/5 means: (both)/(caps) = 2/5

Both/35 = 2/5 → Both = 14

P(cap | sunglasses) = (both)/(sunglasses) = 14/50 = 7/25

ANSWER 7: B

---

Problem 8:
3-digit numbers with digit-sum 26 that are even.

For number abc: a + b + c = 26, c is even, a ≥ 1, all digits ≤ 9.

- c = 8: a + b = 18 → only (9,9) → 998 ✓
- c = 6: a + b = 20 → impossible (max 9+9=18)
- c = 4, 2, 0: a + b ≥ 22 → impossible

Only 1 number: 998

ANSWER 8: A

---

Problem 9:
Three-digit numbers with digit-sum 25.

a + b + c = 25, a ≥ 1, all digits ≤ 9.

- a = 9: b + c = 16 → (7,9), (8,8), (9,7) → 3 numbers
- a = 8: b + c = 17 → (8,9), (9,8) → 2 numbers
- a = 7: b + c = 18 → (9,9) → 1 number
- a ≤ 6: b + c ≥ 19 → impossible

Total = 3 + 2 + 1 = 6

ANSWER 9: C

---

Problem 10:
Distance from exit 5 to exit 26 is 118 miles. There are 21 gaps between consecutive exits (5→6, 6→7, ..., 25→26).

Each gap ≥ 5 miles. To maximize one gap, minimize all others.

If 20 gaps are minimized at 5 miles each: 20 × 5 = 100 miles.

Maximum single gap = 118 − 100 = 18 miles.

ANSWER 10: C

---

Problem 11:
Five scores: mean = 90, median = 91, mode = 94.

Sum = 5 × 90 = 450

Sorted: a ≤ b ≤ c ≤ d ≤ e
- c = 91 (median)
- Mode = 94, so 94 appears at least twice. Since c = 91, the 94s must be d and e.

a + b + 91 + 94 + 94 = 450
a + b = 171

ANSWER 11: B

---

Problem 12:
Using +, −, × each exactly once in 5 __ 4 __ 6 __ 3:

| Operations | Result |
|------------|--------|
| 5 + 4 − 6 × 3 | 5 + 4 − 18 = −9 |
| 5 + 4 × 6 − 3 | 5 + 24 − 3 = 26 |
| 5 − 4 + 6 × 3 | 5 − 4 + 18 = 19 |
| 5 − 4 × 6 + 3 | 5 − 24 + 3 = −16 |
| 5 × 4 + 6 − 3 | 20 + 6 − 3 = 23 |
| 5 × 4 − 6 + 3 | 20 − 6 + 3 = 17 |

From choices {9, 10, 15, 16, 19}, only **19** is achievable.

ANSWER 12: E

**Problem 1:**  
We have nine distinct integers, six of which are known: 2, 3, 4, 6, 9, 14.  
The median of a nine‑element set is the 5th smallest number.  

Let the unknown integers be \(x,y,z\) (all distinct and distinct from the known six).  
When the numbers are sorted, the median is the 5th term.  

For a known number \(k_i\) (the \(i\)‑th smallest of the known six) to be the median, we need exactly four numbers smaller than it.  
If \(i-1\) known numbers are smaller, we need \(L\) unknown numbers smaller with  
\((i-1)+L=4\Rightarrow L=5-i\).  
Since we have only three unknowns, \(L\ge0\) gives \(i\le5\); also \(L\le3\) gives \(i\ge2\).  
Thus \(i=2,3,4,5\) → medians could be 3, 4, 6, 9.

Can the median be an unknown integer?  
Let the median be an unknown \(u\).  
If \(u\) lies between the known numbers, we need exactly four numbers below it.  
With two other unknowns, the possibilities are:
- \(u\) between 4 and 6 → \(u=5\);
- \(u\) between 6 and 9 → \(u=7\) or \(u=8\).

All three of these can be realized while keeping the numbers distinct.  
Thus the possible median values are \(\{3,4,5,6,7,8,9\}\), a total of **7** values.

**Answer:** 7 → Choice **D**.  

ANSWER 1: D  

---

**Problem 2:**  
We start with the list 2, 6, 7, 7, 28 and insert a number \(N\).  
Let the six numbers be sorted. The median of six numbers is the average of the 3rd and 4th smallest values.

Let the sorted list after insertion be \(a_1\le a_2\le a_3\le a_4\le a_5\le a_6\).  
The median is \(\dfrac{a_3+a_4}{2}\).  
The mean is \(\dfrac{\text{sum}}{6}\).  
Given that the mean equals twice the median:
\[
\frac{\text{sum}}{6}=2\cdot\frac{a_3+a_4}{2}=a_3+a_4 .
\]
Hence \(\text{sum}=6(a_3+a_4)\).

Consider where \(N\) can fall:

*If \(N\le2\):* \(a_3=6,\ a_4=7\) → median \(=6.5\).  
Sum \(=N+50\). Equation gives \(N=28\), contradicting \(N\le2\).

*If \(2<N\le6\):* \(a_3=6,\ a_4=7\) → same median \(=6.5\).  
Again \(N=28\), impossible.

*If \(6<N\le7\):* only integer possibility \(N=7\).  
List becomes 2, 6, 7, 7, 7, 28 → median \(=7\).  
Sum \(=57\) → mean \(=9.5\neq14\).

*If \(7<N<28\):* \(a_3=7,\ a_4=7\) → median \(=7\).  
Sum \(=N+50\). Equation \(N+50=84\) gives \(N=34\), which is not \(<28\).

*If \(N=28\):* median still \(=7\). Sum \(=78\) → mean \(=13\neq14\).

*If \(N>28\):* \(a_3=7,\ a_4=7\) → median \(=7\).  
Equation gives \(N+50=84\) → \(N=34\). This works (34 > 28).

Thus the only possible value is **34**.

**Answer:** 34 → Choice **E**.  

ANSWER 2: E  

---

**Problem 3:**  
A four‑digit integer with all digits even can use digits \{0,2,4,6,8\} for each place, except the thousands digit cannot be 0.  

Number of such integers: \(4\times5\times5\times5=500\).

A number is divisible by 4 iff its last two digits form a number divisible by 4.  
List all two‑digit endings with even digits (25 possibilities).  

Checking each yields 15 endings that are multiples of 4:  
00, 04, 08, 20, 24, 28, 40, 44, 48, 60, 64, 68, 80, 84, 88.

Thus the proportion is \(\frac{15}{25}=\frac{3}{5}\).

**Answer:** \(\frac{3}{5}\) → Choice **D**.  

ANSWER 3: D  

---

**Problem 4:**  
6 % tax on \$20 = \(0.06\times20 = \$1.20\).  
6.5 % tax on \$20 = \(0.065\times20 = \$1.30\).  

Difference = \$1.30 − \$1.20 = **\$0.10**.

**Answer:** \$0.10 → Choice **B**.  

ANSWER 4: B  

---

**Problem 5:**  
Sum from 1901 to 1993 (inclusive):  
Number of terms = \(1993-1901+1 = 93\).  
Sum = \(\frac{(1901+1993)\times93}{2}= \frac{3894\times93}{2}=181\,071\).

Sum from 101 to 193 (inclusive):  
Number of terms = \(193-101+1 = 93\).  
Sum = \(\frac{(101+193)\times93}{2}= \frac{294\times93}{2}=13\,671\).

Difference = \(181\,071-13\,671 = 167\,400\).

**Answer:** 167 400 → Choice **A**.  

ANSWER 5: A  

---

**Problem 6:**  
The sign pattern for the numbers \(n\) is:
- \(n\equiv1\pmod4\) → +,
- \(n\equiv2\pmod4\) → −,
- \(n\equiv3\pmod4\) → −,
- \(n\equiv0\pmod4\) → +.

Thus each block of four consecutive integers \(\{4k+1,4k+2,4k+3,4k+4\}\) contributes  
\((4k+1)-(4k+2)-(4k+3)+(4k+4)=0\).

From 1 through 1996 there are \(1996/4=499\) complete blocks, so the total sum is **0**.

**Answer:** 0 → Choice **C**.  

ANSWER 6: C  

---

**Problem 7:**  
Let \(S\) = people wearing sunglasses, \(|S|=50\).  
Let \(C\) = people wearing caps, \(|C|=35\).  
Let \(x=|S\cap C|\).

Given \(P(S\mid C)=\frac{x}{35}= \frac{2}{5}\).  
Thus \(x = \frac{2}{5}\times35 = 14\).

We need \(P(C\mid S)=\frac{x}{50}= \frac{14}{50}= \frac{7}{25}\).

**Answer:** \(\frac{7}{25}\) → Choice **B**.  

ANSWER 7: B  

---

**Problem 8:**  
A three‑digit number with digit sum 26 must use digits that sum to 26.  
The only triple of single‑digit numbers with sum 26 is \(\{9,9,8\}\).

The number is even only if its units digit is 8.  
Among the permutations of 9, 9, 8, only 998 ends in 8.  

Thus exactly **1** such number exists.

**Answer:** 1 → Choice **A**.  

ANSWER 8: A  

---

**Problem 9:**  
Let the digits be \(a,b,c\) (hundreds, tens, units) with  
\(1\le a\le9,\ 0\le b,c\le9,\ a+b+c=25\).

Set \(a'=a-1\) (so \(0\le a'\le8\)). Then  
\(a'+b+c=24\).

Count unrestricted non‑negative solutions: \(\binom{24+3-1}{2}= \binom{26}{2}=325\).

Subtract solutions where a digit exceeds its bound:
- \(a'\ge9\): treat \(a''=a'-9\) → \(a''+b+c=15\) → \(\binom{17}{2}=136\).
- \(b\ge10\): \(b'=b-10\) → \(a'+b'+c=14\) → \(\binom{16}{2}=120\).
- \(c\ge10\): similarly 120.

Intersections:
- \(a'\ge9,\ b\ge10\): \(a''+b'+c=5\) → \(\binom{7}{2}=21\).
- \(a'\ge9,\ c\ge10\): 21.
- \(b\ge10,\ c\ge10\): \(a'+b'+c'=4\) → \(\binom{6}{2}=15\).

No triple intersection (would require sum –5).

By inclusion–exclusion, invalid = \(136+120+120-21-21-15 = 319\).  
Valid = \(325-319 = 6\).

**Answer:** 6 → Choice **C**.  

ANSWER 9: C  

---

**Problem 10:**  
Exits numbered 5 through 26 give 22 exits, hence 21 intervals.  
Let the distances between consecutive exits be \(d_1,\dots,d_{21}\).  
We know \(\sum d_i = 118\) miles, and each \(d_i\ge5\).

To maximize the largest interval, make the othe

Problem 1:  
We are given a set $ R $ of nine distinct integers, six of which are $ 2, 3, 4, 6, 9, 14 $. The median of a set of nine numbers is the fifth number when sorted. To find the number of possible values for the median, we consider the possible placements of the three unknown numbers relative to the known ones. The known numbers sorted are $ 2, 3, 4, 6, 9, 14 $. The three unknown numbers can be placed in various positions to shift the median. The possible medians occur when the additional numbers are placed such that the fifth number in the sorted list can be $ 4, 5, 6, 7, 8, $ or $ 9 $. This gives six possible values for the median.  
ANSWER 1: D (corrected to C? Wait, no: the possible medians are 4, 5, 6, 7, 8, 9 — that’s 6 values, so C)  
**Wait, let's re-evaluate:** The known numbers are 2,3,4,6,9,14. Adding three numbers. The median is the 5th number. To get 4 as median: three numbers ≤4, but only 2,3,4 are available — but they are distinct, so we can add numbers like 1,5? Wait no, the three unknown numbers can be placed in different positions. For example:  
- If all three unknown numbers are ≤4: but only three such numbers exist (2,3,4) but they are already in the set. So we cannot add more. Thus, the unknown numbers must be placed in other positions.  
Actually, the unknown numbers can be:  
- All less than 2: median could be 4 if three numbers are added <4, but only 2,3,4 are already present. We cannot add numbers less than 2 without repeating? Wait, the set has distinct integers. So we can add numbers like 1,0,-1, but the problem doesn’t restrict to positive. However, the known numbers are positive. But the problem says "distinct integers", so negatives are allowed. But the median values: if we add numbers less than 2, the sorted order becomes [new numbers, 2,3,4,6,9,14]. With three new numbers, the sorted list has 9 numbers. The fifth is the median. If we add three numbers less than 2, the sorted list is [a,b,c,2,3,4,6,9,14] — no, wait, adding three numbers less than 2: the sorted list would be [a,b,c,2,3,4,6,9,14] — but a,b,c are less than 2, so the first three, then 2 is fourth, 3 is fifth? Wait, no: the original known numbers are 2,3,4,6,9,14. Adding three numbers, say x,y,z. The full sorted list will be the combination. To get median=4, we need the fifth number to be 4. So in the sorted list of 9, the fifth must be 4. That requires that four numbers are ≤4 and four are ≥4? No: the fifth is the middle. For the fifth to be 4, we need at least five numbers ≤4? No: in a sorted list of 9, the fifth is the median. For the fifth to be 4, we need at least five numbers ≤4? Actually, no: the fifth number is the one where four are less and four are greater? No: in a sorted list of 9, the fifth is the median, meaning four are ≤ median and four are ≥ median. But since numbers can be equal, but here all distinct. So for 4 to be the median, we need exactly four numbers less than 4 and four numbers greater than 4? But 4 is in the set. So the sorted list must have four numbers <4, then 4, then four numbers >4. But the known numbers include 2,3,4,6,9,14. So 2,3 are <4, and 6,9,14 are >4. So currently, without unknowns, we have two <4 and three >4. Adding three unknowns. To have four <4 and four >4, we need to add two numbers <4 and one number >4? But then the sorted list would be [two new <4, 2,3,4, then the new >4, then 6,9,14] — but that’s only three numbers after 4? Wait, no: the sorted list would be [new <4 (two), 2,3,4, new >4, 6,9,14] — that’s 9 numbers? 2+3+1+3=9? Wait, no: the two new <4, then 2,3,4 (three numbers), then the new >4, then 6,9,14 (three numbers). So total: 2+3+1+3=9. The fifth number is 4. So yes, we can achieve median=4 by adding two numbers <4 and one number >4. Similarly, we can achieve:  
- Median=5: by adding numbers such that the fifth number is 5. But 5 is not in the set. So we need the fifth number to be 5. This requires that four numbers are <5 and four are >5. But 5 is not present. So we need to add 5? But the set must consist of distinct integers, and 5 is not in the given. So we can add 5. Then the sorted list would include 5. For example, add 5, and two other numbers. To have four <5 and four >5: currently, without unknowns, numbers <5: 2,3,4 (three), numbers >5: 6,9,14 (three). So to have four <5, we need to add one number <5. To have four >5, we need to add one number >5. But we are adding three numbers. So we can add 5, and say 1 and 10. Then sorted: 1,2,3,4,5,6,9,10,14. Median is 5. So median=5 is achievable. Similarly, median=6: we can add numbers such that the fifth is 6. Currently, numbers <6: 2,3,4 (three), numbers >6: 9,14 (two). So to have four <6 and four >6, we need to add one number <6 and two numbers >6. For example, add 5,10,15. Then sorted: 2,3,4,5,6,9,10,14,15. Median=6. Similarly, median=7: add numbers so fifth is 7. Currently, <7: 2,3,4,6 (four), >7: 9,14 (two). So to have four <7 and four >7, we need to add zero <7 (since already four) and two >7. But we are adding three numbers. So we can add 7,8,15. Then sorted: 2,3,4,6,7,8,9,14,15. Median=7. Similarly, median=8: add 8, and two numbers, say 10,11. Then sorted: 2,3,4,6,8,9,10,11,14. Median=8? The fifth is 8. But to have four <8 and four >8: currently, without unknowns, <8: 2,3,4,6 (four), >8:9,14 (two). So to have four >8, we need to add two numbers >8. But we are adding three numbers. So we can add 8,9,10? But 9 is already there. Add 8,10,11. Then sorted: 2,3,4,6,8,9,10,11,14. The fifth is 8. So yes. Similarly, median=9: add numbers so fifth is 9. Currently, <9: 2,3,4,6 (four), >9:14 (one). So to have four >9, we need to add three numbers >9. But we are adding three numbers. So add 10,11,12. Then sorted: 2,3,4,6,9,10,11,12,14. Median=9. So possible medians: 4,5,6,7,8,9. That’s six values.  
ANSWER 1: C  

Problem 2:  
We are given the list 2,6,7,7,28. A number $ N $ is added. The mean becomes twice the median. The original list has five numbers. After adding $ N $, there are 

### Problem 1:
We have a set of 9 distinct integers, so the median (middle value when sorted) is the 5th element. The 6 known numbers sorted are [2,3,4,6,9,14], with 3 additional distinct integers to add. To find possible median values:
- The minimum possible median is 3: add 3 numbers smaller than 2, sorted set becomes [a,b,c,2,3,4,6,9,14], 5th term=3.
- The maximum possible median is 9: add 3 numbers larger than 14, sorted set becomes [2,3,4,6,9,14,x,y,z], 5th term=9.
- All integers between 3 and 9 are valid medians: 3,4,5,6,7,8,9, for 7 total possible values.
Tempting wrong choices: any count less than 7, from missing intermediate medians that can be formed by adding integers between the known values.
ANSWER 1: D

---

### Problem 2:
Original list sum: 2+6+7+7+28=50. Adding N gives a list of 6 elements, sum=50+N, mean=(50+N)/6. The median of 6 elements is the average of the 3rd and 4th sorted terms. The problem states mean=2×median, so 50+N=12×median.
- If N>28, sorted list is [2,6,7,7,28,N], median=(7+7)/2=7. Then 50+N=12×7=84 → N=34, which fits.
Testing other options confirms no other N satisfies the condition.
ANSWER 2: E

---

### Problem 3:
A number is divisible by 4 if its last two digits form a number divisible by 4. For 4-digit numbers with all even digits:
- Total count: 4 (non-zero thousands digits:2,4,6,8) ×5×5×5 (all even digits for other places)=500.
- Valid last two-digit pairs (both even, divisible by 4): For any even tens digit, 10×even is divisible by 4, so only units digits 0,4,8 work, giving 3 valid units digits per tens digit: 5×3=15 valid pairs.
- Fraction: (4×5×15)/500=300/500=3/5.
ANSWER 3: D

---

### Problem 4:
The tax rate difference is 6.5%-6%=0.5%. 0.5% of $20 is 0.005×20=$0.10.
ANSWER 4: B

---

### Problem 5:
Both sums have the same number of terms: 1993-1901+1=93 terms, 193-101+1=93 terms. Subtract term-by-term: (1901-101)+(1902-102)+...+(1993-193)=1800×93=167400.
ANSWER 5: A

---

### Problem 6:
Group terms in sets of 4: (1-2-3+4)+(5-6-7+8)+... each group sums to 0. There are 1996/4=499 groups, total sum=0.
ANSWER 6: C

---

### Problem 7:
Number of people with both caps and sunglasses: 35 cap-wearers × 2/5=14. Probability a sunglass-wearer has a cap:14/50=7/25.
ANSWER 7: B

---

### Problem 8:
An even 3-digit number with digit sum 26 must have an even units digit. The only possible even units digit is 8 (any smaller even digit leaves a sum of the first two digits >18, impossible). The only number is 998, so 1 valid number.
ANSWER 8: A

---

### Problem 9:
Three-digit numbers with digit sum 25 are all permutations of (9,9,7) and (9,8,8). There are 3 permutations of (9,9,7) and 3 permutations of (9,8,8), for 6 total numbers.
ANSWER 9: C

---

### Problem 10:
Between exit 5 and 26, there are 21 gaps. To maximize one gap, minimize the other 20 gaps at 5 miles each: 20×5=100 miles. Maximum gap=118-100=18 miles.
ANSWER 10: C

---

### Problem 11:
Total sum of 5 scores=5×90=450. Median=91, so sorted scores [a,b,91,c,d]. Mode=94, so 94 must appear at least twice, so c=d=94, sum c+d=188. Sum of the two lowest scores=450-91-188=171.
ANSWER 11: B

---

### Problem 12:
Using +,-,* exactly once, one valid calculation: 5-4+6×3=1+18=19, which uses all three operations and matches option E.
ANSWER 12: E

### Problem 1:
We need to find the number of possible median values for a set of 9 distinct integers, 6 of which are 2, 3, 4, 6, 9, 14. The median of 9 numbers is the 5th element when sorted in ascending order. We add 3 distinct integers (not in the given set) and analyze possible medians:
1.  **Case 1: All 3 new numbers < 2**: Sorted list is [a,b,c,2,3,4,6,9,14], median = 3.
2.  **Case 2: 2 new numbers < 2, 1 new number ≥4**: Sorted list is [a,b,2,3,4,...], median =4.
3.  **Case 3: 1 new number <2**:
    - If we add 5 (the only integer between 4 and 6): Sorted list is [a,2,3,4,5,6,...], median=5.
    - If we do not add 5: Sorted list is [a,2,3,4,6,...], median=6.
4.  **Case 4: 0 new numbers <2 (all new numbers ≥5)**:
    - Add 5: Sorted list [2,3,4,5,6,...], median=6.
    - Add a number between 6 and9 (7 or 8): Sorted list [2,3,4,6,7,...] or [2,3,4,6,8,...], median=7 or 8.
    - Add all new numbers ≥9: Sorted list [2,3,4,6,9,...], median=9.
The only impossible medians are 2 (needs 5 numbers ≤2, only 4 max) and values >9 (existing numbers already have 5 elements ≤9). The possible medians are 3,4,5,6,7,8,9: 7 values.
ANSWER 1: D
---
### Problem 2:
Original list: [2,6,7,7,28], sum = 2+6+7+7+28=50. Insert N to make 6 elements. The median of 6 elements is the average of the 3rd and 4th sorted elements. The mean is $\frac{50+N}{6}$, which equals twice the median:
$$\frac{50+N}{6} = 2 \times \text{median} \implies 50 + N = 12 \times \text{median} \implies N = 12 \times \text{median} -50.$$
Test possible median values from the sorted list with N:
- If N ≤7: Sorted list has 3rd=7, 4th=7, median=7. Then $N=12*7 -50=84-50=34$, which is >7, valid.
- Other median values (e.g., 6.5) give N=28, which does not fit the required sorted position.
Verify N=34: Sorted list [2,6,7,7,28,34], median=(7+7)/2=7, mean=(50+34)/6=14, which is twice 7.
ANSWER 2: E
---
### Problem 3:
Positive 4-digit numbers with all even digits:
- Thousands place: 2,4,6,8 (4 choices, no 0).
- Hundreds, tens, units: 0,2,4,6,8 (5 choices each).
Total numbers: $4 \times 5 \times 5 \times5=500$.
A number is divisible by 4 iff its last two digits form a number divisible by 4. For even tens digits (0,2,4,6,8), a two-digit number $10a + b$ is divisible by 4 iff $b$ (units digit) is 0,4,8 (since $10a \equiv 0 \mod4$ for even a). So valid endings: 5 (tens choices) ×3 (units choices)=15.
Total valid numbers: $4 \times5 \times5 \times3=300$. Fraction: $\frac{300}{500}=\frac{3}{5}$.
ANSWER 3: D
---
### Problem 4:
Difference between 6.5% and 6% tax on $20:
$$0.065*20 - 0.06*20 = (0.065-0.06)*20=0.005*20=0.10.$$
ANSWER 4: B
---
### Problem 5:
Both sums have 93 terms (1993-1901+1=93, 193-101+1=93). Each term in the first sum is 1800 more than the corresponding term in the second sum:
Total difference = $93 \times 1800 = 167400$.
ANSWER 5: A
---
### Problem 6:
The sign pattern repeats every 4 terms: $+ - - +$, and each block of 4 sums to 0:
$$(1-2-3+4)+(5-6-7+8)+\dots+(1993-1994-1995+1996)=0.$$
1996 is divisible by 4, so there are 499 such blocks, total sum 0.
ANSWER 6: C
---
### Problem 7:
Let $S$ = sunglass wearers ($|S|=50$), $C$ = cap wearers ($|C|=35$). $P(S|C)=\frac{|S∩C|}{|C|}=\frac{2}{5}$, so $|S∩C|=\frac{2}{5}*35=14$.
The probability a sunglass wearer wears a cap is $P(C|S)=\frac{|S∩C|}{|S|}=\frac{14}{50}=\frac{7}{25}$.
ANSWER 7: B
---
### Problem 8:
3-digit numbers with digit sum 26: maximum digit sum is 27 (999), so 26 is 1 less than maximum, meaning digits are permutations of (9,9,8). The only even permutation is 998 (units digit 8). So 1 number.
ANSWER 8: A
---
### Problem 9:
3-digit numbers with digit sum 25: 25 is 2 less than 27, so digits are permutations of (9,9,7) and (9,8,8):
- Permutations of (9,9,7): 997,979,799 (3 numbers).
- Permutations of (9,8,8): 988,898,889 (3 numbers).
Total 6 numbers.
ANSWER 9: C
---
### Problem 10:
Between exit 5 and 26, there are $26-5=21$ gaps, total distance 118 miles. Each gap ≥5 miles. To maximize one gap, minimize the other 20: $20*5=100$, so maximum gap = $118-100=18$.
ANSWER 10: C
---
### Problem 11:
Five sorted scores: $a≤b≤c≤d≤e$. Median $c=91$, mean 90, so total sum $5*90=450$, so $a+b+d+e=450-91=359$. Mode is 94, so 94 appears at least twice. Since $c=91$, 94 can only be in $d$ and $e$, so $d=e=94$. Then $a+b=359-94-94=171$. This sum is fixed (valid examples exist, e.g., 81+90=171).
ANSWER 11: B
---
### Problem 12:
Permute +, -, × in 5 __4 __6 __3, using standard order of operations (multiply first):
- $5 -4 +6×3 = 5-4+18=19$, which is an option.
All other permutations give values not in the choices.
ANSWER 12: E

I'll solve each problem step by step.

---

**Problem 1:**

We have 6 elements: {2, 3, 4, 6, 9, 14}, and need to add 3 more distinct integers to make 9 total.

The median of 9 numbers is the 5th element when arranged in order.

Currently we have 6 numbers. By strategically placing 3 new numbers, we can shift which value becomes the median.

Let me consider what values can be the median (5th position):
- If we add 3 numbers all ≤ 2, the sorted list has our 6 original numbers in positions 4-9, so median is 4
- If we add numbers strategically between and around the original set, we can make the median be: 2, 3, 4, 6, 9, 14, or a new number
- If we add 3 numbers all ≥ 14, the sorted list has our 6 original numbers in positions 1-6, so median is 9

Testing systematically:
- Median = 2: add three numbers < 2 (e.g., -2, -1, 0) ✓
- Median = 3: add two < 2, one between 3 and 4 ✓
- Median = 4: add three < 2, or arranged appropriately ✓
- Median = 6: add one or two strategically ✓
- Median = 9: add three > 14, or arranged appropriately ✓
- Median = 14: add three > 14 ✓
- Median = a new value between existing numbers ✓

The possible medians are: 2, 3, 4, 6, 9, 14, and any new integer we insert. We can get 7 different values.

**ANSWER 1: D**

---

**Problem 2:**

Original list: 2, 6, 7, 7, 28. Sum = 50.

When N is inserted, we have 6 numbers. Let's find the median of 6 numbers (average of 3rd and 4th when sorted).

Sum of all 6 numbers = 50 + N
Mean = (50 + N)/6

We need: Mean = 2 × Median

Let me test by considering where N falls:
- If N ≤ 2: sorted list is N, 2, 6, 7, 7, 28. Median = (6+7)/2 = 6.5. Mean = 2×6.5 = 13, so 50+N = 78, N = 28. Contradiction.
- If 2 < N ≤ 6: sorted list is 2, N, 6, 7, 7, 28. Median = (6+7)/2 = 6.5. Mean = 13, so N = 28. Contradiction.
- If 6 < N ≤ 7: sorted list is 2, 6, N, 7, 7, 28. Median = (N+7)/2. Mean = 2×(N+7)/2 = N+7. So (50+N)/6 = N+7, giving 50+N = 6N+42, so 5N = 8, N = 1.6. Not in range.
- If 7 < N ≤ 28: sorted list is 2, 6, 7, N, 7, 28 or 2, 6, 7, 7, N, 28. Median = (7+N)/2. Mean = 2×(7+N)/2 = 7+N. So (50+N)/6 = 7+N, giving 50+N = 42+6N, so 5N = 8, N = 1.6. Not valid.
- If N > 28: sorted list is 2, 6, 7, 7, 28, N. Median = (7+7)/2 = 7. Mean = 14. So 50+N = 84, N = 34. ✓

**ANSWER 2: E**

---

**Problem 3:**

Four-digit numbers with all even digits: first digit ∈ {2,4,6,8}, others ∈ {0,2,4,6,8}.

Total count: 4 × 5 × 5 × 5 = 500

For divisibility by 4, the last two digits must form a number divisible by 4.

Last two digits from {00, 02, 04, 06, 08, 20, 22, 24, 26, 28, 40, 42, 44, 46, 48, 60, 62, 64, 66, 68, 80, 82, 84, 86, 88}

Divisible by 4: 00, 04, 08, 20, 24, 28, 40, 44, 48, 60, 64, 68, 80, 84, 88 (15 out of 25)

Count divisible by 4: 4 × 5 × 15 = 300

But wait, the second digit is 5 choices, third is which determines divisibility by 4.

Actually: 4 × 5 × (number of even two-digit combos divisible by 4) = 4 × 5 × 15 = 300

Hmm, let me recalculate. First digit: 4 choices. Second digit: 5 choices. Last two digits must be divisible by 4: check all pairs from {0,2,4,6,8}×{0,2,4,6,8} giving 15 valid pairs.

Count = 4 × 5 × 15 = 300. Fraction = 300/500 = 3/5.

**ANSWER 3: D**

---

**Problem 4:**

6.5% tax on $20: 0.065 × 20 = $1.30
6% tax on $20: 0.06 × 20 = $1.20
Difference: $1.30 - $1.20 = $0.10

**ANSWER 4: B**

---

**Problem 5:**

Sum from 1901 to 1993: This is 93 consecutive integers starting at 1901.
Sum = (1901 + 1993)/2 × 93 = 1947 × 93 = 181,071

Sum from 101 to 193: This is 93 consecutive integers starting at 101.
Sum = (101 + 193)/2 × 93 = 147 × 93 = 13,671

Difference = 181,071 - 13,671 = 167,400

**ANSWER 5: A**

---

**Problem 6:**

Pattern: 1 - 2 - 3 + 4 + 5 - 6 - 7 + 8 + 9 - 10 - 11 + ...

Group by 4s: (1-2-3+4) + (5-6-7+8) + (9-10-11+12) + ... + (1993-1994-1995+1996)

Each group of 4: (4k+1) - (4k+2) - (4k+3) + (4k+4) = 4k+1-4k-2-4k-3+4k+4 = 0

From 1 to 1996, we have 1996/4 = 499 complete groups, each summing to 0.

Total = 0

**ANSWER 6: C**

---

**Problem 7:**

Let x = people wearing both sunglasses and caps.

P(sunglasses | cap) = x/35 = 2/5, so x = 14.

P(cap | sunglasses) = 14/50 = 7/25.

**ANSWER 7: B**

---

**Problem 8:**

Three-digit even numbers with digit sum 26.

For an even number, last digit ∈ {0, 2, 4, 6, 8}.

If last digit = 0: first two digits sum to 26. Max is 9+9=18. Impossible.
If last digit = 2: first two digits sum to 24. Only 9+9=18 < 24. Impossible.
If last digit = 4: first two digits sum to 22. Impossible (max 18).
If last digit = 6: first two digits sum to 20. Impossible.
If last digit = 8: first two digits sum to 18. This requires 9+9=18. Number: 998. ✓

Only 1 number.

**ANSWER 8: A**

---

**Problem 9:**

Three-digit numbers abc where a+b+c=25, with a ∈ {1,...,9}, b,c ∈ {0,...,9}.

a+b+c = 25 means b+c = 25-a.

For valid solutions: 0 ≤ b+c ≤ 18, so 7 ≤ a ≤ 9.

- a=7: b+c=18. Solutions: (9,9) → 1 way
- a=8: b+c=17. Solutions: (8,9), (9,8) → 2 ways
- a=9: b+c=16. Solutions: (7,9), (8,8), (9,7) → 3 ways

Total = 1 + 2 + 3 = 6

**ANSWER 9: C**

---

**Problem 10:**

Exits from 5th to 26th = 26-5 = 21 segments between consecutive exits.

Each segment ≥ 5 miles. Total distance = 118 miles.

If 20 segments are 5 miles each: 20 × 5 = 100 miles.
Remaining for the largest segment: 118 - 100 = 18 miles.

**ANSWER 10: C**

---

**Problem 11:**

Five scores: mean = 90 (sum = 450), median = 91, mode = 94.

Arranged in order: a, b, c, d, e where c = 91 (median).

Mode = 94 means 94 appears at least twice. Since median is 91, we must have 94 appearing in {d, e}.

Case 1: d = e = 94. Then a + b + 91 + 94 + 94 = 450, so a + b = 171.

Check consistency: we need a ≤ b ≤ 91 and mode is 94 (appears twice). If a or b equals 94, mode wouldn't be unique. So a, b < 91 works. ✓

**ANSWER 11: B**

---

**Problem 12:**

We need to place +, −, × in three blanks: 5 __ 4 __ 6 __ 3

Trying combinations:
- 5 + 4 × 6 − 3 = 5 + 24 − 3 = 26
- 5 × 4 + 6 − 3 = 20 + 6 − 3 = 23
- 5 × 4 − 6 + 3 = 20 − 6 + 3 = 17
- 5 + 4 − 6 × 3 = 5 + 4 − 18 = −9
- 5 − 4 

Problem 1:

We are given a set of 9 distinct integers. Six of them are
\[
2,3,4,6,9,14.
\]
We want the number of possible values of the median of the 9 numbers.

For 9 numbers, the median is the 5th number when the list is ordered.

Now look at the six given numbers in sorted order:
\[
2,3,4,6,9,14.
\]
Since there are 3 other distinct integers, the median can be any number that ends up in the 5th position after inserting those 3 numbers.

Let’s see what values are possible:

- If we place 3 new numbers all very small, then the 5th number could be 4.
- It could also be 6, or 9, or even 14, depending on how many new numbers are inserted before each.
- It cannot be 2 or 3, because there would need to be at least 4 numbers less than or equal to it, but only 3 extra numbers are available and the given numbers already limit this.
- It cannot be any number not among the given six if the median must be one of the existing numbers? Let’s check carefully: since the 3 extra numbers are distinct integers and can be placed anywhere, the 5th number could indeed be one of the inserted numbers, but then it would have to lie between the 4th and 6th numbers. Since the six given numbers already occupy several possible positions, the median can also be a new integer. But because the set is of distinct integers and the question asks how many possible values, we should count all possible medians that can occur.

A cleaner way: the median of 9 distinct integers is the middle element. With 6 fixed numbers, the 3 added numbers can be arranged so that the median is any integer between the 4th smallest and 6th smallest of the fixed numbers, inclusive, but here the possible distinct medians are:
\[
4,5,6,7,8,9
\]
Wait — can all of these really happen? Let’s test.

- Median 4: choose 3 new numbers less than 4.
- Median 5: choose 3 new numbers so that 5 is between 4 and 6 and becomes 5th.
- Median 6: choose 3 new numbers with 6 as middle.
- Median 7, 8, 9 similarly.

So the possible medians are the integers from 4 through 9, which is 6 values.

However, we must match the provided choices. The intended count is the number of possible median values among the listed integers and inserted integers. Since the only restriction is distinct integers, every integer from 4 through 9 can occur.

That gives 6 possible values.

ANSWER 1: C

---

Problem 2:

We insert a number \(N\) into the list
\[
2,6,7,7,28.
\]
Then the mean becomes twice the median.

Let’s analyze the median after inserting \(N\) into 6 numbers total.

The original sorted list is:
\[
2,6,7,7,28.
\]
There are 5 numbers. After inserting one number, there are 6 numbers, so the median is the average of the 3rd and 4th numbers in the sorted list.

We try the choices.

If \(N=7\), the list becomes
\[
2,6,7,7,7,28.
\]
Median:
\[
\frac{7+7}{2}=7.
\]
Mean:
\[
\frac{2+6+7+7+7+28}{6}=\frac{57}{6}=9.5.
\]
Twice the median would be 14, not 9.5, so not 7.

Try \(N=14\):
\[
2,6,7,7,14,28.
\]
Median:
\[
\frac{7+7}{2}=7.
\]
Mean:
\[
\frac{2+6+7+7+14+28}{6}=\frac{64}{6}=\frac{32}{3},
\]
not 14.

Try \(N=20\):
\[
2,6,7,7,20,28.
\]
Median is still 7.
Mean:
\[
\frac{2+6+7+7+20+28}{6}=\frac{70}{6}=\frac{35}{3},
\]
not 14.

Try \(N=28\):
\[
2,6,7,7,28,28.
\]
Median is 7.
Mean:
\[
\frac{2+6+7+7+28+28}{6}=\frac{78}{6}=13,
\]
not 14.

Try \(N=34\):
\[
2,6,7,7,28,34.
\]
Median is 7.
Mean:
\[
\frac{2+6+7+7+28+34}{6}=\frac{84}{6}=14.
\]
Twice the median is
\[
2\cdot 7=14.
\]
This works.

So \(N=34\).

ANSWER 2: E

---

Problem 3:

We consider all positive four-digit integers whose digits are all even.

Even digits are
\[
0,2,4,6,8.
\]
A four-digit number cannot start with 0, so the thousands digit has 4 choices:
\[
2,4,6,8.
\]
Each of the other three digits has 5 choices, so the total number is
\[
4\cdot 5^3.
\]

Now determine how many are divisible by 4.

A number is divisible by 4 if its last two digits form a number divisible by 4.

So we only need to count ordered pairs of even digits \((t,u)\) from \(\{0,2,4,6,8\}\) such that \(10t+u\) is divisible by 4.

Check the possible last two digits:

- \(00\) yes
- \(02\) no
- \(04\) yes
- \(06\) no
- \(08\) yes

- \(20\) yes
- \(22\) no
- \(24\) yes
- \(26\) no
- \(28\) yes

- \(40\) yes
- \(42\) no
- \(44\) yes
- \(46\) no
- \(48\) yes

- \(60\) yes
- \(62\) no
- \(64\) yes
- \(66\) no
- \(68\) yes

- \(80\) yes
- \(82\) no
- \(84\) yes
- \(86\) no
- \(88\) yes

For each tens digit, 3 of the 5 possibilities work, so there are
\[
5\cdot 3=15
\]
valid last-two-digit pairs.

Since the first two digits can be anything allowed:
- thousands digit: 4 choices
- hundreds digit: 5 choices

Total divisible by 4:
\[
4\cdot 5\cdot 15
\]
Total numbers:
\[
4\cdot 5^3 = 4\cdot 125
\]
So the fraction is
\[
\frac{4\cdot 5\cdot 15}{4\cdot 5^3}=\frac{15}{25}=\frac35.
\]

ANSWER 3: D

---

Problem 4:

An item costs \$20 before tax.

Difference between 6.5% tax and 6% tax is
\[
0.5\% = 0.005.
\]
So the difference in tax is
\[
0.005 \times 20 = 0.1
\]
dollar, which is \(\$0.10\).

ANSWER 4: B

---

Problem 5:

We need to compute
\[
(1901+1902+\cdots+1993) - (101+102+\cdots+193).
\]

Notice each term in the second sum is exactly 1800 less than the corresponding term in the first sum:
\[
1901-101=1800,\quad 1902-102=1800,\ \ldots,\ 1993-193=1800.
\]
How many terms are there?

From 1901 to 1993 inclusive:
\[
1993-1901+1=93
\]
terms.

So the difference is
\[
93 \cdot 1800 = 167400.
\]

ANSWER 5: A

---

Problem 6:

We evaluate
\[
1-2-3+4+5-6-7+8+9-10-11+\cdots+1992+1993-1994-1995+1996.
\]

Look at the pattern in blocks of 4:
\[
(1-2-3+4),\ (5-6-7+8),\ (9-10-11+12),\ \ldots
\]
Each block of 4 consecutive integers has sum
\[
n-(n+1)-(n+2)+(n+3)=0
\]
when \(n\equiv 1 \pmod 4\).

So all full blocks cancel.

Now check the last block:
The pattern continues through \(1992+1993-1994-1995+1996\), which is not a full 4-term block. But we can group from 1 to 1996 in blocks of 4:
\[
(1-2-3+4)+(5-6-7+8)+\cdots+(1993-1994-1995+1996).
\]
Each block sums to 0.

Therefore the entire 

Problem 1:  
We have 9 distinct integers. Six of them are fixed: 2, 3, 4, 6, 9, 14. The other 3 are distinct integers of our choice.  
With 9 numbers, the median is the 5th number when sorted.

So the median depends only on how many of the fixed numbers fall below/at various positions, and where we insert the 3 new numbers.

Let’s analyze possible 5th elements.

### Step 1: Sort the fixed numbers
Fixed numbers in increasing order:
2, 3, 4, 6, 9, 14

The median (5th) could be one of these fixed numbers, or one of the new ones.

### Casework by choosing what the 5th number is

#### Can the median be 2?
For 2 to be the median, we need at least 4 numbers ≤ 2, but only 2 itself is ≤ 2 among the fixed set (and new numbers would have to be distinct integers ≤ 2).  
That means we’d need 4 numbers less than or equal to 2, including 2, so we’d need 3 additional distinct integers ≤ 2 (other than 2). That’s possible (e.g., 0, −1, 1).  
But then 2 would be the 5th, meaning exactly 4 numbers are <2. Can we guarantee? Actually, since we can choose any 3 extra integers, we can make exactly 4 numbers less than 2 by picking 3 integers <2 plus none from fixed besides 2. That would make 2 the 5th. However, note 2 is already one of the 9 distinct integers, so it can be the median.  
So potentially 2 is possible.

But the problem asks for the number of possible median values, and the choices are between 4 and 8, so likely only medians among the fixed list can work. Let’s check more carefully: the fixed numbers below 2 is only none (except 2). To make 2 the 5th, we need 4 numbers smaller than 2. We only control 3 extra numbers, so we can only create at most 3 numbers smaller than 2. Therefore, we cannot reach 4 smaller numbers.  
✅ Conclusion: median cannot be 2.

Similarly, we test each fixed candidate using “how many fixed numbers are below it”.

### General method
For a number \(x\) to be the median (5th), there must be exactly 4 numbers less than \(x\), and at least one number equal to \(x\). If \(x\) is one of the fixed numbers, we can count how many fixed numbers are less than \(x\), then see if the remaining “needed” smaller numbers can come from the 3 new integers.

Let’s compute fixed counts less than each candidate:

- Below 2: 0  
- Below 3: 1 (just 2)  
- Below 4: 2 (2,3)  
- Below 6: 3 (2,3,4)  
- Below 9: 4 (2,3,4,6)  
- Below 14: 5 (2,3,4,6,9)

We need exactly 4 numbers less than the median.

Now see possibilities:

#### Median = 3
Fixed numbers below 3: 1.  
We need 4 total below 3, so we must add 3 new numbers < 3. That’s possible.  
✅ So 3 can be median.

#### Median = 4
Fixed below 4: 2. Need 2 more from new numbers < 4. Possible.  
✅ 4 can be median.

#### Median = 6
Fixed below 6: 3. Need 1 more from new numbers < 6. Possible.  
✅ 6 can be median.

#### Median = 9
Fixed below 9: 4 already. Then all 4 smaller numbers are fixed, and the median can be 9 regardless of where new numbers go as long as none are between 9 and 14 in a way that changes ordering?  
We only need 9 to be the 5th, meaning at least 4 numbers are <9 (we have exactly 4 fixed ones). If we add any new numbers <9, that would make more than 4 numbers less than 9 and break it. So we must choose the 3 new integers all > 9. That’s possible.  
✅ 9 can be median.

#### Median = 14
Fixed below 14: 5 already. But we need exactly 4 below the median. Impossible.  
❌ 14 cannot be median.

#### Median = 2
Fixed below 2: 0, need 4 from new numbers <2, but we only have 3 new numbers. Impossible.  
❌ 2 cannot be median.

So the only fixed medians are: 3, 4, 6, 9 → 4 values.

Are there possible medians that are not among the fixed numbers (i.e., one of the 3 new integers)?  
Yes, if we choose all new integers appropriately.

Let the median be a new number \(y\). Then among the fixed numbers, some number of them must be less than \(y\). Since there are only 6 fixed numbers, the possible counts of fixed elements less than \(y\) range from 0 to 6.

But we need exactly 4 numbers total less than the median. Since the median is new, all 4 numbers less than it must come from:
- some fixed numbers less than it, plus
- some of the other two new numbers (because one new number is the median itself).

We can make any new median that sits in the “gaps” between fixed numbers such that exactly 4 elements end up below it.

The median \(y\) must be positioned so that:
- the number of fixed elements less than \(y\) is either 1,2,3, or 4 (because we then use the remaining smaller ones from the other new numbers, limited to 0,1,2, or 3).
But we must also ensure distinctness and that \(y\) lies in the right interval.

There are gaps:
(-∞,2), (2,3), (3,4), (4,6), (6,9), (9,14), (14,∞)

Try each gap for possible median:

- If \(y\in(2,3)\): fixed less than y = 1 (only 2). Need 3 more < y from the other new numbers. Possible.  
So any integer y in (2,3) means y=?? none (only integer between 2 and 3 doesn’t exist). So no.
- \(y\in(3,4)\): fixed less than y = 2. Need 2 more < y. Possible. But no integer between 3 and 4. none.
- \(y\in(4,6)\): fixed less than y = 3 (2,3,4). Need 1 more from new numbers <y. Possible. Integers in (4,6): only 5. So y=5 works.
- \(y\in(6,9)\): fixed less than y = 4 (2,3,4,6). Need 0 more. Then both other new numbers must be >y. Possible. Integers in (6,9): 7,8 → both could be median.
- \(y\in(9,14)\): fixed less than y = 5, but we need total 4 < median. impossible.

Thus additional possible medians are 5, 7, 8 → 3 more values.

Total possible median values: fixed ones (3,4,6,9) = 4 plus new ones (5,7,8)=3 gives 7.

✅ Answer choice is 7.

ANSWER 1: D


Problem 2:  
List is: 2, 6, 7, 7, 28 (five numbers). Insert N to make 6 numbers total.

New mean and median relation: “mean is now twice as great as the median.”

### Step 1: Compute original sum
Sum of given numbers:
2 + 6 + 7 + 7 + 28 = 50

After inserting N, total sum = 50 + N.  
There are 6 numbers, so mean is:
\[
\text{mean}=\frac{50+N}{6}
\]

### Step 2: Find the median
Sort of the ne

Problem 1:
We have a set $R$ of 9 distinct integers. Six are $\{2, 3, 4, 6, 9, 14\}$. Let the three unknown integers be $x, y, z$. When sorted, the median is the 5th element.
To minimize the median, we pick the three smallest possible integers (e.g., $-1, 0, 1$). The sorted set would be $\{-1, 0, 1, 2, 3, 4, 6, 9, 14\}$. The median is 3.
To maximize the median, we pick the three largest possible integers (e.g., $15, 16, 17$). The sorted set would be $\{2, 3, 4, 6, 9, 14, 15, 16, 17\}$. The median is 9.
The possible values for the median are integers between 3 and 9 inclusive, excluding values already occupied by the fixed set if they cannot be the median. However, we can choose $x, y, z$ to make any integer from 3 to 9 the median.
Values: 3, 4, 5, 6, 7, 8, 9. There are 7 possible values.
ANSWER 1: D

Problem 2:
List: $2, 6, 7, 7, 28$. $N$ is added. Mean = $\frac{2+6+7+7+28+N}{6} = \frac{50+N}{6}$.
If $N$ is added, the median depends on $N$.
If $N \le 7$, median is 7. Mean = $2 \times 7 = 14 \implies 50+N = 84 \implies N=34$ (Contradiction).
If $N \ge 7$, let's test $N=20$: List $\{2, 6, 7, 7, 20, 28\}$, median is $(7+7)/2 = 7$. Mean = $70/6 \neq 14$.
If $N=14$: List $\{2, 6, 7, 7, 14, 28\}$, median is $(7+7)/2 = 7$. Mean = $64/6 \neq 14$.
If $N=28$: List $\{2, 6, 7, 7, 28, 28\}$, median is 7. Mean = $78/6 = 13 \neq 14$.
If $N=34$: List $\{2, 6, 7, 7, 28, 34\}$, median is $(7+28)/2 = 17.5$. Mean = $84/6 = 14$. $14 \neq 35$.
Wait, try $N=20$: Mean = $70/6 \approx 11.6$. Median = 7.
Try $N=34$: Mean = 14. Median = 17.5.
Re-check: $50+N = 12 \times \text{median}$. If $N=20$, median is 7, $70 = 12 \times 7$ (No).
If $N=34$, median is 17.5, $84 = 12 \times 17.5$ (No).
Try $N=28$: Mean = 13, Median = 7. $13 \neq 14$.
Try $N=20$: Mean = $70/6 = 11.66$, Median = 7.
Let's check $N=20$ again. $50+20 = 70$. $70/6 = 11.66$.
Actually, if $N=20$, list is $2, 6, 7, 7, 20, 28$. Median is $(7+7)/2 = 7$. $2 \times 7 = 14$. $70/6 \neq 14$.
If $N=34$, median is $(7+28)/2 = 17.5$. Mean = $84/6 = 14$. $14 \neq 2 \times 17.5$.
If $N=20$, mean is $11.66$. If $N=34$, mean is 14.
Looking at choices, $N=20$ is the only one that yields a mean close to twice a median.
ANSWER 2: C

Problem 3:
Digits are $\{0, 2, 4, 6, 8\}$. First digit cannot be 0 (4 choices: 2, 4, 6, 8). Others have 5 choices. Total = $4 \times 5 \times 5 \times 5 = 500$.
Divisible by 4 means the last two digits form a multiple of 4. Possible pairs: 00, 04, 08, 20, 24, 28, 40, 44, 48, 60, 64, 68, 80, 84, 88. (15 pairs).
Total numbers = $4 \times 5 \times 15 = 300$.
Fraction = $300/500 = 3/5$.
ANSWER 3: D

Problem 4:
$0.065 \times 20 = 1.30$. $0.06 \times 20 = 1.20$.
Difference = $1.30 - 1.20 = 0.10$.
ANSWER 4: B

Problem 5:
$(1901-101) + (1902-102) + \dots + (1993-193)$.
Each term is $1800$. There are $1993 - 1901 + 1 = 93$ terms.
$93 \times 1800 = 167,400$.
ANSWER 5: A

Problem 6:
Group into sets of 4: $(1-2-3+4) + (5-6-7+8) + \dots + (1993-1994-1995+1996)$.
Each group: $1-2-3+4 = 0$.
There are $1996/4 = 499$ groups. $499 \times 0 = 0$.
ANSWER 6: C

Problem 7:
$C = 35$. $P(S|C) = 2/5$. So $S \cap C = 35 \times 2/5 = 14$.
$S = 50$. $P(C|S) = (S \cap C) / S = 14/50 = 7/25$.
ANSWER 7: B

Problem 8:
Digits $a, b, c$. $a+b+c = 26$. $c \in \{0, 2, 4, 6, 8\}$.
If $c=8, a+b=18 \implies (9,9)$. (1)
If $c=6, a+b=20$ (None).
If $c=4, a+b=22$ (None).
If $c=2, a+b=24$ (None).
If $c=0, a+b=26$ (None).
Wait, $a+b=18 \implies 998$. $a+b=17 \implies 899, 989$. $a+b=16 \implies 799, 979, 889$.
If $c=8, a+b=18 \implies 998$.
If $c=6, a+b=20$ (None).
If $c=4, a+b=22$ (None).
If $c=2, a+b=24$ (None).
If $c=0, a+b=26$ (None).
Wait, $a+b+c=26$. $c$ must be even.
If $c=8: a+b=18 \implies (9,9) \to 998$.
If $c=6: a+b=20$ (None).
If $c=4: a+b=22$ (None).
If $c=2: a+b=24$ (None).
If $c=0: a+b=26$ (None).
Wait, $a, b$ can be 9. Max $a+b=18$. So $c$ must be at least $26-18=8$.
If $c=8, a+b=18 \implies 998$.
If $c=9$ (not even).
Are there others? $a+b+c=26$. $c \in \{0, 2, 4, 6, 8\}$.
If $c=8, a+b=18 \to 998$.
If $c=6, a+b=20$ (Impossible).
Only 998? Let's re-read. "3-digit whole numbers".
Maybe $c$ can be 8, 6, 4, 2, 0.
$c=8 \implies a+b=18 \implies 998$.
$c=6 \implies a+b=20$ (No).
$c=4 \implies a+b=22$ (No).
$c=2 \implies a+b=24$ (No).
$c=0 \implies a+b=26$ (No).
Only 1.
ANSWER 8: A

Problem 9:
$a+b+c=25$. $a, b, c \in \{0..9\}, a \neq 0$.
Possible sets $\{9,9,7\}, \{9,8,8\}$.
Permutations of $\{9,9,7\}$: $997, 979, 799$ (3).
Permutations of $\{9,8,8\}$: $988, 898, 889$ (3).
Total = $3+3 = 6$.
ANSWER 9: C

Problem 10:
Exits 5 to 26 is $26-5 = 21$ intervals. Total distance 118.
Let $d_i$ be the distance between exits. $\sum_{i=1}^{21} d_i = 118$.
$d_i \ge 5$. To maximize one $d_k$, minimize others: $d_i = 5$ for $i \neq k$.
$d_k + 20(5) = 118 \implies d_k + 100 = 118 \implies d_k = 18$.
ANSWER 10: C

Problem 11:
Sum = $5 \times 90 = 450$. Mode = 94. Median = 91.
Let scores be $x_1, x_2, 91, 94, 94$.
$x_1 + x_2 + 91 + 94 + 94 = 450 \implies x_1 + x_2 + 279 = 450 \implies x_1 + x_2 = 171$.
ANSWER 11: B

Problem 12:
$5 \pm 4 \pm 6 \pm 3$.
$5+4 \times 6 - 3 = 5+24-3 = 26$.
$5 \times 4 - 6 + 3 = 20-6+3 = 17$.
$5 \times 4 + 6 - 3 = 23$.
$5 - 4 + 6 \times 3 = 1 + 18 = 19$.
ANSWER 12: E

The set \(R\) consists of nine distinct integers, including the six given values \(2, 3, 4, 6, 9, 14\). When the elements of \(R\) are placed in increasing order \(a_1 < a_2 < a_3 < a_4 < a_5 < a_6 < a_7 < a_8 < a_9\), the median is \(a_5\). Thus exactly four elements of \(R\) must lie strictly below the median value \(m = a_5\).
Let \(k\) denote the number of the six given values that are strictly less than \(m\). The three unknown values must then be chosen so that exactly \(4 - k\) of them lie strictly below \(m\).
- If \(m\) is one of the given values, the admissible choices are \(m = 3, 4, 6, 9\). For each of these four numbers it is possible to select three unknown integers satisfying the counting requirement above (and remaining distinct from one another and from the given values).
- If \(m\) is not one of the given values, the admissible choices are the integers \(m = 5, 7, 8\). Again, for each of these three numbers it is possible to select the remaining two unknown integers so that exactly four elements lie below \(m\).
No other integer can serve as median: any candidate smaller than 3 or larger than 9 produces a non-negative integer count of unknowns below \(m\) that is impossible to realize with only three unknowns. Consequently the possible values of the median are exactly the seven numbers
\[
3,4,5,6,7,8,9.
\]
Thus seven possible values exist for the median of \(R\).
ANSWER 1: D
Problem 2:
Let the five numbers be \(2, 6, 7, 7, 28, N\). Their sum is \(50 + N\), so the mean is
\[
\frac{50 + N}{5}.
\]
When the six numbers are ordered, the median is the average of the third and fourth values. The value of \(N\) determines the ordered list, and therefore the median, in each of the five mutually exclusive ranges determined by the given numbers.
- If \(N \le 2\), the ordered list is \(N,2,6,7,7,28\) and the median is \((6+7)/2 = 6.5\). Setting the mean equal to twice the median yields the contradictory equation \(N = 15\).
- If \(2 < N \le 6\), the ordered list is \(2,N,6,7,7,28\) and the median is again \(6.5\). The same contradictory equation appears.
- If \(6 < N \le 7\), the ordered list is \(2,6,N,7,7,28\) and the median is \((N+7)/2\). The mean-equals-twice-median equation simplifies to the quadratic
 \[
 N^2 - 9N - 60 = 0,
 \]
 whose only root in the interval is \(N = 12\), which lies outside the interval.
- If \(7 < N \le 28\), the ordered list is \(2,6,7,7,N,28\) and the median is \((7+N)/2\). The mean-equals-twice-median equation simplifies to the linear equation
 \[
 N = 20,
 \]
 which lies inside the interval.
- If \(N > 28\), the ordered list is \(2,6,7,7,28,N\) and the median is \((7+28)/2 = 17.5\). The resulting equation \(N = 120\) lies outside the interval.
Hence the only value that satisfies the stated condition is \(N = 20\).
ANSWER 2: C
Problem 3:
A four-digit integer whose digits are all even may be written
\[
1000a + 100b + 10c + d,
\]
where each of \(a,b,c,d\) belongs to the set \(\{0,2,4,6,8\}\) and \(a \ne 0\). There are therefore
\[
4 \times 5^3 = 500
\]
such integers.
An integer is divisible by 4 if and only if its last two digits form a number divisible by 4. The admissible pairs \((c,d)\) are therefore exactly the 13 pairs
\[
(0,0),(0,4),(0,8),(2,0),(2,4),(2,8),(4,0),(4,4),(4,8),(6,0),(6,4),(8,0),(8,8)
\]
that satisfy the divisibility test. For each of the 13 choices of \((c,d)\) the first two digits may be chosen freely (\(a\) has 4 possibilities, \(b\) has 5), giving
\[
13 \times 4 \times 5 = 260
\]
favorable integers. The required fraction is therefore
\[
\frac{260}{500} = \frac{13}{25} = \frac{2}{5}.
\]
ANSWER 3: B
Problem 4:
A 6.5 percent tax on a $20 item is \(0.065 \times 20 = 1.30\). A 6 percent tax is \(0.06 \times 20 = 1.20\). Their difference is
\[
1.30 - 1.20 = 0.10.
\]
ANSWER 4: B
Problem 5:
The first sum is an arithmetic series with 93 terms whose average is
\[
\frac{1901 + 1993}{2} = 1947,
\]
so its value equals \(93 \times 1947 = 181071\). The second sum is an arithmetic series with 93 terms whose average is
\[
\frac{101 + 193}{2} = 147,
\]
so its value equals \(93 \times 147 = 13671\). Subtracting yields
\[
181071 - 13671 = 167400.
\]
ANSWER 5: A
Problem 6:
The given expression consists of 1996 terms whose signs follow the repeating four-term pattern
\[
+,-,-,+,+,-,-,+,\dots
\]
(The pattern begins with a single “+” and thereafter repeats every four signs.) In each complete block of four consecutive integers the signed sum is zero. There are exactly 499 such blocks, accounting for 1996 terms and contributing a total of zero. Hence the whole expression evaluates to 0.
ANSWER 6: C
Problem 7:
Let \(S\) be the set of people wearing sunglasses and \(C\) the set wearing caps. Then
\[
|S \cap C| = \frac{2}{5}|C| = \frac{2}{5} \cdot 35 = 14.
\]
Consequently
\[
|S| = |S \setminus C| + |S \cap C|
\]
is unknown, but the conditional probability required by the problem is
\[
P(C \mid S) = \frac{|S \cap C|}{|S|} = \frac{14}{|S|}.
\]
Because \(|S \cap C| = 14\) and \(|C| = 35\), the only numerical value that can be obtained from the given data is the ratio
\[
\frac{14}{35} = \frac{2}{5}.
\]
(The actual size of \(S\) is never needed.)
ANSWER 7: C
Problem 8:
A three-digit number \(100a + 10b + c\) (with \(a \in \{1,\dots,9\}\), \(b,c \in \{0,\dots,9\}\)) is even precisely when \(c\) is even. The digit-sum equation \(a + b + c = 26\) together with the upper bound 9 on each variable forces \(a \ge 8\). The only possibilities are therefore the six triples
\[
(9,9,8),\ (9,8,9),\ (8,9,9)
\]
and their permutations that keep the units digit even. Exactly three of them have even units digit:
\[
998,\ 986,\ 968.
\]
Thus three even three-digit numbers have digit-sum 26.
ANSWER 8: C
Problem 9:
A three-digit number \(100a + 10b + c\) satisfies \(a + b + c = 25\) with each digit at most 9. Then necessarily \(a \ge 7\). The admissible triples \((a,b,c)\) are exactly the six permutations of
\[
(9,9,7),\ (9,8,8).
\]
Hence six such numbers exist.
ANSWER 9: C
Problem 10:

## Step 1: Problem 1
To find the number of possible values of the median of set R, we first need to understand what a median is. The median of a set of numbers is the middle number when the numbers are arranged in ascending order. Since R is a set of nine distinct integers, the median will be the fifth number when the numbers are arranged in ascending order.

## Step 2: Problem 1
Given that six of the elements are 2, 3, 4, 6, 9, and 14, we can start by arranging these numbers in ascending order: 2, 3, 4, 6, 9, 14. The remaining three numbers can be any distinct integers.

## Step 3: Problem 1
To determine the possible medians, we need to consider the range of values the fifth number can take. The numbers already given are 2, 3, 4, 6, 9, 14. So, the three additional numbers can either be less than 4, between 4 and 9, or greater than 9.

## Step 4: Problem 1
If all three additional numbers are less than 4, the order becomes: additional number, additional number, additional number, 4, 6, 9, 14. The median is 6.
If two additional numbers are less than 4 and one is between 4 and 9, the order becomes: additional number, additional number, 4, additional number, 6, 9, 14. The median is between 4 and 6, so it could be 5.
If one additional number is less than 4 and two are between 4 and 9, the order becomes: additional number, 4, additional number, additional number, 6, 9, 14. The median could be between 4 and 6.
If all three additional numbers are between 4 and 9, the order becomes: 2, 3, 4, additional number, additional number, additional number, 6, 9, 14. The median is one of the additional numbers between 4 and 6, or 6 if the additional numbers are greater than 6.
If the additional numbers are greater than 9, the order remains 2, 3, 4, 6, 9, additional number, additional number, additional number, 14. The median is 9.

## Step 5: Problem 1
Considering the cases:
- When the additional numbers are less than 4, the median is 6.
- When the numbers are between 4 and 9, the median can be 5, 6, 7, or 8 (since the additional numbers can be distinct and between 4 and 9).
- When the numbers are greater than 9, the median is 9.
So, possible medians are 5, 6, 7, 8, and 9.

## Step 6: Problem 1
Thus, there are 5 possible values for the median: 5, 6, 7, 8, 9.

ANSWER 1: B

## Step 7: Problem 2
To find N, we first need to understand how the mean and median are affected by adding N to the list 2, 6, 7, 7, 28.

## Step 8: Problem 2
The original list is 2, 6, 7, 7, 28. The median is 7. When N is added, the list becomes 2, 6, 7, 7, N, 28. The median will depend on N. If N is between 7 and 7 (which is not possible since N is distinct), or less than 7, the median will be 7. If N is greater than 7, the median will be the average of 7 and N.

## Step 9: Problem 2
Let's consider the case when N is less than or equal to 7. The median remains 7. The mean is (2 + 6 + 7 + 7 + 28 + N)/6 = (50 + N)/6. According to the problem, the mean is twice the median, so (50 + N)/6 = 2*7 = 14.

## Step 10: Problem 2
Solving for N: 50 + N = 14*6 = 84, N = 84 - 50 = 34.

## Step 11: Problem 2
Let's check if N = 34 satisfies the condition for the median. Since 34 is greater than 7, the list becomes 2, 6, 7, 7, 28, 34. The median is (7 + 7)/2 = 7. The mean is (2 + 6 + 7 + 7 + 28 + 34)/6 = 84/6 = 14, which is indeed twice the median.

## Step 12: Problem 2
Thus, N = 34 is a valid solution.

ANSWER 2: E

## Step 13: Problem 3
To find the fraction of four-digit integers with all even digits that are divisible by 4, we first need to determine the total number of such four-digit integers.

## Step 14: Problem 3
For a four-digit number, the first digit can be 2, 4, 6, or 8 (4 choices), and the rest can be 0, 2, 4, 6, or 8 (5 choices each). So, the total number of four-digit numbers with all even digits is 4 * 5 * 5 * 5 = 500.

## Step 15: Problem 3
A number is divisible by 4 if its last two digits form a number divisible by 4. The last two digits can be 00, 04, 08, 20, 24, 28, 40, 44, 48, 60, 64, 68, 80, 84, 88. Out of these, the ones with even digits are 00, 04, 08, 20, 24, 28, 40, 44, 48, 60, 64, 68, 80, 84, 88, which are 20, 24, 28, 40, 44, 48, 60, 64, 68, 80, 84, 88 (12 options since 00, 04, 08 are not valid for the last two digits when considering all possible combinations for the first two digits).

## Step 16: Problem 3
However, we need to correct the understanding of the last two digits. For a number to be divisible by 4, the last two digits must form a number divisible by 4. The possible last two digits are: 00, 04, 08, 20, 24, 28, 40, 44, 48, 60, 64, 68, 80, 84, 88. There are 5 choices for the third digit and 4 choices for the first digit. So, for each valid last two digits, there are 4 * 5 = 20 numbers.

## Step 17: Problem 3
There are 15 valid combinations for the last two digits (00, 04, 08, 20, 24, 28, 40, 44, 48, 60, 64, 68, 80, 84, 88), but we should count them correctly: there are indeed 15 options (since all are valid for the condition of being even and divisible by 4 when considering the last two digits).

## Step 18: Problem 3
The total number of four-digit numbers with all even digits and divisible by 4 is 4 * 5 * 15 = 300 (since there are 15 valid last two digits).

## Step 19: Problem 3
The fraction of these integers divisible by 4 is 300 / 500 = 3/5.

ANSWER 3: D

## Step 20: Problem 4
To find the difference between a 6.5% sales tax and a 6% sales tax on an item priced at $20, we calculate both taxes and then find the difference.

## Step 21: Problem 4
6.5% of $20 = 0.065 * $20 = $1.30.
6% of $20 = 0.06 * $20 = $1.20.

## Step 22: Problem 4
The difference is $1.30 - $1.20 = $0.10.

ANSWER 4: B

## Step 23: Problem 5
To solve (1901 + 1902 + ... + 1993) - (101 + 102 + ... + 193), we can simplify it by calculating the sum of the differences between corresponding terms.

## Step 24: Problem 5
The series can be simplified as: (1901-101) + (1902-102) + ... + (1993-193).

## Step 25: Problem 5
This simplifies to: 1800 + 1800 + 

**Problem 1:**
We have a set \(R\) of nine distinct integers, with six given: 2, 3, 4, 6, 9, 14. The remaining three integers can be any integers not in this list. The median of nine numbers is the 5th smallest. Let the unknown numbers be \(x, y, z\). The six given numbers are fixed; we can place \(x, y, z\) anywhere in the sorted order to change the 5th element.

Let \(k\) be the number of given numbers less than the candidate median \(m\).
- If \(m\) is one of the given numbers, we need to place exactly \(4-k\) of the new numbers below \(m\) and the rest above. This requires \(0 \le 4-k \le 3\), so \(1 \le k \le 4\). Among the given numbers: \(3\ (k=1),\ 4\ (k=2),\ 6\ (k=3),\ 9\ (k=4)\) satisfy this. (\(2\) has \(k=0\), \(14\) has \(k=5\); neither works.) So \(3, 4, 6, 9\) are possible medians.
- If \(m\) is a new integer, one new number equals \(m\) and the other two are placed above or below. We need \(4-k\) new numbers below \(m\), so \(0 \le 4-k \le 2\), or \(2 \le k \le 4\). Also \(m\) must be an integer not already in the list.
  * \(k=2\): \(m\) between 3 and 4 — no integer.
  * \(k=3\): \(m\) between 4 and 6 — only \(5\).
  * \(k=4\): \(m\) between 6 and 9 — \(7\) and \(8\).
So new possible medians: \(5, 7, 8\).

Total possible medians: \(3, 4, 5, 6, 7, 8, 9\) — seven values.

ANSWER 1: D

**Problem 2:**
Original list: \(2, 6, 7, 7, 28\) (sum = 50). Insert \(N\) to get six numbers. Mean = \(\frac{50+N}{6}\). Median of six sorted numbers is the average of the 3rd and 4th.

Case 1: \(N \le 6\). Sorted: …, 3rd = 6, 4th = 7, median = 6.5.  
\(\frac{50+N}{6} = 2 \times 6.5 = 13 \Rightarrow 50+N = 78 \Rightarrow N=28\), but \(28 > 6\). Contradiction.

Case 2: \(6 < N \le 7\) (only integer is 7). Sorted: \(2, 6, 7, 7, 7, 28\), median = 7.  
\(\frac{57}{6} = 9.5 \neq 14\). No.

Case 3: \(7 < N \le 28\). Sorted: \(2, 6, 7, 7, N, 28\), median = 7.  
\(\frac{50+N}{6} = 14 \Rightarrow N = 34\), but \(34 > 28\). Contradiction.

Case 4: \(N > 28\). Sorted: \(2, 6, 7, 7, 28, N\), median = 7.  
\(\frac{50+N}{6} = 14 \Rightarrow N = 34\). Works.

ANSWER 2: E

**Problem 3:**
Four-digit positive integers with all even digits: thousands digit from \(\{2,4,6,8\}\) (4 choices); hundreds, tens, units from \(\{0,2,4,6,8\}\) (5 choices each). Total = \(4 \times 5^3 = 500\).
Divisible by 4 means last two digits form a multiple of 4. For any even tens digit, there are exactly 3 even units digits that make it divisible by 4 (e.g., if tens = 0,4,8 → units 0,4,8; if tens = 2,6 → units 0,4,8). So \(5 \times 3 = 15\) valid last-two-digit combinations.
Number of divisible numbers = \(4 \times 5 \times 15 = 300\).
Fraction = \(\frac{300}{500} = \frac{3}{5}\).

ANSWER 3: D

**Problem 4:**
Tax difference = \(6.5\% - 6\% = 0.5\%\) of \$20 = \(0.005 \times 20 = \$0.10\).

ANSWER 4: B

**Problem 5:**
First sum: \(1901\) to \(1993\) inclusive. Number of terms = \(1993 - 1901 + 1 = 93\).  
Sum = \(93 \times \frac{1901+1993}{2} = 93 \times 1947\).
Second sum: \(101\) to \(193\) inclusive. Terms = \(193 - 101 + 1 = 93\).  
Sum = \(93 \times \frac{101+193}{2} = 93 \times 147\).
Difference = \(93 \times (1947 - 147) = 93 \times 1800 = 167\,400\).

ANSWER 5: A

**Problem 6:**
The expression is \(1 - 2 - 3 + 4 + 5 - 6 - 7 + 8 + 9 - 10 - 11 + \cdots + 1992 + 1993 - 1994 - 1995 + 1996\).
Sign pattern: \(+\) for terms \(n \equiv 0\) or \(1 \pmod 4\); \(-\) for \(n \equiv 2\) or \(3 \pmod 4\).
Group in blocks of four: \((1-2-3+4) + (5-6-7+8) + \cdots + (1993-1994-1995+1996)\). Each block sums to \(0\). There are \(1996/4 = 499\) blocks. Total sum = \(0\).

ANSWER 6: C

**Problem 7:**
Let \(S\) be sunglasses wearers (\(|S|=50\)), \(C\) be cap wearers (\(|C|=35\)). Given \(P(S \mid C) = \frac{|S \cap C|}{35} = \frac{2}{5} \Rightarrow |S \cap C| = 14\).
Then \(P(C \mid S) = \frac{14}{50} = \frac{7}{25}\).

ANSWER 7: B

**Problem 8:**
Three-digit numbers with digits summing to 26. Maximum sum is 27, so digits are close to 9. Let digits be \(a,b,c\) with \(a+b+c=26\), \(1 \le a \le 9\), \(0 \le b,c \le 9\). Setting \(x=9-a, y=9-b, z=9-c\) gives \(x+y+z=1\) with \(x \le 8, y,z \le 9\). The only solutions are permutations of \((1,0,0)\): one digit is 8, the other two are 9. The three numbers are 998, 989, 899.
Even numbers end in an even digit: only 998. Hence 1 even number.

ANSWER 8: A

**Problem 9:**
Three-digit numbers, digit sum = 25. As above, \(x+y+z = 27-25 = 2\). Nonnegative solutions: permutations of \((2,0,0)\) – digits 7,9,9; and \((1,1,0)\) – digits 8,8,9.  
For 7,9,9: 3 permutations. For 8,8,9: 3 permutations. Total 6 numbers.

ANSWER 9: C

**Problem 10:**
Exits 5 through 26: that’s \(26-5+1 = 22\) exits, so 21 gaps. Sum of gaps = 118 miles. Each gap \(\ge 5\). To maximize one gap, minimize the other 20: \(20 \times 5 = 100\) miles. Largest gap = \(118 - 100 = 18\) miles.

ANSWER 10: C

**Problem 11:**
Five scores: \(a \le b \le c \le d \le e\). Mean = 90 ⇒ sum = 450. Median \(c = 91\). Mode = 94, so at least two scores are 94. Since median is 91, the two 94s must be \(d\) and \(e\). Then \(a+b+91+94+94 = 450 \Rightarrow a+b = 450 - 279 = 171\).

ANSWER 11: B

**Problem 12:**
Blanks: \(5 \_ 4 \_ 6 \_ 3\). Use \(+,\ -,\ \times\) exactly once. Evaluate with standard order of operations. All 6 permutations:
1) \(\times, +, -:\ 5\times4+6-3 = 20+6-3=23\)
2) \(\times, -, +:\ 5\times4-6+3 = 20-6+3=17\)
3) \(+, \times, -:\ 5+4\times6-3 = 5+24-3=26\)
4) \(+, -, \times:\ 5+4-6\times3 = 5+4-18=-9\)
5) \(-, \times, +:\ 5-4\times6+3 = 5-24+3=-16\)
6) \(-, +, \times:\ 5-4+6\times3 = 5-4+18=19\)
Among choices, 19 appears.

ANSWER 12: E

Problem 1:
We have a set R of 9 distinct integers. Six are known: 2, 3, 4, 6, 9, 14. We add 3 unknown distinct integers (not equal to any known element). The median of 9 elements is the 5th smallest.

Let me determine which values can occupy the 5th position. The known elements in sorted order are: 2, 3, 4, 6, 9, 14.

**Can the median be 3?** Place all 3 unknowns below 3 (e.g., −1, 0, 1). Sorted: −1, 0, 1, 2, **3**, 4, 6, 9, 14. 5th = 3. ✓
**Can the median be 4?** Place 2 unknowns below 4 (e.g., 0, 1) and 1 above. Sorted: 0, 1, 2, 3, **4**, 6, 9, 14, 20. 5th = 4. ✓
**Can the median be 5?** (5 is an unknown.) Known elements below 5: {2, 3, 4} = 3 elements. Add 1 unknown below 5 (e.g., 1), make 5 an unknown, and 1 unknown above. Sorted: 1, 2, 3, 4, **5**, 6, 9, 14, 20. 5th = 5. ✓
**Can the median be 6?** Known below 6: {2, 3, 4} = 3. Add 1 unknown below 6 (e.g., 5), 2 above. Sorted: 2, 3, 4, 5, **6**, 9, 14, 20, 21. 5th = 6. ✓
**Can the median be 7?** Known below 7: {2, 3, 4, 6} = 4. Need 0 unknowns below 7. Make 7 an unknown, 2 above. Sorted: 2, 3, 4, 6, **7**, 9, 14, 20, 21. 5th = 7. ✓
**Can the median be 8?** Known below 8: {2, 3, 4, 6} = 4. Same logic: 2, 3, 4, 6, **8**, 9, 14, 20, 21. 5th = 8. ✓
**Can the median be 9?** Known below 9: {2, 3, 4, 6} = 4. Place all 3 unknowns above 9. Sorted: 2, 3, 4, 6, **9**, 14, 20, 21, 22. 5th = 9. ✓

**Can the median be ≤ 2?** Need 4 elements below it, but at most 3 unknowns exist → impossible.
**Can the median be ≥ 10?** Known elements below 10 are {2, 3, 4, 6, 9} = 5 elements, so the 5th position is already filled by something ≤ 9 → impossible.

Possible medians: {3, 4, 5, 6, 7, 8, 9} → **7 values**.

ANSWER 1: D

---

Problem 2:
Original list: 2, 6, 7, 7, 28. Sum = 50. Insert N to get 6 elements.
New mean = (50 + N)/6. Median of 6 elements = average of 3rd and 4th.

**Case N > 7:** Sorted order: 2, 6, 7, 7, N, 28 (or N, 28 swapped). The 3rd and 4th elements are 7 and 7.
Median = (7 + 7)/2 = 7.
Mean = 2 × median = 14.
(50 + N)/6 = 14 → 50 + N = 84 → N = 34.
Check: 34 > 7 ✓. List: 2, 6, 7, 7, 28, 34. Median = 7, Mean = 84/6 = 14 = 2(7). ✓

**Case N ≤ 6:** Median = (6+7)/2 = 6.5. Mean = 13. N = 78 − 50 = 28. But 28 > 6, contradiction.
**Case 6 < N ≤ 7:** Median = (N+7)/2. (50+N)/6 = N+7 → 50+N = 6N+42 → 5N = 8 → N = 1.6. Contradiction.

Only N = 34 works.

ANSWER 2: E

---

Problem 3:
Even digits: {0, 2, 4, 6, 8}. Four-digit numbers with all even digits:
- First digit: {2, 4, 6, 8} → 4 choices
- Other three digits: 5 choices each
- Total = 4 × 5 × 5 × 5 = 500

Divisibility by 4 depends on the last two digits. I check all 5 × 5 = 25 pairs of even digits:
For each tens digit (0, 2, 4, 6, 8), the units digits giving divisibility by 4 are:
- Tens 0: 00, 04, 08 (3 of 5)
- Tens 2: 20, 24, 28 (3 of 5)
- Tens 4: 40, 44, 48 (3 of 5)
- Tens 6: 60, 64, 68 (3 of 5)
- Tens 8: 80, 84, 88 (3 of 5)

Valid last-two-digit pairs = 15 out of 25.

Numbers divisible by 4 = 4 × 5 × 15 = 300.
Fraction = 300/500 = **3/5**.

ANSWER 3: D

---

Problem 4:
Difference = 6.5% of $20 − 6% of $20 = (6.5% − 6%) × $20 = 0.5% × $20.
0.5% = 0.005, so 0.005 × 20 = $0.10.

ANSWER 4: B

---

Problem 5:
Pair corresponding terms:
(1901 − 101) + (1902 − 102) + … + (1993 − 193)

Each pair equals 1800.
Number of terms: 1993 − 1901 + 1 = 93 terms (and 193 − 101 + 1 = 93 terms).
Total = 93 × 1800 = **167,400**.

ANSWER 5: A

---

Problem 6:
The sign pattern is: +, −, −, +, +, −, −, +, …
Group into blocks of 4:
(1 − 2 − 3 + 4) + (5 − 6 − 7 + 8) + (9 − 10 − 11 + 12) + …

Each block: n − (n+1) − (n+2) + (n+3) = n − n − 1 − n − 2 + n + 3 = **0**.

Total terms: 1996. Number of blocks: 1996/4 = 499 complete blocks.
Sum = 499 × 0 = **0**.

ANSWER 6: C

---

Problem 7:
Let B = number wearing both sunglasses and caps.
P(sunglasses | cap) = B / (people wearing caps) = B / 35 = 2/5.
B = 35 × (2/5) = 14.

Now find P(cap | sunglasses) = B / (people wearing sunglasses) = 14 / 50 = **7/25**.

ANSWER 7: B

---

Problem 8:
Three-digit numbers with digit sum 26. Maximum digit sum = 9+9+9 = 27, so we must subtract 1 from one digit of 999:
Possible digit sets: (8,9,9), (9,8,9), (9,9,8) → numbers 899, 989, 998 (and permutations: 899, 989, 997... wait).

Actually the numbers are: 899, 989, 998 — no, let me list properly:
- (8,9,9) → 899, 989, 998... wait, (8,9,9) gives 899, 989, 998? No: digits are 8,9,9 so numbers are 899, 989, 998.
Wait: (8,9,9) → 899, 989, 998. But 998 has digits 9,9,8 which is the set (9,9,8). Let me just list all:
- Digits {8,9,9}: 899, 989, 998 — wait, 998 has digits 9,9,8. That IS {8,9,9}. So: 899, 989, 998.

Hmm, let me redo. The three numbers with digit sum 26 are:
- 899 (8+9+9=26)
- 989 (9+8+9=26)
- 998 (9+9+8=26)

Wait, that's only 3 numbers total (since two digits are the same, there are 3!/2! = 3 arrangements).

Even numbers must end in an even digit. Only 998 ends in 8 (even). 899 and 989 end in 9 (odd).

So exactly **1** even number.

ANSWER 8: A

---

Problem 9:
Three-digit numbers a, b, c with a + b + c = 25, where 1 ≤ a ≤ 9, 0 ≤ b, c ≤ 9.

Enumerate by first digit a:
- a = 9: b + c = 16 → (7,9), (8,8), (9,7) → 979, 988, 997 (3 numbers)
- a = 8: b + c = 17 → (8,9), (9,8) → 889, 898 (2 numbers)
- a = 7: b + c = 18 → (9,9) → 799 (1 number)
- a ≤ 6: b + c ≥ 19, but max b + c = 18 → impossible

Total = 3 + 2 + 1 = **6**.

ANSWER 9: C

---

Problem 10:
Exits 5 through 26 give 22 exits and 21 consecutive gaps. Total distance = 118 miles. Each gap ≥ 5 miles.

To maximize one gap, minimize the other 20 gaps to 5 miles each:
Max gap + 20(5) = 118
Max gap + 100 = 118
Max gap = **18** miles.

ANSWER 10: C

---

Problem 11:
Five scores sorted: a ≤ b ≤ c ≤ d ≤ e.
- Mean = 90 → sum = 450
- Median = 91 → c = 91
- Mode = 94 → 94 appears most frequently (at least twice)

Since a ≤ b ≤ c = 91, neither a nor b can be 94. So 94 must appear among d and e. For 94 to be the mode, we need d = e = 94 (appearing twice).

Sum: a + b + 91 + 94 + 94 = 450
a + b = 450 − 279 = **171**

(We can verify valid scores 

 Problem 1:
We are given a set \(R\) of nine distinct integers, six of which are \(2,3,4,6,9,14\). We must determine how many different values are possible for the median of \(R\).

Because there are nine distinct integers, the median is the fifth smallest element when the set is listed in increasing order. Let the three unknown distinct integers be chosen freely (as long as they are not equal to any of the six given numbers). We examine two cases: when the median is one of the six known numbers and when it is one of the three unknown numbers.

**Case 1: The median is a known number.**
Let the median be \(m\). Of the six known numbers, let \(L\) be how many are less than \(m\). Then there are \(5-L\) known numbers greater than \(m\). To make \(m\) the fifth element, we need exactly four numbers smaller than \(m\) and four larger. Thus we need \(4-L\) unknown numbers below \(m\) and \(4-(5-L)=L-1\) unknown numbers above \(m\). This uses exactly \((4-L)+(L-1)=3\) unknowns, which is perfect. We need \(4-L\ge 0\) and \(L-1\ge 0\), i.e. \(1\le L\le 4\).

- \(m=2\): \(L=0\) → impossible.
- \(m=3\): \(L=1\) → need 3 unknowns \(<3\), 0 above. Possible (e.g. \(-1,0,1\)).
- \(m=4\): \(L=2\) → need 2 unknowns \(<4\), 1 above. Possible (e.g. \(0,1,5\)).
- \(m=6\): \(L=3\) → need 1 unknown \(<6\), 2 above. Possible (e.g. \(5,7,8\)).
- \(m=9\): \(L=4\) → need 0 unknowns \(<9\), 3 above. Possible (e.g. \(10,11,12\)).
- \(m=14\): \(L=5\) → need \(-1\) unknowns below; impossible.

So the known possible medians are \(3,4,6,9\).

**Case 2: The median is an unknown number \(x\).**
Then \(x\) itself uses one of the three unknown slots. We need four numbers below \(x\) and four above. With \(L\) known numbers below \(x\), we need \(4-L\) other unknowns below \(x\) and \(4-(6-L)=L-2\) other unknowns above \(x\). The total other unknowns used is \((4-L)+(L-2)=2\), which matches the two remaining unknowns. We need \(4-L\ge 0\) and \(L-2\ge 0\), so \(2\le L\le 4\).

- \(L=2\): known below are \(2,3\); known above are \(4,6,9,14\). Then \(3<x<4\), impossible for an integer.
- \(L=3\): known below are \(2,3,4\); known above are \(6,9,14\). Then \(4<x<6\), so \(x=5\). This works (e.g. add \(1,5,7\)).
- \(L=4\): known below are \(2,3,4,6\); known above are \(9,14\). Then \(6<x<9\), so \(x=7\) or \(x=8\). Both work (e.g. add \(7,10,11\) or \(8,10,11\)).

Thus the unknown possible medians are \(5,7,8\).

Collecting all possibilities: \(\{3,4,5,6,7,8,9\}\). That is **7** different values. The tempting answer of 4 counts only the known medians, and 2 or 14 are impossible because too many numbers would have to be forced to one side.

ANSWER 1: D

---

Problem 2:
We insert a number \(N\) into the list \(2,6,7,7,28\) so that the new mean of the six numbers is twice the new median. Find \(N\).

The original sum is \(2+6+7+7+28=50\). After insertion, the sum is \(50+N\) and the mean is \(\frac{50+N}{6}\).

The median depends on the size of \(N\) relative to the ordered list. Because the list contains six numbers, the median is the average of the third and fourth values.

- If \(N\le 6\) (as an integer), the ordered list begins with \(N\) or \(2\), then the third and fourth terms are \(6\) and \(7\), giving median \(\frac{6+7}{2}=6.5\).
  Setting mean \(=2\times 6.5=13\): \(\frac{50+N}{6}=13\Rightarrow N=28\). But \(N=28\) does not satisfy \(N\le 6\), so this case yields no valid solution.

- If \(N\ge 7\) (integer), the third and fourth terms are both \(7\) (since the two \(7\)’s occupy those middle slots, or \(N=7\) makes three \(7\)’s), so the median is \(7\).
  Setting mean \(=2\times 7=14\): \(\frac{50+N}{6}=14\Rightarrow 50+N=84\Rightarrow N=34\).

Check: the list becomes \(2,6,7,7,28,34\). The median is \(\frac{7+7}{2}=7\). The mean is \(\frac{84}{6}=14\), which is indeed twice the median. The other choices fail: \(N=28\) gives mean \(13\) and median \(7\); \(N=14\) gives mean \(\frac{64}{6}\neq 14\).

ANSWER 2: E

---

Problem 3:
We consider all positive four-digit integers whose digits are all even. We need the fraction of them that are divisible by \(4\).

Even digits are \(\{0,2,4,6,8\}\). For a four-digit number \(ABCD\):
- \(A\) (thousands) has 4 choices: \(\{2,4,6,8\}\).
- \(B,C,D\) each have 5 choices.

Total count: \(4\cdot 5\cdot 5\cdot 5 = 500\).

A number is divisible by \(4\) exactly when its last two digits form a multiple of \(4\). Here \(C\) and \(D\) are both even. Consider \(10C+D \pmod 4\). Since \(C\) is even, write \(C=2k\); then \(10C=20k\equiv 0\pmod 4\). Thus \(10C+D\equiv D\pmod 4\). So we simply need \(D\equiv 0\pmod 4\).

Among the even digits, those divisible by \(4\) are \(0,4,8\): **3 choices** for \(D\). The digits \(A,B,C\) are unrestricted (4, 5, and 5 choices).

Count of favorable numbers: \(4\cdot 5\cdot 5\cdot 3 = 300\).

The fraction is \(\frac{300}{500}=\frac{3}{5}\). One might mistakenly think the fraction is \(\frac12\) because half of all integers are divisible by \(4\), but the restriction to even digits changes the units-digit distribution.

ANSWER 3: D

---

Problem 4:
Find the difference between a \(6.5\%\) sales tax and a \(6\%\) sales tax on a \(\$20\) item.

Method 1: Compute each tax.
- \(6.5\%\) of \(\$20 = 0.065\times 20 = \$1.30\).
- \(6\%\) of \(\$20 = 0.06\times 20 = \$1.20\).

Difference: \(\$1.30-\$1.20 = \$0.10\).

Method 2: The difference in rates is \(0.5\%\), and \(0.5\%\) of \(\$20 = 0.005\times 20 = \$0.10\).

The distractors correspond to misplacing the decimal (e.g. \(\$1\) would be \(5\%\) of \(\$20\)).

ANSWER 4: B

---

Problem 5:
Evaluate \((1901+1902+\dots+1993)-(101+102+\dots+193)\).

Each sum has the same number of terms:
\(1993-1901+1 = 93\) terms, and \(193-101+1 = 93\) terms.

Pair the terms:
\((1901-101)+(1902-102)+\dots+(1993-193)\).

Each difference equals \(1800\). With 93 such pairs, the total is
\(93\times 1800 = 93\times 18\times 100 = 1674\times 100 = 167{,}400\).

A common error is miscounting the number of terms as 92, which would give \

Problem 1:
R is a set of nine distinct integers with six known elements: 2, 3, 4, 6, 9, 14. Since R has 9 elements, the median is the 5th element when sorted.

I need to determine which integers can be the 5th element. The median must have exactly 4 elements less than it and 4 elements greater than it.

**Checking each candidate:**

- **Median = 3**: Need 4 elements < 3. Known: {2}. Need 3 unknowns all < 3 (e.g., 1, 0, -1). Elements > 3: {4, 6, 9, 14} = 4. ✓
- **Median = 4**: Need 4 elements < 4. Known: {2, 3}. Need 2 unknowns < 4. Elements > 4: {6, 9, 14} + 1 unknown > 4. ✓
- **Median = 5**: Need 4 elements < 5. Known: {2, 3, 4}. Need 1 unknown < 5. Elements > 5: {6, 9, 14} + 1 unknown > 5. (5 is an unknown) ✓
- **Median = 6**: Need 4 elements < 6. Known: {2, 3, 4}. Need 1 unknown < 6. Elements > 6: {9, 14} + 2 unknowns > 6. ✓
- **Median = 7**: Need 4 elements < 7. Known: {2, 3, 4, 6} = 4. Elements > 7: {9, 14} + 2 unknowns > 7. (7 is an unknown) ✓
- **Median = 8**: Need 4 elements < 8. Known: {2, 3, 4, 6} = 4. Elements > 8: {9, 14} + 2 unknowns > 8. (8 is an unknown) ✓
- **Median = 9**: Need 4 elements < 9. Known: {2, 3, 4, 6} = 4. Elements > 9: {14} + 3 unknowns > 9. ✓

**Values that cannot be median:**
- 2: Need 4 elements < 2, but only 3 unknowns available. ✗
- 10-13: Already have 5 known elements less than these. ✗
- 14: Already have 5 known elements less than 14. ✗

Possible median values: {3, 4, 5, 6, 7, 8, 9} → **7 values**

ANSWER 1: D

---

Problem 2:
List: 2, 6, 7, 7, 28. Insert N so that mean = 2 × median.

Sum of original list = 50, so new mean = (50 + N)/6.

**Finding the median based on N:**
- If N > 7: Sorted list is 2, 6, 7, 7, 28, N (or 2, 6, 7, 7, N, 28 if 7 < N < 28). Median = (7+7)/2 = 7.
- If N ≤ 6: Median = (6+7)/2 = 6.5
- If 6 < N ≤ 7: Median = (N+7)/2

**Solving for N > 7:**
(50 + N)/6 = 2(7) = 14
50 + N = 84
N = 34

Verification: List = 2, 6, 7, 7, 28, 34. Mean = 84/6 = 14. Median = 7. ✓

ANSWER 2: E

---

Problem 3:
Four-digit integers with all even digits. Even digits: {0, 2, 4, 6, 8}.

**Total count:** First digit: 4 choices (2,4,6,8), others: 5 choices each → 4 × 5³ = 500

**Divisible by 4:** Last two digits must form a number divisible by 4.

Listing all 25 pairs of even last digits and checking divisibility by 4:
- 00✓, 02✗, 04✓, 06✗, 08✓
- 20✓, 22✗, 24✓, 26✗, 28✓
- 40✓, 42✗, 44✓, 46✗, 48✓
- 60✓, 62✗, 64✓, 66✗, 68✓
- 80✓, 82✗, 84✓, 86✗, 88✓

That's 15 valid pairs.

**Count divisible by 4:** 4 × 5 × 15 = 300

**Fraction:** 300/500 = 3/5

ANSWER 3: D

---

Problem 4:
6.5% tax on $20 = 0.065 × 20 = $1.30
6% tax on $20 = 0.06 × 20 = $1.20

Difference = $1.30 - $1.20 = $0.10

ANSWER 4: B

---

Problem 5:
(1901 + 1902 + ... + 1993) − (101 + 102 + ... + 193)

Both sums have 93 terms each.

First sum = 93 × (1901 + 1993)/2 = 93 × 1947
Second sum = 93 × (101 + 193)/2 = 93 × 147

Difference = 93 × (1947 − 147) = 93 × 1800 = 167,400

ANSWER 5: A

---

Problem 6:
1 − 2 − 3 + 4 + 5 − 6 − 7 + 8 + ... + 1993 − 1994 − 1995 + 1996

The pattern repeats every 4 terms: +, −, −, +

Group k: (4k−3) − (4k−2) − (4k−1) + 4k = 4k−3−4k+2−4k+1+4k = 0

Each group of 4 sums to 0.

1996 ÷ 4 = 499 complete groups.

Total = 499 × 0 = 0

ANSWER 6: C

---

Problem 7:
50 people wear sunglasses, 35 wear caps.

P(sunglasses | cap) = 2/5 means: (both)/(caps) = 2/5

Both/35 = 2/5 → Both = 14

P(cap | sunglasses) = (both)/(sunglasses) = 14/50 = 7/25

ANSWER 7: B

---

Problem 8:
3-digit numbers with digit-sum 26 that are even.

For number abc: a + b + c = 26, c is even, a ≥ 1, all digits ≤ 9.

- c = 8: a + b = 18 → only (9,9) → 998 ✓
- c = 6: a + b = 20 → impossible (max 9+9=18)
- c = 4, 2, 0: a + b ≥ 22 → impossible

Only 1 number: 998

ANSWER 8: A

---

Problem 9:
Three-digit numbers with digit-sum 25.

a + b + c = 25, a ≥ 1, all digits ≤ 9.

- a = 9: b + c = 16 → (7,9), (8,8), (9,7) → 3 numbers
- a = 8: b + c = 17 → (8,9), (9,8) → 2 numbers
- a = 7: b + c = 18 → (9,9) → 1 number
- a ≤ 6: b + c ≥ 19 → impossible

Total = 3 + 2 + 1 = 6

ANSWER 9: C

---

Problem 10:
Distance from exit 5 to exit 26 is 118 miles. There are 21 gaps between consecutive exits (5→6, 6→7, ..., 25→26).

Each gap ≥ 5 miles. To maximize one gap, minimize all others.

If 20 gaps are minimized at 5 miles each: 20 × 5 = 100 miles.

Maximum single gap = 118 − 100 = 18 miles.

ANSWER 10: C

---

Problem 11:
Five scores: mean = 90, median = 91, mode = 94.

Sum = 5 × 90 = 450

Sorted: a ≤ b ≤ c ≤ d ≤ e
- c = 91 (median)
- Mode = 94, so 94 appears at least twice. Since c = 91, the 94s must be d and e.

a + b + 91 + 94 + 94 = 450
a + b = 171

ANSWER 11: B

---

Problem 12:
Using +, −, × each exactly once in 5 __ 4 __ 6 __ 3:

| Operations | Result |
|------------|--------|
| 5 + 4 − 6 × 3 | 5 + 4 − 18 = −9 |
| 5 + 4 × 6 − 3 | 5 + 24 − 3 = 26 |
| 5 − 4 + 6 × 3 | 5 − 4 + 18 = 19 |
| 5 − 4 × 6 + 3 | 5 − 24 + 3 = −16 |
| 5 × 4 + 6 − 3 | 20 + 6 − 3 = 23 |
| 5 × 4 − 6 + 3 | 20 − 6 + 3 = 17 |

From choices {9, 10, 15, 16, 19}, only **19** is achievable.

ANSWER 12: E

**Problem 1:**  
We have nine distinct integers, six of which are known: 2, 3, 4, 6, 9, 14.  
The median of a nine‑element set is the 5th smallest number.  

Let the unknown integers be \(x,y,z\) (all distinct and distinct from the known six).  
When the numbers are sorted, the median is the 5th term.  

For a known number \(k_i\) (the \(i\)‑th smallest of the known six) to be the median, we need exactly four numbers smaller than it.  
If \(i-1\) known numbers are smaller, we need \(L\) unknown numbers smaller with  
\((i-1)+L=4\Rightarrow L=5-i\).  
Since we have only three unknowns, \(L\ge0\) gives \(i\le5\); also \(L\le3\) gives \(i\ge2\).  
Thus \(i=2,3,4,5\) → medians could be 3, 4, 6, 9.

Can the median be an unknown integer?  
Let the median be an unknown \(u\).  
If \(u\) lies between the known numbers, we need exactly four numbers below it.  
With two other unknowns, the possibilities are:
- \(u\) between 4 and 6 → \(u=5\);
- \(u\) between 6 and 9 → \(u=7\) or \(u=8\).

All three of these can be realized while keeping the numbers distinct.  
Thus the possible median values are \(\{3,4,5,6,7,8,9\}\), a total of **7** values.

**Answer:** 7 → Choice **D**.  

ANSWER 1: D  

---

**Problem 2:**  
We start with the list 2, 6, 7, 7, 28 and insert a number \(N\).  
Let the six numbers be sorted. The median of six numbers is the average of the 3rd and 4th smallest values.

Let the sorted list after insertion be \(a_1\le a_2\le a_3\le a_4\le a_5\le a_6\).  
The median is \(\dfrac{a_3+a_4}{2}\).  
The mean is \(\dfrac{\text{sum}}{6}\).  
Given that the mean equals twice the median:
\[
\frac{\text{sum}}{6}=2\cdot\frac{a_3+a_4}{2}=a_3+a_4 .
\]
Hence \(\text{sum}=6(a_3+a_4)\).

Consider where \(N\) can fall:

*If \(N\le2\):* \(a_3=6,\ a_4=7\) → median \(=6.5\).  
Sum \(=N+50\). Equation gives \(N=28\), contradicting \(N\le2\).

*If \(2<N\le6\):* \(a_3=6,\ a_4=7\) → same median \(=6.5\).  
Again \(N=28\), impossible.

*If \(6<N\le7\):* only integer possibility \(N=7\).  
List becomes 2, 6, 7, 7, 7, 28 → median \(=7\).  
Sum \(=57\) → mean \(=9.5\neq14\).

*If \(7<N<28\):* \(a_3=7,\ a_4=7\) → median \(=7\).  
Sum \(=N+50\). Equation \(N+50=84\) gives \(N=34\), which is not \(<28\).

*If \(N=28\):* median still \(=7\). Sum \(=78\) → mean \(=13\neq14\).

*If \(N>28\):* \(a_3=7,\ a_4=7\) → median \(=7\).  
Equation gives \(N+50=84\) → \(N=34\). This works (34 > 28).

Thus the only possible value is **34**.

**Answer:** 34 → Choice **E**.  

ANSWER 2: E  

---

**Problem 3:**  
A four‑digit integer with all digits even can use digits \{0,2,4,6,8\} for each place, except the thousands digit cannot be 0.  

Number of such integers: \(4\times5\times5\times5=500\).

A number is divisible by 4 iff its last two digits form a number divisible by 4.  
List all two‑digit endings with even digits (25 possibilities).  

Checking each yields 15 endings that are multiples of 4:  
00, 04, 08, 20, 24, 28, 40, 44, 48, 60, 64, 68, 80, 84, 88.

Thus the proportion is \(\frac{15}{25}=\frac{3}{5}\).

**Answer:** \(\frac{3}{5}\) → Choice **D**.  

ANSWER 3: D  

---

**Problem 4:**  
6 % tax on \$20 = \(0.06\times20 = \$1.20\).  
6.5 % tax on \$20 = \(0.065\times20 = \$1.30\).  

Difference = \$1.30 − \$1.20 = **\$0.10**.

**Answer:** \$0.10 → Choice **B**.  

ANSWER 4: B  

---

**Problem 5:**  
Sum from 1901 to 1993 (inclusive):  
Number of terms = \(1993-1901+1 = 93\).  
Sum = \(\frac{(1901+1993)\times93}{2}= \frac{3894\times93}{2}=181\,071\).

Sum from 101 to 193 (inclusive):  
Number of terms = \(193-101+1 = 93\).  
Sum = \(\frac{(101+193)\times93}{2}= \frac{294\times93}{2}=13\,671\).

Difference = \(181\,071-13\,671 = 167\,400\).

**Answer:** 167 400 → Choice **A**.  

ANSWER 5: A  

---

**Problem 6:**  
The sign pattern for the numbers \(n\) is:
- \(n\equiv1\pmod4\) → +,
- \(n\equiv2\pmod4\) → −,
- \(n\equiv3\pmod4\) → −,
- \(n\equiv0\pmod4\) → +.

Thus each block of four consecutive integers \(\{4k+1,4k+2,4k+3,4k+4\}\) contributes  
\((4k+1)-(4k+2)-(4k+3)+(4k+4)=0\).

From 1 through 1996 there are \(1996/4=499\) complete blocks, so the total sum is **0**.

**Answer:** 0 → Choice **C**.  

ANSWER 6: C  

---

**Problem 7:**  
Let \(S\) = people wearing sunglasses, \(|S|=50\).  
Let \(C\) = people wearing caps, \(|C|=35\).  
Let \(x=|S\cap C|\).

Given \(P(S\mid C)=\frac{x}{35}= \frac{2}{5}\).  
Thus \(x = \frac{2}{5}\times35 = 14\).

We need \(P(C\mid S)=\frac{x}{50}= \frac{14}{50}= \frac{7}{25}\).

**Answer:** \(\frac{7}{25}\) → Choice **B**.  

ANSWER 7: B  

---

**Problem 8:**  
A three‑digit number with digit sum 26 must use digits that sum to 26.  
The only triple of single‑digit numbers with sum 26 is \(\{9,9,8\}\).

The number is even only if its units digit is 8.  
Among the permutations of 9, 9, 8, only 998 ends in 8.  

Thus exactly **1** such number exists.

**Answer:** 1 → Choice **A**.  

ANSWER 8: A  

---

**Problem 9:**  
Let the digits be \(a,b,c\) (hundreds, tens, units) with  
\(1\le a\le9,\ 0\le b,c\le9,\ a+b+c=25\).

Set \(a'=a-1\) (so \(0\le a'\le8\)). Then  
\(a'+b+c=24\).

Count unrestricted non‑negative solutions: \(\binom{24+3-1}{2}= \binom{26}{2}=325\).

Subtract solutions where a digit exceeds its bound:
- \(a'\ge9\): treat \(a''=a'-9\) → \(a''+b+c=15\) → \(\binom{17}{2}=136\).
- \(b\ge10\): \(b'=b-10\) → \(a'+b'+c=14\) → \(\binom{16}{2}=120\).
- \(c\ge10\): similarly 120.

Intersections:
- \(a'\ge9,\ b\ge10\): \(a''+b'+c=5\) → \(\binom{7}{2}=21\).
- \(a'\ge9,\ c\ge10\): 21.
- \(b\ge10,\ c\ge10\): \(a'+b'+c'=4\) → \(\binom{6}{2}=15\).

No triple intersection (would require sum –5).

By inclusion–exclusion, invalid = \(136+120+120-21-21-15 = 319\).  
Valid = \(325-319 = 6\).

**Answer:** 6 → Choice **C**.  

ANSWER 9: C  

---

**Problem 10:**  
Exits numbered 5 through 26 give 22 exits, hence 21 intervals.  
Let the distances between consecutive exits be \(d_1,\dots,d_{21}\).  
We know \(\sum d_i = 118\) miles, and each \(d_i\ge5\).

To maximize the largest interval, make the othe

Problem 1:  
We are given a set $ R $ of nine distinct integers, six of which are $ 2, 3, 4, 6, 9, 14 $. The median of a set of nine numbers is the fifth number when sorted. To find the number of possible values for the median, we consider the possible placements of the three unknown numbers relative to the known ones. The known numbers sorted are $ 2, 3, 4, 6, 9, 14 $. The three unknown numbers can be placed in various positions to shift the median. The possible medians occur when the additional numbers are placed such that the fifth number in the sorted list can be $ 4, 5, 6, 7, 8, $ or $ 9 $. This gives six possible values for the median.  
ANSWER 1: D (corrected to C? Wait, no: the possible medians are 4, 5, 6, 7, 8, 9 — that’s 6 values, so C)  
**Wait, let's re-evaluate:** The known numbers are 2,3,4,6,9,14. Adding three numbers. The median is the 5th number. To get 4 as median: three numbers ≤4, but only 2,3,4 are available — but they are distinct, so we can add numbers like 1,5? Wait no, the three unknown numbers can be placed in different positions. For example:  
- If all three unknown numbers are ≤4: but only three such numbers exist (2,3,4) but they are already in the set. So we cannot add more. Thus, the unknown numbers must be placed in other positions.  
Actually, the unknown numbers can be:  
- All less than 2: median could be 4 if three numbers are added <4, but only 2,3,4 are already present. We cannot add numbers less than 2 without repeating? Wait, the set has distinct integers. So we can add numbers like 1,0,-1, but the problem doesn’t restrict to positive. However, the known numbers are positive. But the problem says "distinct integers", so negatives are allowed. But the median values: if we add numbers less than 2, the sorted order becomes [new numbers, 2,3,4,6,9,14]. With three new numbers, the sorted list has 9 numbers. The fifth is the median. If we add three numbers less than 2, the sorted list is [a,b,c,2,3,4,6,9,14] — no, wait, adding three numbers less than 2: the sorted list would be [a,b,c,2,3,4,6,9,14] — but a,b,c are less than 2, so the first three, then 2 is fourth, 3 is fifth? Wait, no: the original known numbers are 2,3,4,6,9,14. Adding three numbers, say x,y,z. The full sorted list will be the combination. To get median=4, we need the fifth number to be 4. So in the sorted list of 9, the fifth must be 4. That requires that four numbers are ≤4 and four are ≥4? No: the fifth is the middle. For the fifth to be 4, we need at least five numbers ≤4? No: in a sorted list of 9, the fifth is the median. For the fifth to be 4, we need at least five numbers ≤4? Actually, no: the fifth number is the one where four are less and four are greater? No: in a sorted list of 9, the fifth is the median, meaning four are ≤ median and four are ≥ median. But since numbers can be equal, but here all distinct. So for 4 to be the median, we need exactly four numbers less than 4 and four numbers greater than 4? But 4 is in the set. So the sorted list must have four numbers <4, then 4, then four numbers >4. But the known numbers include 2,3,4,6,9,14. So 2,3 are <4, and 6,9,14 are >4. So currently, without unknowns, we have two <4 and three >4. Adding three unknowns. To have four <4 and four >4, we need to add two numbers <4 and one number >4? But then the sorted list would be [two new <4, 2,3,4, then the new >4, then 6,9,14] — but that’s only three numbers after 4? Wait, no: the sorted list would be [new <4 (two), 2,3,4, new >4, 6,9,14] — that’s 9 numbers? 2+3+1+3=9? Wait, no: the two new <4, then 2,3,4 (three numbers), then the new >4, then 6,9,14 (three numbers). So total: 2+3+1+3=9. The fifth number is 4. So yes, we can achieve median=4 by adding two numbers <4 and one number >4. Similarly, we can achieve:  
- Median=5: by adding numbers such that the fifth number is 5. But 5 is not in the set. So we need the fifth number to be 5. This requires that four numbers are <5 and four are >5. But 5 is not present. So we need to add 5? But the set must consist of distinct integers, and 5 is not in the given. So we can add 5. Then the sorted list would include 5. For example, add 5, and two other numbers. To have four <5 and four >5: currently, without unknowns, numbers <5: 2,3,4 (three), numbers >5: 6,9,14 (three). So to have four <5, we need to add one number <5. To have four >5, we need to add one number >5. But we are adding three numbers. So we can add 5, and say 1 and 10. Then sorted: 1,2,3,4,5,6,9,10,14. Median is 5. So median=5 is achievable. Similarly, median=6: we can add numbers such that the fifth is 6. Currently, numbers <6: 2,3,4 (three), numbers >6: 9,14 (two). So to have four <6 and four >6, we need to add one number <6 and two numbers >6. For example, add 5,10,15. Then sorted: 2,3,4,5,6,9,10,14,15. Median=6. Similarly, median=7: add numbers so fifth is 7. Currently, <7: 2,3,4,6 (four), >7: 9,14 (two). So to have four <7 and four >7, we need to add zero <7 (since already four) and two >7. But we are adding three numbers. So we can add 7,8,15. Then sorted: 2,3,4,6,7,8,9,14,15. Median=7. Similarly, median=8: add 8, and two numbers, say 10,11. Then sorted: 2,3,4,6,8,9,10,11,14. Median=8? The fifth is 8. But to have four <8 and four >8: currently, without unknowns, <8: 2,3,4,6 (four), >8:9,14 (two). So to have four >8, we need to add two numbers >8. But we are adding three numbers. So we can add 8,9,10? But 9 is already there. Add 8,10,11. Then sorted: 2,3,4,6,8,9,10,11,14. The fifth is 8. So yes. Similarly, median=9: add numbers so fifth is 9. Currently, <9: 2,3,4,6 (four), >9:14 (one). So to have four >9, we need to add three numbers >9. But we are adding three numbers. So add 10,11,12. Then sorted: 2,3,4,6,9,10,11,12,14. Median=9. So possible medians: 4,5,6,7,8,9. That’s six values.  
ANSWER 1: C  

Problem 2:  
We are given the list 2,6,7,7,28. A number $ N $ is added. The mean becomes twice the median. The original list has five numbers. After adding $ N $, there are 

### Problem 1:
We have a set of 9 distinct integers, so the median (middle value when sorted) is the 5th element. The 6 known numbers sorted are [2,3,4,6,9,14], with 3 additional distinct integers to add. To find possible median values:
- The minimum possible median is 3: add 3 numbers smaller than 2, sorted set becomes [a,b,c,2,3,4,6,9,14], 5th term=3.
- The maximum possible median is 9: add 3 numbers larger than 14, sorted set becomes [2,3,4,6,9,14,x,y,z], 5th term=9.
- All integers between 3 and 9 are valid medians: 3,4,5,6,7,8,9, for 7 total possible values.
Tempting wrong choices: any count less than 7, from missing intermediate medians that can be formed by adding integers between the known values.
ANSWER 1: D

---

### Problem 2:
Original list sum: 2+6+7+7+28=50. Adding N gives a list of 6 elements, sum=50+N, mean=(50+N)/6. The median of 6 elements is the average of the 3rd and 4th sorted terms. The problem states mean=2×median, so 50+N=12×median.
- If N>28, sorted list is [2,6,7,7,28,N], median=(7+7)/2=7. Then 50+N=12×7=84 → N=34, which fits.
Testing other options confirms no other N satisfies the condition.
ANSWER 2: E

---

### Problem 3:
A number is divisible by 4 if its last two digits form a number divisible by 4. For 4-digit numbers with all even digits:
- Total count: 4 (non-zero thousands digits:2,4,6,8) ×5×5×5 (all even digits for other places)=500.
- Valid last two-digit pairs (both even, divisible by 4): For any even tens digit, 10×even is divisible by 4, so only units digits 0,4,8 work, giving 3 valid units digits per tens digit: 5×3=15 valid pairs.
- Fraction: (4×5×15)/500=300/500=3/5.
ANSWER 3: D

---

### Problem 4:
The tax rate difference is 6.5%-6%=0.5%. 0.5% of $20 is 0.005×20=$0.10.
ANSWER 4: B

---

### Problem 5:
Both sums have the same number of terms: 1993-1901+1=93 terms, 193-101+1=93 terms. Subtract term-by-term: (1901-101)+(1902-102)+...+(1993-193)=1800×93=167400.
ANSWER 5: A

---

### Problem 6:
Group terms in sets of 4: (1-2-3+4)+(5-6-7+8)+... each group sums to 0. There are 1996/4=499 groups, total sum=0.
ANSWER 6: C

---

### Problem 7:
Number of people with both caps and sunglasses: 35 cap-wearers × 2/5=14. Probability a sunglass-wearer has a cap:14/50=7/25.
ANSWER 7: B

---

### Problem 8:
An even 3-digit number with digit sum 26 must have an even units digit. The only possible even units digit is 8 (any smaller even digit leaves a sum of the first two digits >18, impossible). The only number is 998, so 1 valid number.
ANSWER 8: A

---

### Problem 9:
Three-digit numbers with digit sum 25 are all permutations of (9,9,7) and (9,8,8). There are 3 permutations of (9,9,7) and 3 permutations of (9,8,8), for 6 total numbers.
ANSWER 9: C

---

### Problem 10:
Between exit 5 and 26, there are 21 gaps. To maximize one gap, minimize the other 20 gaps at 5 miles each: 20×5=100 miles. Maximum gap=118-100=18 miles.
ANSWER 10: C

---

### Problem 11:
Total sum of 5 scores=5×90=450. Median=91, so sorted scores [a,b,91,c,d]. Mode=94, so 94 must appear at least twice, so c=d=94, sum c+d=188. Sum of the two lowest scores=450-91-188=171.
ANSWER 11: B

---

### Problem 12:
Using +,-,* exactly once, one valid calculation: 5-4+6×3=1+18=19, which uses all three operations and matches option E.
ANSWER 12: E

### Problem 1:
We need to find the number of possible median values for a set of 9 distinct integers, 6 of which are 2, 3, 4, 6, 9, 14. The median of 9 numbers is the 5th element when sorted in ascending order. We add 3 distinct integers (not in the given set) and analyze possible medians:
1.  **Case 1: All 3 new numbers < 2**: Sorted list is [a,b,c,2,3,4,6,9,14], median = 3.
2.  **Case 2: 2 new numbers < 2, 1 new number ≥4**: Sorted list is [a,b,2,3,4,...], median =4.
3.  **Case 3: 1 new number <2**:
    - If we add 5 (the only integer between 4 and 6): Sorted list is [a,2,3,4,5,6,...], median=5.
    - If we do not add 5: Sorted list is [a,2,3,4,6,...], median=6.
4.  **Case 4: 0 new numbers <2 (all new numbers ≥5)**:
    - Add 5: Sorted list [2,3,4,5,6,...], median=6.
    - Add a number between 6 and9 (7 or 8): Sorted list [2,3,4,6,7,...] or [2,3,4,6,8,...], median=7 or 8.
    - Add all new numbers ≥9: Sorted list [2,3,4,6,9,...], median=9.
The only impossible medians are 2 (needs 5 numbers ≤2, only 4 max) and values >9 (existing numbers already have 5 elements ≤9). The possible medians are 3,4,5,6,7,8,9: 7 values.
ANSWER 1: D
---
### Problem 2:
Original list: [2,6,7,7,28], sum = 2+6+7+7+28=50. Insert N to make 6 elements. The median of 6 elements is the average of the 3rd and 4th sorted elements. The mean is $\frac{50+N}{6}$, which equals twice the median:
$$\frac{50+N}{6} = 2 \times \text{median} \implies 50 + N = 12 \times \text{median} \implies N = 12 \times \text{median} -50.$$
Test possible median values from the sorted list with N:
- If N ≤7: Sorted list has 3rd=7, 4th=7, median=7. Then $N=12*7 -50=84-50=34$, which is >7, valid.
- Other median values (e.g., 6.5) give N=28, which does not fit the required sorted position.
Verify N=34: Sorted list [2,6,7,7,28,34], median=(7+7)/2=7, mean=(50+34)/6=14, which is twice 7.
ANSWER 2: E
---
### Problem 3:
Positive 4-digit numbers with all even digits:
- Thousands place: 2,4,6,8 (4 choices, no 0).
- Hundreds, tens, units: 0,2,4,6,8 (5 choices each).
Total numbers: $4 \times 5 \times 5 \times5=500$.
A number is divisible by 4 iff its last two digits form a number divisible by 4. For even tens digits (0,2,4,6,8), a two-digit number $10a + b$ is divisible by 4 iff $b$ (units digit) is 0,4,8 (since $10a \equiv 0 \mod4$ for even a). So valid endings: 5 (tens choices) ×3 (units choices)=15.
Total valid numbers: $4 \times5 \times5 \times3=300$. Fraction: $\frac{300}{500}=\frac{3}{5}$.
ANSWER 3: D
---
### Problem 4:
Difference between 6.5% and 6% tax on $20:
$$0.065*20 - 0.06*20 = (0.065-0.06)*20=0.005*20=0.10.$$
ANSWER 4: B
---
### Problem 5:
Both sums have 93 terms (1993-1901+1=93, 193-101+1=93). Each term in the first sum is 1800 more than the corresponding term in the second sum:
Total difference = $93 \times 1800 = 167400$.
ANSWER 5: A
---
### Problem 6:
The sign pattern repeats every 4 terms: $+ - - +$, and each block of 4 sums to 0:
$$(1-2-3+4)+(5-6-7+8)+\dots+(1993-1994-1995+1996)=0.$$
1996 is divisible by 4, so there are 499 such blocks, total sum 0.
ANSWER 6: C
---
### Problem 7:
Let $S$ = sunglass wearers ($|S|=50$), $C$ = cap wearers ($|C|=35$). $P(S|C)=\frac{|S∩C|}{|C|}=\frac{2}{5}$, so $|S∩C|=\frac{2}{5}*35=14$.
The probability a sunglass wearer wears a cap is $P(C|S)=\frac{|S∩C|}{|S|}=\frac{14}{50}=\frac{7}{25}$.
ANSWER 7: B
---
### Problem 8:
3-digit numbers with digit sum 26: maximum digit sum is 27 (999), so 26 is 1 less than maximum, meaning digits are permutations of (9,9,8). The only even permutation is 998 (units digit 8). So 1 number.
ANSWER 8: A
---
### Problem 9:
3-digit numbers with digit sum 25: 25 is 2 less than 27, so digits are permutations of (9,9,7) and (9,8,8):
- Permutations of (9,9,7): 997,979,799 (3 numbers).
- Permutations of (9,8,8): 988,898,889 (3 numbers).
Total 6 numbers.
ANSWER 9: C
---
### Problem 10:
Between exit 5 and 26, there are $26-5=21$ gaps, total distance 118 miles. Each gap ≥5 miles. To maximize one gap, minimize the other 20: $20*5=100$, so maximum gap = $118-100=18$.
ANSWER 10: C
---
### Problem 11:
Five sorted scores: $a≤b≤c≤d≤e$. Median $c=91$, mean 90, so total sum $5*90=450$, so $a+b+d+e=450-91=359$. Mode is 94, so 94 appears at least twice. Since $c=91$, 94 can only be in $d$ and $e$, so $d=e=94$. Then $a+b=359-94-94=171$. This sum is fixed (valid examples exist, e.g., 81+90=171).
ANSWER 11: B
---
### Problem 12:
Permute +, -, × in 5 __4 __6 __3, using standard order of operations (multiply first):
- $5 -4 +6×3 = 5-4+18=19$, which is an option.
All other permutations give values not in the choices.
ANSWER 12: E

I'll solve each problem step by step.

---

**Problem 1:**

We have 6 elements: {2, 3, 4, 6, 9, 14}, and need to add 3 more distinct integers to make 9 total.

The median of 9 numbers is the 5th element when arranged in order.

Currently we have 6 numbers. By strategically placing 3 new numbers, we can shift which value becomes the median.

Let me consider what values can be the median (5th position):
- If we add 3 numbers all ≤ 2, the sorted list has our 6 original numbers in positions 4-9, so median is 4
- If we add numbers strategically between and around the original set, we can make the median be: 2, 3, 4, 6, 9, 14, or a new number
- If we add 3 numbers all ≥ 14, the sorted list has our 6 original numbers in positions 1-6, so median is 9

Testing systematically:
- Median = 2: add three numbers < 2 (e.g., -2, -1, 0) ✓
- Median = 3: add two < 2, one between 3 and 4 ✓
- Median = 4: add three < 2, or arranged appropriately ✓
- Median = 6: add one or two strategically ✓
- Median = 9: add three > 14, or arranged appropriately ✓
- Median = 14: add three > 14 ✓
- Median = a new value between existing numbers ✓

The possible medians are: 2, 3, 4, 6, 9, 14, and any new integer we insert. We can get 7 different values.

**ANSWER 1: D**

---

**Problem 2:**

Original list: 2, 6, 7, 7, 28. Sum = 50.

When N is inserted, we have 6 numbers. Let's find the median of 6 numbers (average of 3rd and 4th when sorted).

Sum of all 6 numbers = 50 + N
Mean = (50 + N)/6

We need: Mean = 2 × Median

Let me test by considering where N falls:
- If N ≤ 2: sorted list is N, 2, 6, 7, 7, 28. Median = (6+7)/2 = 6.5. Mean = 2×6.5 = 13, so 50+N = 78, N = 28. Contradiction.
- If 2 < N ≤ 6: sorted list is 2, N, 6, 7, 7, 28. Median = (6+7)/2 = 6.5. Mean = 13, so N = 28. Contradiction.
- If 6 < N ≤ 7: sorted list is 2, 6, N, 7, 7, 28. Median = (N+7)/2. Mean = 2×(N+7)/2 = N+7. So (50+N)/6 = N+7, giving 50+N = 6N+42, so 5N = 8, N = 1.6. Not in range.
- If 7 < N ≤ 28: sorted list is 2, 6, 7, N, 7, 28 or 2, 6, 7, 7, N, 28. Median = (7+N)/2. Mean = 2×(7+N)/2 = 7+N. So (50+N)/6 = 7+N, giving 50+N = 42+6N, so 5N = 8, N = 1.6. Not valid.
- If N > 28: sorted list is 2, 6, 7, 7, 28, N. Median = (7+7)/2 = 7. Mean = 14. So 50+N = 84, N = 34. ✓

**ANSWER 2: E**

---

**Problem 3:**

Four-digit numbers with all even digits: first digit ∈ {2,4,6,8}, others ∈ {0,2,4,6,8}.

Total count: 4 × 5 × 5 × 5 = 500

For divisibility by 4, the last two digits must form a number divisible by 4.

Last two digits from {00, 02, 04, 06, 08, 20, 22, 24, 26, 28, 40, 42, 44, 46, 48, 60, 62, 64, 66, 68, 80, 82, 84, 86, 88}

Divisible by 4: 00, 04, 08, 20, 24, 28, 40, 44, 48, 60, 64, 68, 80, 84, 88 (15 out of 25)

Count divisible by 4: 4 × 5 × 15 = 300

But wait, the second digit is 5 choices, third is which determines divisibility by 4.

Actually: 4 × 5 × (number of even two-digit combos divisible by 4) = 4 × 5 × 15 = 300

Hmm, let me recalculate. First digit: 4 choices. Second digit: 5 choices. Last two digits must be divisible by 4: check all pairs from {0,2,4,6,8}×{0,2,4,6,8} giving 15 valid pairs.

Count = 4 × 5 × 15 = 300. Fraction = 300/500 = 3/5.

**ANSWER 3: D**

---

**Problem 4:**

6.5% tax on $20: 0.065 × 20 = $1.30
6% tax on $20: 0.06 × 20 = $1.20
Difference: $1.30 - $1.20 = $0.10

**ANSWER 4: B**

---

**Problem 5:**

Sum from 1901 to 1993: This is 93 consecutive integers starting at 1901.
Sum = (1901 + 1993)/2 × 93 = 1947 × 93 = 181,071

Sum from 101 to 193: This is 93 consecutive integers starting at 101.
Sum = (101 + 193)/2 × 93 = 147 × 93 = 13,671

Difference = 181,071 - 13,671 = 167,400

**ANSWER 5: A**

---

**Problem 6:**

Pattern: 1 - 2 - 3 + 4 + 5 - 6 - 7 + 8 + 9 - 10 - 11 + ...

Group by 4s: (1-2-3+4) + (5-6-7+8) + (9-10-11+12) + ... + (1993-1994-1995+1996)

Each group of 4: (4k+1) - (4k+2) - (4k+3) + (4k+4) = 4k+1-4k-2-4k-3+4k+4 = 0

From 1 to 1996, we have 1996/4 = 499 complete groups, each summing to 0.

Total = 0

**ANSWER 6: C**

---

**Problem 7:**

Let x = people wearing both sunglasses and caps.

P(sunglasses | cap) = x/35 = 2/5, so x = 14.

P(cap | sunglasses) = 14/50 = 7/25.

**ANSWER 7: B**

---

**Problem 8:**

Three-digit even numbers with digit sum 26.

For an even number, last digit ∈ {0, 2, 4, 6, 8}.

If last digit = 0: first two digits sum to 26. Max is 9+9=18. Impossible.
If last digit = 2: first two digits sum to 24. Only 9+9=18 < 24. Impossible.
If last digit = 4: first two digits sum to 22. Impossible (max 18).
If last digit = 6: first two digits sum to 20. Impossible.
If last digit = 8: first two digits sum to 18. This requires 9+9=18. Number: 998. ✓

Only 1 number.

**ANSWER 8: A**

---

**Problem 9:**

Three-digit numbers abc where a+b+c=25, with a ∈ {1,...,9}, b,c ∈ {0,...,9}.

a+b+c = 25 means b+c = 25-a.

For valid solutions: 0 ≤ b+c ≤ 18, so 7 ≤ a ≤ 9.

- a=7: b+c=18. Solutions: (9,9) → 1 way
- a=8: b+c=17. Solutions: (8,9), (9,8) → 2 ways
- a=9: b+c=16. Solutions: (7,9), (8,8), (9,7) → 3 ways

Total = 1 + 2 + 3 = 6

**ANSWER 9: C**

---

**Problem 10:**

Exits from 5th to 26th = 26-5 = 21 segments between consecutive exits.

Each segment ≥ 5 miles. Total distance = 118 miles.

If 20 segments are 5 miles each: 20 × 5 = 100 miles.
Remaining for the largest segment: 118 - 100 = 18 miles.

**ANSWER 10: C**

---

**Problem 11:**

Five scores: mean = 90 (sum = 450), median = 91, mode = 94.

Arranged in order: a, b, c, d, e where c = 91 (median).

Mode = 94 means 94 appears at least twice. Since median is 91, we must have 94 appearing in {d, e}.

Case 1: d = e = 94. Then a + b + 91 + 94 + 94 = 450, so a + b = 171.

Check consistency: we need a ≤ b ≤ 91 and mode is 94 (appears twice). If a or b equals 94, mode wouldn't be unique. So a, b < 91 works. ✓

**ANSWER 11: B**

---

**Problem 12:**

We need to place +, −, × in three blanks: 5 __ 4 __ 6 __ 3

Trying combinations:
- 5 + 4 × 6 − 3 = 5 + 24 − 3 = 26
- 5 × 4 + 6 − 3 = 20 + 6 − 3 = 23
- 5 × 4 − 6 + 3 = 20 − 6 + 3 = 17
- 5 + 4 − 6 × 3 = 5 + 4 − 18 = −9
- 5 − 4 

Problem 1:

We are given a set of 9 distinct integers. Six of them are
\[
2,3,4,6,9,14.
\]
We want the number of possible values of the median of the 9 numbers.

For 9 numbers, the median is the 5th number when the list is ordered.

Now look at the six given numbers in sorted order:
\[
2,3,4,6,9,14.
\]
Since there are 3 other distinct integers, the median can be any number that ends up in the 5th position after inserting those 3 numbers.

Let’s see what values are possible:

- If we place 3 new numbers all very small, then the 5th number could be 4.
- It could also be 6, or 9, or even 14, depending on how many new numbers are inserted before each.
- It cannot be 2 or 3, because there would need to be at least 4 numbers less than or equal to it, but only 3 extra numbers are available and the given numbers already limit this.
- It cannot be any number not among the given six if the median must be one of the existing numbers? Let’s check carefully: since the 3 extra numbers are distinct integers and can be placed anywhere, the 5th number could indeed be one of the inserted numbers, but then it would have to lie between the 4th and 6th numbers. Since the six given numbers already occupy several possible positions, the median can also be a new integer. But because the set is of distinct integers and the question asks how many possible values, we should count all possible medians that can occur.

A cleaner way: the median of 9 distinct integers is the middle element. With 6 fixed numbers, the 3 added numbers can be arranged so that the median is any integer between the 4th smallest and 6th smallest of the fixed numbers, inclusive, but here the possible distinct medians are:
\[
4,5,6,7,8,9
\]
Wait — can all of these really happen? Let’s test.

- Median 4: choose 3 new numbers less than 4.
- Median 5: choose 3 new numbers so that 5 is between 4 and 6 and becomes 5th.
- Median 6: choose 3 new numbers with 6 as middle.
- Median 7, 8, 9 similarly.

So the possible medians are the integers from 4 through 9, which is 6 values.

However, we must match the provided choices. The intended count is the number of possible median values among the listed integers and inserted integers. Since the only restriction is distinct integers, every integer from 4 through 9 can occur.

That gives 6 possible values.

ANSWER 1: C

---

Problem 2:

We insert a number \(N\) into the list
\[
2,6,7,7,28.
\]
Then the mean becomes twice the median.

Let’s analyze the median after inserting \(N\) into 6 numbers total.

The original sorted list is:
\[
2,6,7,7,28.
\]
There are 5 numbers. After inserting one number, there are 6 numbers, so the median is the average of the 3rd and 4th numbers in the sorted list.

We try the choices.

If \(N=7\), the list becomes
\[
2,6,7,7,7,28.
\]
Median:
\[
\frac{7+7}{2}=7.
\]
Mean:
\[
\frac{2+6+7+7+7+28}{6}=\frac{57}{6}=9.5.
\]
Twice the median would be 14, not 9.5, so not 7.

Try \(N=14\):
\[
2,6,7,7,14,28.
\]
Median:
\[
\frac{7+7}{2}=7.
\]
Mean:
\[
\frac{2+6+7+7+14+28}{6}=\frac{64}{6}=\frac{32}{3},
\]
not 14.

Try \(N=20\):
\[
2,6,7,7,20,28.
\]
Median is still 7.
Mean:
\[
\frac{2+6+7+7+20+28}{6}=\frac{70}{6}=\frac{35}{3},
\]
not 14.

Try \(N=28\):
\[
2,6,7,7,28,28.
\]
Median is 7.
Mean:
\[
\frac{2+6+7+7+28+28}{6}=\frac{78}{6}=13,
\]
not 14.

Try \(N=34\):
\[
2,6,7,7,28,34.
\]
Median is 7.
Mean:
\[
\frac{2+6+7+7+28+34}{6}=\frac{84}{6}=14.
\]
Twice the median is
\[
2\cdot 7=14.
\]
This works.

So \(N=34\).

ANSWER 2: E

---

Problem 3:

We consider all positive four-digit integers whose digits are all even.

Even digits are
\[
0,2,4,6,8.
\]
A four-digit number cannot start with 0, so the thousands digit has 4 choices:
\[
2,4,6,8.
\]
Each of the other three digits has 5 choices, so the total number is
\[
4\cdot 5^3.
\]

Now determine how many are divisible by 4.

A number is divisible by 4 if its last two digits form a number divisible by 4.

So we only need to count ordered pairs of even digits \((t,u)\) from \(\{0,2,4,6,8\}\) such that \(10t+u\) is divisible by 4.

Check the possible last two digits:

- \(00\) yes
- \(02\) no
- \(04\) yes
- \(06\) no
- \(08\) yes

- \(20\) yes
- \(22\) no
- \(24\) yes
- \(26\) no
- \(28\) yes

- \(40\) yes
- \(42\) no
- \(44\) yes
- \(46\) no
- \(48\) yes

- \(60\) yes
- \(62\) no
- \(64\) yes
- \(66\) no
- \(68\) yes

- \(80\) yes
- \(82\) no
- \(84\) yes
- \(86\) no
- \(88\) yes

For each tens digit, 3 of the 5 possibilities work, so there are
\[
5\cdot 3=15
\]
valid last-two-digit pairs.

Since the first two digits can be anything allowed:
- thousands digit: 4 choices
- hundreds digit: 5 choices

Total divisible by 4:
\[
4\cdot 5\cdot 15
\]
Total numbers:
\[
4\cdot 5^3 = 4\cdot 125
\]
So the fraction is
\[
\frac{4\cdot 5\cdot 15}{4\cdot 5^3}=\frac{15}{25}=\frac35.
\]

ANSWER 3: D

---

Problem 4:

An item costs \$20 before tax.

Difference between 6.5% tax and 6% tax is
\[
0.5\% = 0.005.
\]
So the difference in tax is
\[
0.005 \times 20 = 0.1
\]
dollar, which is \(\$0.10\).

ANSWER 4: B

---

Problem 5:

We need to compute
\[
(1901+1902+\cdots+1993) - (101+102+\cdots+193).
\]

Notice each term in the second sum is exactly 1800 less than the corresponding term in the first sum:
\[
1901-101=1800,\quad 1902-102=1800,\ \ldots,\ 1993-193=1800.
\]
How many terms are there?

From 1901 to 1993 inclusive:
\[
1993-1901+1=93
\]
terms.

So the difference is
\[
93 \cdot 1800 = 167400.
\]

ANSWER 5: A

---

Problem 6:

We evaluate
\[
1-2-3+4+5-6-7+8+9-10-11+\cdots+1992+1993-1994-1995+1996.
\]

Look at the pattern in blocks of 4:
\[
(1-2-3+4),\ (5-6-7+8),\ (9-10-11+12),\ \ldots
\]
Each block of 4 consecutive integers has sum
\[
n-(n+1)-(n+2)+(n+3)=0
\]
when \(n\equiv 1 \pmod 4\).

So all full blocks cancel.

Now check the last block:
The pattern continues through \(1992+1993-1994-1995+1996\), which is not a full 4-term block. But we can group from 1 to 1996 in blocks of 4:
\[
(1-2-3+4)+(5-6-7+8)+\cdots+(1993-1994-1995+1996).
\]
Each block sums to 0.

Therefore the entire 

Problem 1:  
We have 9 distinct integers. Six of them are fixed: 2, 3, 4, 6, 9, 14. The other 3 are distinct integers of our choice.  
With 9 numbers, the median is the 5th number when sorted.

So the median depends only on how many of the fixed numbers fall below/at various positions, and where we insert the 3 new numbers.

Let’s analyze possible 5th elements.

### Step 1: Sort the fixed numbers
Fixed numbers in increasing order:
2, 3, 4, 6, 9, 14

The median (5th) could be one of these fixed numbers, or one of the new ones.

### Casework by choosing what the 5th number is

#### Can the median be 2?
For 2 to be the median, we need at least 4 numbers ≤ 2, but only 2 itself is ≤ 2 among the fixed set (and new numbers would have to be distinct integers ≤ 2).  
That means we’d need 4 numbers less than or equal to 2, including 2, so we’d need 3 additional distinct integers ≤ 2 (other than 2). That’s possible (e.g., 0, −1, 1).  
But then 2 would be the 5th, meaning exactly 4 numbers are <2. Can we guarantee? Actually, since we can choose any 3 extra integers, we can make exactly 4 numbers less than 2 by picking 3 integers <2 plus none from fixed besides 2. That would make 2 the 5th. However, note 2 is already one of the 9 distinct integers, so it can be the median.  
So potentially 2 is possible.

But the problem asks for the number of possible median values, and the choices are between 4 and 8, so likely only medians among the fixed list can work. Let’s check more carefully: the fixed numbers below 2 is only none (except 2). To make 2 the 5th, we need 4 numbers smaller than 2. We only control 3 extra numbers, so we can only create at most 3 numbers smaller than 2. Therefore, we cannot reach 4 smaller numbers.  
✅ Conclusion: median cannot be 2.

Similarly, we test each fixed candidate using “how many fixed numbers are below it”.

### General method
For a number \(x\) to be the median (5th), there must be exactly 4 numbers less than \(x\), and at least one number equal to \(x\). If \(x\) is one of the fixed numbers, we can count how many fixed numbers are less than \(x\), then see if the remaining “needed” smaller numbers can come from the 3 new integers.

Let’s compute fixed counts less than each candidate:

- Below 2: 0  
- Below 3: 1 (just 2)  
- Below 4: 2 (2,3)  
- Below 6: 3 (2,3,4)  
- Below 9: 4 (2,3,4,6)  
- Below 14: 5 (2,3,4,6,9)

We need exactly 4 numbers less than the median.

Now see possibilities:

#### Median = 3
Fixed numbers below 3: 1.  
We need 4 total below 3, so we must add 3 new numbers < 3. That’s possible.  
✅ So 3 can be median.

#### Median = 4
Fixed below 4: 2. Need 2 more from new numbers < 4. Possible.  
✅ 4 can be median.

#### Median = 6
Fixed below 6: 3. Need 1 more from new numbers < 6. Possible.  
✅ 6 can be median.

#### Median = 9
Fixed below 9: 4 already. Then all 4 smaller numbers are fixed, and the median can be 9 regardless of where new numbers go as long as none are between 9 and 14 in a way that changes ordering?  
We only need 9 to be the 5th, meaning at least 4 numbers are <9 (we have exactly 4 fixed ones). If we add any new numbers <9, that would make more than 4 numbers less than 9 and break it. So we must choose the 3 new integers all > 9. That’s possible.  
✅ 9 can be median.

#### Median = 14
Fixed below 14: 5 already. But we need exactly 4 below the median. Impossible.  
❌ 14 cannot be median.

#### Median = 2
Fixed below 2: 0, need 4 from new numbers <2, but we only have 3 new numbers. Impossible.  
❌ 2 cannot be median.

So the only fixed medians are: 3, 4, 6, 9 → 4 values.

Are there possible medians that are not among the fixed numbers (i.e., one of the 3 new integers)?  
Yes, if we choose all new integers appropriately.

Let the median be a new number \(y\). Then among the fixed numbers, some number of them must be less than \(y\). Since there are only 6 fixed numbers, the possible counts of fixed elements less than \(y\) range from 0 to 6.

But we need exactly 4 numbers total less than the median. Since the median is new, all 4 numbers less than it must come from:
- some fixed numbers less than it, plus
- some of the other two new numbers (because one new number is the median itself).

We can make any new median that sits in the “gaps” between fixed numbers such that exactly 4 elements end up below it.

The median \(y\) must be positioned so that:
- the number of fixed elements less than \(y\) is either 1,2,3, or 4 (because we then use the remaining smaller ones from the other new numbers, limited to 0,1,2, or 3).
But we must also ensure distinctness and that \(y\) lies in the right interval.

There are gaps:
(-∞,2), (2,3), (3,4), (4,6), (6,9), (9,14), (14,∞)

Try each gap for possible median:

- If \(y\in(2,3)\): fixed less than y = 1 (only 2). Need 3 more < y from the other new numbers. Possible.  
So any integer y in (2,3) means y=?? none (only integer between 2 and 3 doesn’t exist). So no.
- \(y\in(3,4)\): fixed less than y = 2. Need 2 more < y. Possible. But no integer between 3 and 4. none.
- \(y\in(4,6)\): fixed less than y = 3 (2,3,4). Need 1 more from new numbers <y. Possible. Integers in (4,6): only 5. So y=5 works.
- \(y\in(6,9)\): fixed less than y = 4 (2,3,4,6). Need 0 more. Then both other new numbers must be >y. Possible. Integers in (6,9): 7,8 → both could be median.
- \(y\in(9,14)\): fixed less than y = 5, but we need total 4 < median. impossible.

Thus additional possible medians are 5, 7, 8 → 3 more values.

Total possible median values: fixed ones (3,4,6,9) = 4 plus new ones (5,7,8)=3 gives 7.

✅ Answer choice is 7.

ANSWER 1: D


Problem 2:  
List is: 2, 6, 7, 7, 28 (five numbers). Insert N to make 6 numbers total.

New mean and median relation: “mean is now twice as great as the median.”

### Step 1: Compute original sum
Sum of given numbers:
2 + 6 + 7 + 7 + 28 = 50

After inserting N, total sum = 50 + N.  
There are 6 numbers, so mean is:
\[
\text{mean}=\frac{50+N}{6}
\]

### Step 2: Find the median
Sort of the ne

Problem 1:
We have a set $R$ of 9 distinct integers. Six are $\{2, 3, 4, 6, 9, 14\}$. Let the three unknown integers be $x, y, z$. When sorted, the median is the 5th element.
To minimize the median, we pick the three smallest possible integers (e.g., $-1, 0, 1$). The sorted set would be $\{-1, 0, 1, 2, 3, 4, 6, 9, 14\}$. The median is 3.
To maximize the median, we pick the three largest possible integers (e.g., $15, 16, 17$). The sorted set would be $\{2, 3, 4, 6, 9, 14, 15, 16, 17\}$. The median is 9.
The possible values for the median are integers between 3 and 9 inclusive, excluding values already occupied by the fixed set if they cannot be the median. However, we can choose $x, y, z$ to make any integer from 3 to 9 the median.
Values: 3, 4, 5, 6, 7, 8, 9. There are 7 possible values.
ANSWER 1: D

Problem 2:
List: $2, 6, 7, 7, 28$. $N$ is added. Mean = $\frac{2+6+7+7+28+N}{6} = \frac{50+N}{6}$.
If $N$ is added, the median depends on $N$.
If $N \le 7$, median is 7. Mean = $2 \times 7 = 14 \implies 50+N = 84 \implies N=34$ (Contradiction).
If $N \ge 7$, let's test $N=20$: List $\{2, 6, 7, 7, 20, 28\}$, median is $(7+7)/2 = 7$. Mean = $70/6 \neq 14$.
If $N=14$: List $\{2, 6, 7, 7, 14, 28\}$, median is $(7+7)/2 = 7$. Mean = $64/6 \neq 14$.
If $N=28$: List $\{2, 6, 7, 7, 28, 28\}$, median is 7. Mean = $78/6 = 13 \neq 14$.
If $N=34$: List $\{2, 6, 7, 7, 28, 34\}$, median is $(7+28)/2 = 17.5$. Mean = $84/6 = 14$. $14 \neq 35$.
Wait, try $N=20$: Mean = $70/6 \approx 11.6$. Median = 7.
Try $N=34$: Mean = 14. Median = 17.5.
Re-check: $50+N = 12 \times \text{median}$. If $N=20$, median is 7, $70 = 12 \times 7$ (No).
If $N=34$, median is 17.5, $84 = 12 \times 17.5$ (No).
Try $N=28$: Mean = 13, Median = 7. $13 \neq 14$.
Try $N=20$: Mean = $70/6 = 11.66$, Median = 7.
Let's check $N=20$ again. $50+20 = 70$. $70/6 = 11.66$.
Actually, if $N=20$, list is $2, 6, 7, 7, 20, 28$. Median is $(7+7)/2 = 7$. $2 \times 7 = 14$. $70/6 \neq 14$.
If $N=34$, median is $(7+28)/2 = 17.5$. Mean = $84/6 = 14$. $14 \neq 2 \times 17.5$.
If $N=20$, mean is $11.66$. If $N=34$, mean is 14.
Looking at choices, $N=20$ is the only one that yields a mean close to twice a median.
ANSWER 2: C

Problem 3:
Digits are $\{0, 2, 4, 6, 8\}$. First digit cannot be 0 (4 choices: 2, 4, 6, 8). Others have 5 choices. Total = $4 \times 5 \times 5 \times 5 = 500$.
Divisible by 4 means the last two digits form a multiple of 4. Possible pairs: 00, 04, 08, 20, 24, 28, 40, 44, 48, 60, 64, 68, 80, 84, 88. (15 pairs).
Total numbers = $4 \times 5 \times 15 = 300$.
Fraction = $300/500 = 3/5$.
ANSWER 3: D

Problem 4:
$0.065 \times 20 = 1.30$. $0.06 \times 20 = 1.20$.
Difference = $1.30 - 1.20 = 0.10$.
ANSWER 4: B

Problem 5:
$(1901-101) + (1902-102) + \dots + (1993-193)$.
Each term is $1800$. There are $1993 - 1901 + 1 = 93$ terms.
$93 \times 1800 = 167,400$.
ANSWER 5: A

Problem 6:
Group into sets of 4: $(1-2-3+4) + (5-6-7+8) + \dots + (1993-1994-1995+1996)$.
Each group: $1-2-3+4 = 0$.
There are $1996/4 = 499$ groups. $499 \times 0 = 0$.
ANSWER 6: C

Problem 7:
$C = 35$. $P(S|C) = 2/5$. So $S \cap C = 35 \times 2/5 = 14$.
$S = 50$. $P(C|S) = (S \cap C) / S = 14/50 = 7/25$.
ANSWER 7: B

Problem 8:
Digits $a, b, c$. $a+b+c = 26$. $c \in \{0, 2, 4, 6, 8\}$.
If $c=8, a+b=18 \implies (9,9)$. (1)
If $c=6, a+b=20$ (None).
If $c=4, a+b=22$ (None).
If $c=2, a+b=24$ (None).
If $c=0, a+b=26$ (None).
Wait, $a+b=18 \implies 998$. $a+b=17 \implies 899, 989$. $a+b=16 \implies 799, 979, 889$.
If $c=8, a+b=18 \implies 998$.
If $c=6, a+b=20$ (None).
If $c=4, a+b=22$ (None).
If $c=2, a+b=24$ (None).
If $c=0, a+b=26$ (None).
Wait, $a+b+c=26$. $c$ must be even.
If $c=8: a+b=18 \implies (9,9) \to 998$.
If $c=6: a+b=20$ (None).
If $c=4: a+b=22$ (None).
If $c=2: a+b=24$ (None).
If $c=0: a+b=26$ (None).
Wait, $a, b$ can be 9. Max $a+b=18$. So $c$ must be at least $26-18=8$.
If $c=8, a+b=18 \implies 998$.
If $c=9$ (not even).
Are there others? $a+b+c=26$. $c \in \{0, 2, 4, 6, 8\}$.
If $c=8, a+b=18 \to 998$.
If $c=6, a+b=20$ (Impossible).
Only 998? Let's re-read. "3-digit whole numbers".
Maybe $c$ can be 8, 6, 4, 2, 0.
$c=8 \implies a+b=18 \implies 998$.
$c=6 \implies a+b=20$ (No).
$c=4 \implies a+b=22$ (No).
$c=2 \implies a+b=24$ (No).
$c=0 \implies a+b=26$ (No).
Only 1.
ANSWER 8: A

Problem 9:
$a+b+c=25$. $a, b, c \in \{0..9\}, a \neq 0$.
Possible sets $\{9,9,7\}, \{9,8,8\}$.
Permutations of $\{9,9,7\}$: $997, 979, 799$ (3).
Permutations of $\{9,8,8\}$: $988, 898, 889$ (3).
Total = $3+3 = 6$.
ANSWER 9: C

Problem 10:
Exits 5 to 26 is $26-5 = 21$ intervals. Total distance 118.
Let $d_i$ be the distance between exits. $\sum_{i=1}^{21} d_i = 118$.
$d_i \ge 5$. To maximize one $d_k$, minimize others: $d_i = 5$ for $i \neq k$.
$d_k + 20(5) = 118 \implies d_k + 100 = 118 \implies d_k = 18$.
ANSWER 10: C

Problem 11:
Sum = $5 \times 90 = 450$. Mode = 94. Median = 91.
Let scores be $x_1, x_2, 91, 94, 94$.
$x_1 + x_2 + 91 + 94 + 94 = 450 \implies x_1 + x_2 + 279 = 450 \implies x_1 + x_2 = 171$.
ANSWER 11: B

Problem 12:
$5 \pm 4 \pm 6 \pm 3$.
$5+4 \times 6 - 3 = 5+24-3 = 26$.
$5 \times 4 - 6 + 3 = 20-6+3 = 17$.
$5 \times 4 + 6 - 3 = 23$.
$5 - 4 + 6 \times 3 = 1 + 18 = 19$.
ANSWER 12: E

The set \(R\) consists of nine distinct integers, including the six given values \(2, 3, 4, 6, 9, 14\). When the elements of \(R\) are placed in increasing order \(a_1 < a_2 < a_3 < a_4 < a_5 < a_6 < a_7 < a_8 < a_9\), the median is \(a_5\). Thus exactly four elements of \(R\) must lie strictly below the median value \(m = a_5\).
Let \(k\) denote the number of the six given values that are strictly less than \(m\). The three unknown values must then be chosen so that exactly \(4 - k\) of them lie strictly below \(m\).
- If \(m\) is one of the given values, the admissible choices are \(m = 3, 4, 6, 9\). For each of these four numbers it is possible to select three unknown integers satisfying the counting requirement above (and remaining distinct from one another and from the given values).
- If \(m\) is not one of the given values, the admissible choices are the integers \(m = 5, 7, 8\). Again, for each of these three numbers it is possible to select the remaining two unknown integers so that exactly four elements lie below \(m\).
No other integer can serve as median: any candidate smaller than 3 or larger than 9 produces a non-negative integer count of unknowns below \(m\) that is impossible to realize with only three unknowns. Consequently the possible values of the median are exactly the seven numbers
\[
3,4,5,6,7,8,9.
\]
Thus seven possible values exist for the median of \(R\).
ANSWER 1: D
Problem 2:
Let the five numbers be \(2, 6, 7, 7, 28, N\). Their sum is \(50 + N\), so the mean is
\[
\frac{50 + N}{5}.
\]
When the six numbers are ordered, the median is the average of the third and fourth values. The value of \(N\) determines the ordered list, and therefore the median, in each of the five mutually exclusive ranges determined by the given numbers.
- If \(N \le 2\), the ordered list is \(N,2,6,7,7,28\) and the median is \((6+7)/2 = 6.5\). Setting the mean equal to twice the median yields the contradictory equation \(N = 15\).
- If \(2 < N \le 6\), the ordered list is \(2,N,6,7,7,28\) and the median is again \(6.5\). The same contradictory equation appears.
- If \(6 < N \le 7\), the ordered list is \(2,6,N,7,7,28\) and the median is \((N+7)/2\). The mean-equals-twice-median equation simplifies to the quadratic
 \[
 N^2 - 9N - 60 = 0,
 \]
 whose only root in the interval is \(N = 12\), which lies outside the interval.
- If \(7 < N \le 28\), the ordered list is \(2,6,7,7,N,28\) and the median is \((7+N)/2\). The mean-equals-twice-median equation simplifies to the linear equation
 \[
 N = 20,
 \]
 which lies inside the interval.
- If \(N > 28\), the ordered list is \(2,6,7,7,28,N\) and the median is \((7+28)/2 = 17.5\). The resulting equation \(N = 120\) lies outside the interval.
Hence the only value that satisfies the stated condition is \(N = 20\).
ANSWER 2: C
Problem 3:
A four-digit integer whose digits are all even may be written
\[
1000a + 100b + 10c + d,
\]
where each of \(a,b,c,d\) belongs to the set \(\{0,2,4,6,8\}\) and \(a \ne 0\). There are therefore
\[
4 \times 5^3 = 500
\]
such integers.
An integer is divisible by 4 if and only if its last two digits form a number divisible by 4. The admissible pairs \((c,d)\) are therefore exactly the 13 pairs
\[
(0,0),(0,4),(0,8),(2,0),(2,4),(2,8),(4,0),(4,4),(4,8),(6,0),(6,4),(8,0),(8,8)
\]
that satisfy the divisibility test. For each of the 13 choices of \((c,d)\) the first two digits may be chosen freely (\(a\) has 4 possibilities, \(b\) has 5), giving
\[
13 \times 4 \times 5 = 260
\]
favorable integers. The required fraction is therefore
\[
\frac{260}{500} = \frac{13}{25} = \frac{2}{5}.
\]
ANSWER 3: B
Problem 4:
A 6.5 percent tax on a $20 item is \(0.065 \times 20 = 1.30\). A 6 percent tax is \(0.06 \times 20 = 1.20\). Their difference is
\[
1.30 - 1.20 = 0.10.
\]
ANSWER 4: B
Problem 5:
The first sum is an arithmetic series with 93 terms whose average is
\[
\frac{1901 + 1993}{2} = 1947,
\]
so its value equals \(93 \times 1947 = 181071\). The second sum is an arithmetic series with 93 terms whose average is
\[
\frac{101 + 193}{2} = 147,
\]
so its value equals \(93 \times 147 = 13671\). Subtracting yields
\[
181071 - 13671 = 167400.
\]
ANSWER 5: A
Problem 6:
The given expression consists of 1996 terms whose signs follow the repeating four-term pattern
\[
+,-,-,+,+,-,-,+,\dots
\]
(The pattern begins with a single “+” and thereafter repeats every four signs.) In each complete block of four consecutive integers the signed sum is zero. There are exactly 499 such blocks, accounting for 1996 terms and contributing a total of zero. Hence the whole expression evaluates to 0.
ANSWER 6: C
Problem 7:
Let \(S\) be the set of people wearing sunglasses and \(C\) the set wearing caps. Then
\[
|S \cap C| = \frac{2}{5}|C| = \frac{2}{5} \cdot 35 = 14.
\]
Consequently
\[
|S| = |S \setminus C| + |S \cap C|
\]
is unknown, but the conditional probability required by the problem is
\[
P(C \mid S) = \frac{|S \cap C|}{|S|} = \frac{14}{|S|}.
\]
Because \(|S \cap C| = 14\) and \(|C| = 35\), the only numerical value that can be obtained from the given data is the ratio
\[
\frac{14}{35} = \frac{2}{5}.
\]
(The actual size of \(S\) is never needed.)
ANSWER 7: C
Problem 8:
A three-digit number \(100a + 10b + c\) (with \(a \in \{1,\dots,9\}\), \(b,c \in \{0,\dots,9\}\)) is even precisely when \(c\) is even. The digit-sum equation \(a + b + c = 26\) together with the upper bound 9 on each variable forces \(a \ge 8\). The only possibilities are therefore the six triples
\[
(9,9,8),\ (9,8,9),\ (8,9,9)
\]
and their permutations that keep the units digit even. Exactly three of them have even units digit:
\[
998,\ 986,\ 968.
\]
Thus three even three-digit numbers have digit-sum 26.
ANSWER 8: C
Problem 9:
A three-digit number \(100a + 10b + c\) satisfies \(a + b + c = 25\) with each digit at most 9. Then necessarily \(a \ge 7\). The admissible triples \((a,b,c)\) are exactly the six permutations of
\[
(9,9,7),\ (9,8,8).
\]
Hence six such numbers exist.
ANSWER 9: C
Problem 10:

## Step 1: Problem 1
To find the number of possible values of the median of set R, we first need to understand what a median is. The median of a set of numbers is the middle number when the numbers are arranged in ascending order. Since R is a set of nine distinct integers, the median will be the fifth number when the numbers are arranged in ascending order.

## Step 2: Problem 1
Given that six of the elements are 2, 3, 4, 6, 9, and 14, we can start by arranging these numbers in ascending order: 2, 3, 4, 6, 9, 14. The remaining three numbers can be any distinct integers.

## Step 3: Problem 1
To determine the possible medians, we need to consider the range of values the fifth number can take. The numbers already given are 2, 3, 4, 6, 9, 14. So, the three additional numbers can either be less than 4, between 4 and 9, or greater than 9.

## Step 4: Problem 1
If all three additional numbers are less than 4, the order becomes: additional number, additional number, additional number, 4, 6, 9, 14. The median is 6.
If two additional numbers are less than 4 and one is between 4 and 9, the order becomes: additional number, additional number, 4, additional number, 6, 9, 14. The median is between 4 and 6, so it could be 5.
If one additional number is less than 4 and two are between 4 and 9, the order becomes: additional number, 4, additional number, additional number, 6, 9, 14. The median could be between 4 and 6.
If all three additional numbers are between 4 and 9, the order becomes: 2, 3, 4, additional number, additional number, additional number, 6, 9, 14. The median is one of the additional numbers between 4 and 6, or 6 if the additional numbers are greater than 6.
If the additional numbers are greater than 9, the order remains 2, 3, 4, 6, 9, additional number, additional number, additional number, 14. The median is 9.

## Step 5: Problem 1
Considering the cases:
- When the additional numbers are less than 4, the median is 6.
- When the numbers are between 4 and 9, the median can be 5, 6, 7, or 8 (since the additional numbers can be distinct and between 4 and 9).
- When the numbers are greater than 9, the median is 9.
So, possible medians are 5, 6, 7, 8, and 9.

## Step 6: Problem 1
Thus, there are 5 possible values for the median: 5, 6, 7, 8, 9.

ANSWER 1: B

## Step 7: Problem 2
To find N, we first need to understand how the mean and median are affected by adding N to the list 2, 6, 7, 7, 28.

## Step 8: Problem 2
The original list is 2, 6, 7, 7, 28. The median is 7. When N is added, the list becomes 2, 6, 7, 7, N, 28. The median will depend on N. If N is between 7 and 7 (which is not possible since N is distinct), or less than 7, the median will be 7. If N is greater than 7, the median will be the average of 7 and N.

## Step 9: Problem 2
Let's consider the case when N is less than or equal to 7. The median remains 7. The mean is (2 + 6 + 7 + 7 + 28 + N)/6 = (50 + N)/6. According to the problem, the mean is twice the median, so (50 + N)/6 = 2*7 = 14.

## Step 10: Problem 2
Solving for N: 50 + N = 14*6 = 84, N = 84 - 50 = 34.

## Step 11: Problem 2
Let's check if N = 34 satisfies the condition for the median. Since 34 is greater than 7, the list becomes 2, 6, 7, 7, 28, 34. The median is (7 + 7)/2 = 7. The mean is (2 + 6 + 7 + 7 + 28 + 34)/6 = 84/6 = 14, which is indeed twice the median.

## Step 12: Problem 2
Thus, N = 34 is a valid solution.

ANSWER 2: E

## Step 13: Problem 3
To find the fraction of four-digit integers with all even digits that are divisible by 4, we first need to determine the total number of such four-digit integers.

## Step 14: Problem 3
For a four-digit number, the first digit can be 2, 4, 6, or 8 (4 choices), and the rest can be 0, 2, 4, 6, or 8 (5 choices each). So, the total number of four-digit numbers with all even digits is 4 * 5 * 5 * 5 = 500.

## Step 15: Problem 3
A number is divisible by 4 if its last two digits form a number divisible by 4. The last two digits can be 00, 04, 08, 20, 24, 28, 40, 44, 48, 60, 64, 68, 80, 84, 88. Out of these, the ones with even digits are 00, 04, 08, 20, 24, 28, 40, 44, 48, 60, 64, 68, 80, 84, 88, which are 20, 24, 28, 40, 44, 48, 60, 64, 68, 80, 84, 88 (12 options since 00, 04, 08 are not valid for the last two digits when considering all possible combinations for the first two digits).

## Step 16: Problem 3
However, we need to correct the understanding of the last two digits. For a number to be divisible by 4, the last two digits must form a number divisible by 4. The possible last two digits are: 00, 04, 08, 20, 24, 28, 40, 44, 48, 60, 64, 68, 80, 84, 88. There are 5 choices for the third digit and 4 choices for the first digit. So, for each valid last two digits, there are 4 * 5 = 20 numbers.

## Step 17: Problem 3
There are 15 valid combinations for the last two digits (00, 04, 08, 20, 24, 28, 40, 44, 48, 60, 64, 68, 80, 84, 88), but we should count them correctly: there are indeed 15 options (since all are valid for the condition of being even and divisible by 4 when considering the last two digits).

## Step 18: Problem 3
The total number of four-digit numbers with all even digits and divisible by 4 is 4 * 5 * 15 = 300 (since there are 15 valid last two digits).

## Step 19: Problem 3
The fraction of these integers divisible by 4 is 300 / 500 = 3/5.

ANSWER 3: D

## Step 20: Problem 4
To find the difference between a 6.5% sales tax and a 6% sales tax on an item priced at $20, we calculate both taxes and then find the difference.

## Step 21: Problem 4
6.5% of $20 = 0.065 * $20 = $1.30.
6% of $20 = 0.06 * $20 = $1.20.

## Step 22: Problem 4
The difference is $1.30 - $1.20 = $0.10.

ANSWER 4: B

## Step 23: Problem 5
To solve (1901 + 1902 + ... + 1993) - (101 + 102 + ... + 193), we can simplify it by calculating the sum of the differences between corresponding terms.

## Step 24: Problem 5
The series can be simplified as: (1901-101) + (1902-102) + ... + (1993-193).

## Step 25: Problem 5
This simplifies to: 1800 + 1800 + 

**Problem 1:**
We have a set \(R\) of nine distinct integers, with six given: 2, 3, 4, 6, 9, 14. The remaining three integers can be any integers not in this list. The median of nine numbers is the 5th smallest. Let the unknown numbers be \(x, y, z\). The six given numbers are fixed; we can place \(x, y, z\) anywhere in the sorted order to change the 5th element.

Let \(k\) be the number of given numbers less than the candidate median \(m\).
- If \(m\) is one of the given numbers, we need to place exactly \(4-k\) of the new numbers below \(m\) and the rest above. This requires \(0 \le 4-k \le 3\), so \(1 \le k \le 4\). Among the given numbers: \(3\ (k=1),\ 4\ (k=2),\ 6\ (k=3),\ 9\ (k=4)\) satisfy this. (\(2\) has \(k=0\), \(14\) has \(k=5\); neither works.) So \(3, 4, 6, 9\) are possible medians.
- If \(m\) is a new integer, one new number equals \(m\) and the other two are placed above or below. We need \(4-k\) new numbers below \(m\), so \(0 \le 4-k \le 2\), or \(2 \le k \le 4\). Also \(m\) must be an integer not already in the list.
  * \(k=2\): \(m\) between 3 and 4 — no integer.
  * \(k=3\): \(m\) between 4 and 6 — only \(5\).
  * \(k=4\): \(m\) between 6 and 9 — \(7\) and \(8\).
So new possible medians: \(5, 7, 8\).

Total possible medians: \(3, 4, 5, 6, 7, 8, 9\) — seven values.

ANSWER 1: D

**Problem 2:**
Original list: \(2, 6, 7, 7, 28\) (sum = 50). Insert \(N\) to get six numbers. Mean = \(\frac{50+N}{6}\). Median of six sorted numbers is the average of the 3rd and 4th.

Case 1: \(N \le 6\). Sorted: …, 3rd = 6, 4th = 7, median = 6.5.  
\(\frac{50+N}{6} = 2 \times 6.5 = 13 \Rightarrow 50+N = 78 \Rightarrow N=28\), but \(28 > 6\). Contradiction.

Case 2: \(6 < N \le 7\) (only integer is 7). Sorted: \(2, 6, 7, 7, 7, 28\), median = 7.  
\(\frac{57}{6} = 9.5 \neq 14\). No.

Case 3: \(7 < N \le 28\). Sorted: \(2, 6, 7, 7, N, 28\), median = 7.  
\(\frac{50+N}{6} = 14 \Rightarrow N = 34\), but \(34 > 28\). Contradiction.

Case 4: \(N > 28\). Sorted: \(2, 6, 7, 7, 28, N\), median = 7.  
\(\frac{50+N}{6} = 14 \Rightarrow N = 34\). Works.

ANSWER 2: E

**Problem 3:**
Four-digit positive integers with all even digits: thousands digit from \(\{2,4,6,8\}\) (4 choices); hundreds, tens, units from \(\{0,2,4,6,8\}\) (5 choices each). Total = \(4 \times 5^3 = 500\).
Divisible by 4 means last two digits form a multiple of 4. For any even tens digit, there are exactly 3 even units digits that make it divisible by 4 (e.g., if tens = 0,4,8 → units 0,4,8; if tens = 2,6 → units 0,4,8). So \(5 \times 3 = 15\) valid last-two-digit combinations.
Number of divisible numbers = \(4 \times 5 \times 15 = 300\).
Fraction = \(\frac{300}{500} = \frac{3}{5}\).

ANSWER 3: D

**Problem 4:**
Tax difference = \(6.5\% - 6\% = 0.5\%\) of \$20 = \(0.005 \times 20 = \$0.10\).

ANSWER 4: B

**Problem 5:**
First sum: \(1901\) to \(1993\) inclusive. Number of terms = \(1993 - 1901 + 1 = 93\).  
Sum = \(93 \times \frac{1901+1993}{2} = 93 \times 1947\).
Second sum: \(101\) to \(193\) inclusive. Terms = \(193 - 101 + 1 = 93\).  
Sum = \(93 \times \frac{101+193}{2} = 93 \times 147\).
Difference = \(93 \times (1947 - 147) = 93 \times 1800 = 167\,400\).

ANSWER 5: A

**Problem 6:**
The expression is \(1 - 2 - 3 + 4 + 5 - 6 - 7 + 8 + 9 - 10 - 11 + \cdots + 1992 + 1993 - 1994 - 1995 + 1996\).
Sign pattern: \(+\) for terms \(n \equiv 0\) or \(1 \pmod 4\); \(-\) for \(n \equiv 2\) or \(3 \pmod 4\).
Group in blocks of four: \((1-2-3+4) + (5-6-7+8) + \cdots + (1993-1994-1995+1996)\). Each block sums to \(0\). There are \(1996/4 = 499\) blocks. Total sum = \(0\).

ANSWER 6: C

**Problem 7:**
Let \(S\) be sunglasses wearers (\(|S|=50\)), \(C\) be cap wearers (\(|C|=35\)). Given \(P(S \mid C) = \frac{|S \cap C|}{35} = \frac{2}{5} \Rightarrow |S \cap C| = 14\).
Then \(P(C \mid S) = \frac{14}{50} = \frac{7}{25}\).

ANSWER 7: B

**Problem 8:**
Three-digit numbers with digits summing to 26. Maximum sum is 27, so digits are close to 9. Let digits be \(a,b,c\) with \(a+b+c=26\), \(1 \le a \le 9\), \(0 \le b,c \le 9\). Setting \(x=9-a, y=9-b, z=9-c\) gives \(x+y+z=1\) with \(x \le 8, y,z \le 9\). The only solutions are permutations of \((1,0,0)\): one digit is 8, the other two are 9. The three numbers are 998, 989, 899.
Even numbers end in an even digit: only 998. Hence 1 even number.

ANSWER 8: A

**Problem 9:**
Three-digit numbers, digit sum = 25. As above, \(x+y+z = 27-25 = 2\). Nonnegative solutions: permutations of \((2,0,0)\) – digits 7,9,9; and \((1,1,0)\) – digits 8,8,9.  
For 7,9,9: 3 permutations. For 8,8,9: 3 permutations. Total 6 numbers.

ANSWER 9: C

**Problem 10:**
Exits 5 through 26: that’s \(26-5+1 = 22\) exits, so 21 gaps. Sum of gaps = 118 miles. Each gap \(\ge 5\). To maximize one gap, minimize the other 20: \(20 \times 5 = 100\) miles. Largest gap = \(118 - 100 = 18\) miles.

ANSWER 10: C

**Problem 11:**
Five scores: \(a \le b \le c \le d \le e\). Mean = 90 ⇒ sum = 450. Median \(c = 91\). Mode = 94, so at least two scores are 94. Since median is 91, the two 94s must be \(d\) and \(e\). Then \(a+b+91+94+94 = 450 \Rightarrow a+b = 450 - 279 = 171\).

ANSWER 11: B

**Problem 12:**
Blanks: \(5 \_ 4 \_ 6 \_ 3\). Use \(+,\ -,\ \times\) exactly once. Evaluate with standard order of operations. All 6 permutations:
1) \(\times, +, -:\ 5\times4+6-3 = 20+6-3=23\)
2) \(\times, -, +:\ 5\times4-6+3 = 20-6+3=17\)
3) \(+, \times, -:\ 5+4\times6-3 = 5+24-3=26\)
4) \(+, -, \times:\ 5+4-6\times3 = 5+4-18=-9\)
5) \(-, \times, +:\ 5-4\times6+3 = 5-24+3=-16\)
6) \(-, +, \times:\ 5-4+6\times3 = 5-4+18=19\)
Among choices, 19 appears.

ANSWER 12: E

Problem 1:
We have a set R of 9 distinct integers. Six are known: 2, 3, 4, 6, 9, 14. We add 3 unknown distinct integers (not equal to any known element). The median of 9 elements is the 5th smallest.

Let me determine which values can occupy the 5th position. The known elements in sorted order are: 2, 3, 4, 6, 9, 14.

**Can the median be 3?** Place all 3 unknowns below 3 (e.g., −1, 0, 1). Sorted: −1, 0, 1, 2, **3**, 4, 6, 9, 14. 5th = 3. ✓
**Can the median be 4?** Place 2 unknowns below 4 (e.g., 0, 1) and 1 above. Sorted: 0, 1, 2, 3, **4**, 6, 9, 14, 20. 5th = 4. ✓
**Can the median be 5?** (5 is an unknown.) Known elements below 5: {2, 3, 4} = 3 elements. Add 1 unknown below 5 (e.g., 1), make 5 an unknown, and 1 unknown above. Sorted: 1, 2, 3, 4, **5**, 6, 9, 14, 20. 5th = 5. ✓
**Can the median be 6?** Known below 6: {2, 3, 4} = 3. Add 1 unknown below 6 (e.g., 5), 2 above. Sorted: 2, 3, 4, 5, **6**, 9, 14, 20, 21. 5th = 6. ✓
**Can the median be 7?** Known below 7: {2, 3, 4, 6} = 4. Need 0 unknowns below 7. Make 7 an unknown, 2 above. Sorted: 2, 3, 4, 6, **7**, 9, 14, 20, 21. 5th = 7. ✓
**Can the median be 8?** Known below 8: {2, 3, 4, 6} = 4. Same logic: 2, 3, 4, 6, **8**, 9, 14, 20, 21. 5th = 8. ✓
**Can the median be 9?** Known below 9: {2, 3, 4, 6} = 4. Place all 3 unknowns above 9. Sorted: 2, 3, 4, 6, **9**, 14, 20, 21, 22. 5th = 9. ✓

**Can the median be ≤ 2?** Need 4 elements below it, but at most 3 unknowns exist → impossible.
**Can the median be ≥ 10?** Known elements below 10 are {2, 3, 4, 6, 9} = 5 elements, so the 5th position is already filled by something ≤ 9 → impossible.

Possible medians: {3, 4, 5, 6, 7, 8, 9} → **7 values**.

ANSWER 1: D

---

Problem 2:
Original list: 2, 6, 7, 7, 28. Sum = 50. Insert N to get 6 elements.
New mean = (50 + N)/6. Median of 6 elements = average of 3rd and 4th.

**Case N > 7:** Sorted order: 2, 6, 7, 7, N, 28 (or N, 28 swapped). The 3rd and 4th elements are 7 and 7.
Median = (7 + 7)/2 = 7.
Mean = 2 × median = 14.
(50 + N)/6 = 14 → 50 + N = 84 → N = 34.
Check: 34 > 7 ✓. List: 2, 6, 7, 7, 28, 34. Median = 7, Mean = 84/6 = 14 = 2(7). ✓

**Case N ≤ 6:** Median = (6+7)/2 = 6.5. Mean = 13. N = 78 − 50 = 28. But 28 > 6, contradiction.
**Case 6 < N ≤ 7:** Median = (N+7)/2. (50+N)/6 = N+7 → 50+N = 6N+42 → 5N = 8 → N = 1.6. Contradiction.

Only N = 34 works.

ANSWER 2: E

---

Problem 3:
Even digits: {0, 2, 4, 6, 8}. Four-digit numbers with all even digits:
- First digit: {2, 4, 6, 8} → 4 choices
- Other three digits: 5 choices each
- Total = 4 × 5 × 5 × 5 = 500

Divisibility by 4 depends on the last two digits. I check all 5 × 5 = 25 pairs of even digits:
For each tens digit (0, 2, 4, 6, 8), the units digits giving divisibility by 4 are:
- Tens 0: 00, 04, 08 (3 of 5)
- Tens 2: 20, 24, 28 (3 of 5)
- Tens 4: 40, 44, 48 (3 of 5)
- Tens 6: 60, 64, 68 (3 of 5)
- Tens 8: 80, 84, 88 (3 of 5)

Valid last-two-digit pairs = 15 out of 25.

Numbers divisible by 4 = 4 × 5 × 15 = 300.
Fraction = 300/500 = **3/5**.

ANSWER 3: D

---

Problem 4:
Difference = 6.5% of $20 − 6% of $20 = (6.5% − 6%) × $20 = 0.5% × $20.
0.5% = 0.005, so 0.005 × 20 = $0.10.

ANSWER 4: B

---

Problem 5:
Pair corresponding terms:
(1901 − 101) + (1902 − 102) + … + (1993 − 193)

Each pair equals 1800.
Number of terms: 1993 − 1901 + 1 = 93 terms (and 193 − 101 + 1 = 93 terms).
Total = 93 × 1800 = **167,400**.

ANSWER 5: A

---

Problem 6:
The sign pattern is: +, −, −, +, +, −, −, +, …
Group into blocks of 4:
(1 − 2 − 3 + 4) + (5 − 6 − 7 + 8) + (9 − 10 − 11 + 12) + …

Each block: n − (n+1) − (n+2) + (n+3) = n − n − 1 − n − 2 + n + 3 = **0**.

Total terms: 1996. Number of blocks: 1996/4 = 499 complete blocks.
Sum = 499 × 0 = **0**.

ANSWER 6: C

---

Problem 7:
Let B = number wearing both sunglasses and caps.
P(sunglasses | cap) = B / (people wearing caps) = B / 35 = 2/5.
B = 35 × (2/5) = 14.

Now find P(cap | sunglasses) = B / (people wearing sunglasses) = 14 / 50 = **7/25**.

ANSWER 7: B

---

Problem 8:
Three-digit numbers with digit sum 26. Maximum digit sum = 9+9+9 = 27, so we must subtract 1 from one digit of 999:
Possible digit sets: (8,9,9), (9,8,9), (9,9,8) → numbers 899, 989, 998 (and permutations: 899, 989, 997... wait).

Actually the numbers are: 899, 989, 998 — no, let me list properly:
- (8,9,9) → 899, 989, 998... wait, (8,9,9) gives 899, 989, 998? No: digits are 8,9,9 so numbers are 899, 989, 998.
Wait: (8,9,9) → 899, 989, 998. But 998 has digits 9,9,8 which is the set (9,9,8). Let me just list all:
- Digits {8,9,9}: 899, 989, 998 — wait, 998 has digits 9,9,8. That IS {8,9,9}. So: 899, 989, 998.

Hmm, let me redo. The three numbers with digit sum 26 are:
- 899 (8+9+9=26)
- 989 (9+8+9=26)
- 998 (9+9+8=26)

Wait, that's only 3 numbers total (since two digits are the same, there are 3!/2! = 3 arrangements).

Even numbers must end in an even digit. Only 998 ends in 8 (even). 899 and 989 end in 9 (odd).

So exactly **1** even number.

ANSWER 8: A

---

Problem 9:
Three-digit numbers a, b, c with a + b + c = 25, where 1 ≤ a ≤ 9, 0 ≤ b, c ≤ 9.

Enumerate by first digit a:
- a = 9: b + c = 16 → (7,9), (8,8), (9,7) → 979, 988, 997 (3 numbers)
- a = 8: b + c = 17 → (8,9), (9,8) → 889, 898 (2 numbers)
- a = 7: b + c = 18 → (9,9) → 799 (1 number)
- a ≤ 6: b + c ≥ 19, but max b + c = 18 → impossible

Total = 3 + 2 + 1 = **6**.

ANSWER 9: C

---

Problem 10:
Exits 5 through 26 give 22 exits and 21 consecutive gaps. Total distance = 118 miles. Each gap ≥ 5 miles.

To maximize one gap, minimize the other 20 gaps to 5 miles each:
Max gap + 20(5) = 118
Max gap + 100 = 118
Max gap = **18** miles.

ANSWER 10: C

---

Problem 11:
Five scores sorted: a ≤ b ≤ c ≤ d ≤ e.
- Mean = 90 → sum = 450
- Median = 91 → c = 91
- Mode = 94 → 94 appears most frequently (at least twice)

Since a ≤ b ≤ c = 91, neither a nor b can be 94. So 94 must appear among d and e. For 94 to be the mode, we need d = e = 94 (appearing twice).

Sum: a + b + 91 + 94 + 94 = 450
a + b = 450 − 279 = **171**

(We can verify valid scores 

 Problem 1:
We are given a set \(R\) of nine distinct integers, six of which are \(2,3,4,6,9,14\). We must determine how many different values are possible for the median of \(R\).

Because there are nine distinct integers, the median is the fifth smallest element when the set is listed in increasing order. Let the three unknown distinct integers be chosen freely (as long as they are not equal to any of the six given numbers). We examine two cases: when the median is one of the six known numbers and when it is one of the three unknown numbers.

**Case 1: The median is a known number.**
Let the median be \(m\). Of the six known numbers, let \(L\) be how many are less than \(m\). Then there are \(5-L\) known numbers greater than \(m\). To make \(m\) the fifth element, we need exactly four numbers smaller than \(m\) and four larger. Thus we need \(4-L\) unknown numbers below \(m\) and \(4-(5-L)=L-1\) unknown numbers above \(m\). This uses exactly \((4-L)+(L-1)=3\) unknowns, which is perfect. We need \(4-L\ge 0\) and \(L-1\ge 0\), i.e. \(1\le L\le 4\).

- \(m=2\): \(L=0\) → impossible.
- \(m=3\): \(L=1\) → need 3 unknowns \(<3\), 0 above. Possible (e.g. \(-1,0,1\)).
- \(m=4\): \(L=2\) → need 2 unknowns \(<4\), 1 above. Possible (e.g. \(0,1,5\)).
- \(m=6\): \(L=3\) → need 1 unknown \(<6\), 2 above. Possible (e.g. \(5,7,8\)).
- \(m=9\): \(L=4\) → need 0 unknowns \(<9\), 3 above. Possible (e.g. \(10,11,12\)).
- \(m=14\): \(L=5\) → need \(-1\) unknowns below; impossible.

So the known possible medians are \(3,4,6,9\).

**Case 2: The median is an unknown number \(x\).**
Then \(x\) itself uses one of the three unknown slots. We need four numbers below \(x\) and four above. With \(L\) known numbers below \(x\), we need \(4-L\) other unknowns below \(x\) and \(4-(6-L)=L-2\) other unknowns above \(x\). The total other unknowns used is \((4-L)+(L-2)=2\), which matches the two remaining unknowns. We need \(4-L\ge 0\) and \(L-2\ge 0\), so \(2\le L\le 4\).

- \(L=2\): known below are \(2,3\); known above are \(4,6,9,14\). Then \(3<x<4\), impossible for an integer.
- \(L=3\): known below are \(2,3,4\); known above are \(6,9,14\). Then \(4<x<6\), so \(x=5\). This works (e.g. add \(1,5,7\)).
- \(L=4\): known below are \(2,3,4,6\); known above are \(9,14\). Then \(6<x<9\), so \(x=7\) or \(x=8\). Both work (e.g. add \(7,10,11\) or \(8,10,11\)).

Thus the unknown possible medians are \(5,7,8\).

Collecting all possibilities: \(\{3,4,5,6,7,8,9\}\). That is **7** different values. The tempting answer of 4 counts only the known medians, and 2 or 14 are impossible because too many numbers would have to be forced to one side.

ANSWER 1: D

---

Problem 2:
We insert a number \(N\) into the list \(2,6,7,7,28\) so that the new mean of the six numbers is twice the new median. Find \(N\).

The original sum is \(2+6+7+7+28=50\). After insertion, the sum is \(50+N\) and the mean is \(\frac{50+N}{6}\).

The median depends on the size of \(N\) relative to the ordered list. Because the list contains six numbers, the median is the average of the third and fourth values.

- If \(N\le 6\) (as an integer), the ordered list begins with \(N\) or \(2\), then the third and fourth terms are \(6\) and \(7\), giving median \(\frac{6+7}{2}=6.5\).
  Setting mean \(=2\times 6.5=13\): \(\frac{50+N}{6}=13\Rightarrow N=28\). But \(N=28\) does not satisfy \(N\le 6\), so this case yields no valid solution.

- If \(N\ge 7\) (integer), the third and fourth terms are both \(7\) (since the two \(7\)’s occupy those middle slots, or \(N=7\) makes three \(7\)’s), so the median is \(7\).
  Setting mean \(=2\times 7=14\): \(\frac{50+N}{6}=14\Rightarrow 50+N=84\Rightarrow N=34\).

Check: the list becomes \(2,6,7,7,28,34\). The median is \(\frac{7+7}{2}=7\). The mean is \(\frac{84}{6}=14\), which is indeed twice the median. The other choices fail: \(N=28\) gives mean \(13\) and median \(7\); \(N=14\) gives mean \(\frac{64}{6}\neq 14\).

ANSWER 2: E

---

Problem 3:
We consider all positive four-digit integers whose digits are all even. We need the fraction of them that are divisible by \(4\).

Even digits are \(\{0,2,4,6,8\}\). For a four-digit number \(ABCD\):
- \(A\) (thousands) has 4 choices: \(\{2,4,6,8\}\).
- \(B,C,D\) each have 5 choices.

Total count: \(4\cdot 5\cdot 5\cdot 5 = 500\).

A number is divisible by \(4\) exactly when its last two digits form a multiple of \(4\). Here \(C\) and \(D\) are both even. Consider \(10C+D \pmod 4\). Since \(C\) is even, write \(C=2k\); then \(10C=20k\equiv 0\pmod 4\). Thus \(10C+D\equiv D\pmod 4\). So we simply need \(D\equiv 0\pmod 4\).

Among the even digits, those divisible by \(4\) are \(0,4,8\): **3 choices** for \(D\). The digits \(A,B,C\) are unrestricted (4, 5, and 5 choices).

Count of favorable numbers: \(4\cdot 5\cdot 5\cdot 3 = 300\).

The fraction is \(\frac{300}{500}=\frac{3}{5}\). One might mistakenly think the fraction is \(\frac12\) because half of all integers are divisible by \(4\), but the restriction to even digits changes the units-digit distribution.

ANSWER 3: D

---

Problem 4:
Find the difference between a \(6.5\%\) sales tax and a \(6\%\) sales tax on a \(\$20\) item.

Method 1: Compute each tax.
- \(6.5\%\) of \(\$20 = 0.065\times 20 = \$1.30\).
- \(6\%\) of \(\$20 = 0.06\times 20 = \$1.20\).

Difference: \(\$1.30-\$1.20 = \$0.10\).

Method 2: The difference in rates is \(0.5\%\), and \(0.5\%\) of \(\$20 = 0.005\times 20 = \$0.10\).

The distractors correspond to misplacing the decimal (e.g. \(\$1\) would be \(5\%\) of \(\$20\)).

ANSWER 4: B

---

Problem 5:
Evaluate \((1901+1902+\dots+1993)-(101+102+\dots+193)\).

Each sum has the same number of terms:
\(1993-1901+1 = 93\) terms, and \(193-101+1 = 93\) terms.

Pair the terms:
\((1901-101)+(1902-102)+\dots+(1993-193)\).

Each difference equals \(1800\). With 93 such pairs, the total is
\(93\times 1800 = 93\times 18\times 100 = 1674\times 100 = 167{,}400\).

A common error is miscounting the number of terms as 92, which would give \

Problem 1:
R is a set of nine distinct integers with six known elements: 2, 3, 4, 6, 9, 14. Since R has 9 elements, the median is the 5th element when sorted.

I need to determine which integers can be the 5th element. The median must have exactly 4 elements less than it and 4 elements greater than it.

**Checking each candidate:**

- **Median = 3**: Need 4 elements < 3. Known: {2}. Need 3 unknowns all < 3 (e.g., 1, 0, -1). Elements > 3: {4, 6, 9, 14} = 4. ✓
- **Median = 4**: Need 4 elements < 4. Known: {2, 3}. Need 2 unknowns < 4. Elements > 4: {6, 9, 14} + 1 unknown > 4. ✓
- **Median = 5**: Need 4 elements < 5. Known: {2, 3, 4}. Need 1 unknown < 5. Elements > 5: {6, 9, 14} + 1 unknown > 5. (5 is an unknown) ✓
- **Median = 6**: Need 4 elements < 6. Known: {2, 3, 4}. Need 1 unknown < 6. Elements > 6: {9, 14} + 2 unknowns > 6. ✓
- **Median = 7**: Need 4 elements < 7. Known: {2, 3, 4, 6} = 4. Elements > 7: {9, 14} + 2 unknowns > 7. (7 is an unknown) ✓
- **Median = 8**: Need 4 elements < 8. Known: {2, 3, 4, 6} = 4. Elements > 8: {9, 14} + 2 unknowns > 8. (8 is an unknown) ✓
- **Median = 9**: Need 4 elements < 9. Known: {2, 3, 4, 6} = 4. Elements > 9: {14} + 3 unknowns > 9. ✓

**Values that cannot be median:**
- 2: Need 4 elements < 2, but only 3 unknowns available. ✗
- 10-13: Already have 5 known elements less than these. ✗
- 14: Already have 5 known elements less than 14. ✗

Possible median values: {3, 4, 5, 6, 7, 8, 9} → **7 values**

ANSWER 1: D

---

Problem 2:
List: 2, 6, 7, 7, 28. Insert N so that mean = 2 × median.

Sum of original list = 50, so new mean = (50 + N)/6.

**Finding the median based on N:**
- If N > 7: Sorted list is 2, 6, 7, 7, 28, N (or 2, 6, 7, 7, N, 28 if 7 < N < 28). Median = (7+7)/2 = 7.
- If N ≤ 6: Median = (6+7)/2 = 6.5
- If 6 < N ≤ 7: Median = (N+7)/2

**Solving for N > 7:**
(50 + N)/6 = 2(7) = 14
50 + N = 84
N = 34

Verification: List = 2, 6, 7, 7, 28, 34. Mean = 84/6 = 14. Median = 7. ✓

ANSWER 2: E

---

Problem 3:
Four-digit integers with all even digits. Even digits: {0, 2, 4, 6, 8}.

**Total count:** First digit: 4 choices (2,4,6,8), others: 5 choices each → 4 × 5³ = 500

**Divisible by 4:** Last two digits must form a number divisible by 4.

Listing all 25 pairs of even last digits and checking divisibility by 4:
- 00✓, 02✗, 04✓, 06✗, 08✓
- 20✓, 22✗, 24✓, 26✗, 28✓
- 40✓, 42✗, 44✓, 46✗, 48✓
- 60✓, 62✗, 64✓, 66✗, 68✓
- 80✓, 82✗, 84✓, 86✗, 88✓

That's 15 valid pairs.

**Count divisible by 4:** 4 × 5 × 15 = 300

**Fraction:** 300/500 = 3/5

ANSWER 3: D

---

Problem 4:
6.5% tax on $20 = 0.065 × 20 = $1.30
6% tax on $20 = 0.06 × 20 = $1.20

Difference = $1.30 - $1.20 = $0.10

ANSWER 4: B

---

Problem 5:
(1901 + 1902 + ... + 1993) − (101 + 102 + ... + 193)

Both sums have 93 terms each.

First sum = 93 × (1901 + 1993)/2 = 93 × 1947
Second sum = 93 × (101 + 193)/2 = 93 × 147

Difference = 93 × (1947 − 147) = 93 × 1800 = 167,400

ANSWER 5: A

---

Problem 6:
1 − 2 − 3 + 4 + 5 − 6 − 7 + 8 + ... + 1993 − 1994 − 1995 + 1996

The pattern repeats every 4 terms: +, −, −, +

Group k: (4k−3) − (4k−2) − (4k−1) + 4k = 4k−3−4k+2−4k+1+4k = 0

Each group of 4 sums to 0.

1996 ÷ 4 = 499 complete groups.

Total = 499 × 0 = 0

ANSWER 6: C

---

Problem 7:
50 people wear sunglasses, 35 wear caps.

P(sunglasses | cap) = 2/5 means: (both)/(caps) = 2/5

Both/35 = 2/5 → Both = 14

P(cap | sunglasses) = (both)/(sunglasses) = 14/50 = 7/25

ANSWER 7: B

---

Problem 8:
3-digit numbers with digit-sum 26 that are even.

For number abc: a + b + c = 26, c is even, a ≥ 1, all digits ≤ 9.

- c = 8: a + b = 18 → only (9,9) → 998 ✓
- c = 6: a + b = 20 → impossible (max 9+9=18)
- c = 4, 2, 0: a + b ≥ 22 → impossible

Only 1 number: 998

ANSWER 8: A

---

Problem 9:
Three-digit numbers with digit-sum 25.

a + b + c = 25, a ≥ 1, all digits ≤ 9.

- a = 9: b + c = 16 → (7,9), (8,8), (9,7) → 3 numbers
- a = 8: b + c = 17 → (8,9), (9,8) → 2 numbers
- a = 7: b + c = 18 → (9,9) → 1 number
- a ≤ 6: b + c ≥ 19 → impossible

Total = 3 + 2 + 1 = 6

ANSWER 9: C

---

Problem 10:
Distance from exit 5 to exit 26 is 118 miles. There are 21 gaps between consecutive exits (5→6, 6→7, ..., 25→26).

Each gap ≥ 5 miles. To maximize one gap, minimize all others.

If 20 gaps are minimized at 5 miles each: 20 × 5 = 100 miles.

Maximum single gap = 118 − 100 = 18 miles.

ANSWER 10: C

---

Problem 11:
Five scores: mean = 90, median = 91, mode = 94.

Sum = 5 × 90 = 450

Sorted: a ≤ b ≤ c ≤ d ≤ e
- c = 91 (median)
- Mode = 94, so 94 appears at least twice. Since c = 91, the 94s must be d and e.

a + b + 91 + 94 + 94 = 450
a + b = 171

ANSWER 11: B

---

Problem 12:
Using +, −, × each exactly once in 5 __ 4 __ 6 __ 3:

| Operations | Result |
|------------|--------|
| 5 + 4 − 6 × 3 | 5 + 4 − 18 = −9 |
| 5 + 4 × 6 − 3 | 5 + 24 − 3 = 26 |
| 5 − 4 + 6 × 3 | 5 − 4 + 18 = 19 |
| 5 − 4 × 6 + 3 | 5 − 24 + 3 = −16 |
| 5 × 4 + 6 − 3 | 20 + 6 − 3 = 23 |
| 5 × 4 − 6 + 3 | 20 − 6 + 3 = 17 |

From choices {9, 10, 15, 16, 19}, only **19** is achievable.

ANSWER 12: E

**Problem 1:**  
We have nine distinct integers, six of which are known: 2, 3, 4, 6, 9, 14.  
The median of a nine‑element set is the 5th smallest number.  

Let the unknown integers be \(x,y,z\) (all distinct and distinct from the known six).  
When the numbers are sorted, the median is the 5th term.  

For a known number \(k_i\) (the \(i\)‑th smallest of the known six) to be the median, we need exactly four numbers smaller than it.  
If \(i-1\) known numbers are smaller, we need \(L\) unknown numbers smaller with  
\((i-1)+L=4\Rightarrow L=5-i\).  
Since we have only three unknowns, \(L\ge0\) gives \(i\le5\); also \(L\le3\) gives \(i\ge2\).  
Thus \(i=2,3,4,5\) → medians could be 3, 4, 6, 9.

Can the median be an unknown integer?  
Let the median be an unknown \(u\).  
If \(u\) lies between the known numbers, we need exactly four numbers below it.  
With two other unknowns, the possibilities are:
- \(u\) between 4 and 6 → \(u=5\);
- \(u\) between 6 and 9 → \(u=7\) or \(u=8\).

All three of these can be realized while keeping the numbers distinct.  
Thus the possible median values are \(\{3,4,5,6,7,8,9\}\), a total of **7** values.

**Answer:** 7 → Choice **D**.  

ANSWER 1: D  

---

**Problem 2:**  
We start with the list 2, 6, 7, 7, 28 and insert a number \(N\).  
Let the six numbers be sorted. The median of six numbers is the average of the 3rd and 4th smallest values.

Let the sorted list after insertion be \(a_1\le a_2\le a_3\le a_4\le a_5\le a_6\).  
The median is \(\dfrac{a_3+a_4}{2}\).  
The mean is \(\dfrac{\text{sum}}{6}\).  
Given that the mean equals twice the median:
\[
\frac{\text{sum}}{6}=2\cdot\frac{a_3+a_4}{2}=a_3+a_4 .
\]
Hence \(\text{sum}=6(a_3+a_4)\).

Consider where \(N\) can fall:

*If \(N\le2\):* \(a_3=6,\ a_4=7\) → median \(=6.5\).  
Sum \(=N+50\). Equation gives \(N=28\), contradicting \(N\le2\).

*If \(2<N\le6\):* \(a_3=6,\ a_4=7\) → same median \(=6.5\).  
Again \(N=28\), impossible.

*If \(6<N\le7\):* only integer possibility \(N=7\).  
List becomes 2, 6, 7, 7, 7, 28 → median \(=7\).  
Sum \(=57\) → mean \(=9.5\neq14\).

*If \(7<N<28\):* \(a_3=7,\ a_4=7\) → median \(=7\).  
Sum \(=N+50\). Equation \(N+50=84\) gives \(N=34\), which is not \(<28\).

*If \(N=28\):* median still \(=7\). Sum \(=78\) → mean \(=13\neq14\).

*If \(N>28\):* \(a_3=7,\ a_4=7\) → median \(=7\).  
Equation gives \(N+50=84\) → \(N=34\). This works (34 > 28).

Thus the only possible value is **34**.

**Answer:** 34 → Choice **E**.  

ANSWER 2: E  

---

**Problem 3:**  
A four‑digit integer with all digits even can use digits \{0,2,4,6,8\} for each place, except the thousands digit cannot be 0.  

Number of such integers: \(4\times5\times5\times5=500\).

A number is divisible by 4 iff its last two digits form a number divisible by 4.  
List all two‑digit endings with even digits (25 possibilities).  

Checking each yields 15 endings that are multiples of 4:  
00, 04, 08, 20, 24, 28, 40, 44, 48, 60, 64, 68, 80, 84, 88.

Thus the proportion is \(\frac{15}{25}=\frac{3}{5}\).

**Answer:** \(\frac{3}{5}\) → Choice **D**.  

ANSWER 3: D  

---

**Problem 4:**  
6 % tax on \$20 = \(0.06\times20 = \$1.20\).  
6.5 % tax on \$20 = \(0.065\times20 = \$1.30\).  

Difference = \$1.30 − \$1.20 = **\$0.10**.

**Answer:** \$0.10 → Choice **B**.  

ANSWER 4: B  

---

**Problem 5:**  
Sum from 1901 to 1993 (inclusive):  
Number of terms = \(1993-1901+1 = 93\).  
Sum = \(\frac{(1901+1993)\times93}{2}= \frac{3894\times93}{2}=181\,071\).

Sum from 101 to 193 (inclusive):  
Number of terms = \(193-101+1 = 93\).  
Sum = \(\frac{(101+193)\times93}{2}= \frac{294\times93}{2}=13\,671\).

Difference = \(181\,071-13\,671 = 167\,400\).

**Answer:** 167 400 → Choice **A**.  

ANSWER 5: A  

---

**Problem 6:**  
The sign pattern for the numbers \(n\) is:
- \(n\equiv1\pmod4\) → +,
- \(n\equiv2\pmod4\) → −,
- \(n\equiv3\pmod4\) → −,
- \(n\equiv0\pmod4\) → +.

Thus each block of four consecutive integers \(\{4k+1,4k+2,4k+3,4k+4\}\) contributes  
\((4k+1)-(4k+2)-(4k+3)+(4k+4)=0\).

From 1 through 1996 there are \(1996/4=499\) complete blocks, so the total sum is **0**.

**Answer:** 0 → Choice **C**.  

ANSWER 6: C  

---

**Problem 7:**  
Let \(S\) = people wearing sunglasses, \(|S|=50\).  
Let \(C\) = people wearing caps, \(|C|=35\).  
Let \(x=|S\cap C|\).

Given \(P(S\mid C)=\frac{x}{35}= \frac{2}{5}\).  
Thus \(x = \frac{2}{5}\times35 = 14\).

We need \(P(C\mid S)=\frac{x}{50}= \frac{14}{50}= \frac{7}{25}\).

**Answer:** \(\frac{7}{25}\) → Choice **B**.  

ANSWER 7: B  

---

**Problem 8:**  
A three‑digit number with digit sum 26 must use digits that sum to 26.  
The only triple of single‑digit numbers with sum 26 is \(\{9,9,8\}\).

The number is even only if its units digit is 8.  
Among the permutations of 9, 9, 8, only 998 ends in 8.  

Thus exactly **1** such number exists.

**Answer:** 1 → Choice **A**.  

ANSWER 8: A  

---

**Problem 9:**  
Let the digits be \(a,b,c\) (hundreds, tens, units) with  
\(1\le a\le9,\ 0\le b,c\le9,\ a+b+c=25\).

Set \(a'=a-1\) (so \(0\le a'\le8\)). Then  
\(a'+b+c=24\).

Count unrestricted non‑negative solutions: \(\binom{24+3-1}{2}= \binom{26}{2}=325\).

Subtract solutions where a digit exceeds its bound:
- \(a'\ge9\): treat \(a''=a'-9\) → \(a''+b+c=15\) → \(\binom{17}{2}=136\).
- \(b\ge10\): \(b'=b-10\) → \(a'+b'+c=14\) → \(\binom{16}{2}=120\).
- \(c\ge10\): similarly 120.

Intersections:
- \(a'\ge9,\ b\ge10\): \(a''+b'+c=5\) → \(\binom{7}{2}=21\).
- \(a'\ge9,\ c\ge10\): 21.
- \(b\ge10,\ c\ge10\): \(a'+b'+c'=4\) → \(\binom{6}{2}=15\).

No triple intersection (would require sum –5).

By inclusion–exclusion, invalid = \(136+120+120-21-21-15 = 319\).  
Valid = \(325-319 = 6\).

**Answer:** 6 → Choice **C**.  

ANSWER 9: C  

---

**Problem 10:**  
Exits numbered 5 through 26 give 22 exits, hence 21 intervals.  
Let the distances between consecutive exits be \(d_1,\dots,d_{21}\).  
We know \(\sum d_i = 118\) miles, and each \(d_i\ge5\).

To maximize the largest interval, make the othe

Problem 1:  
We are given a set $ R $ of nine distinct integers, six of which are $ 2, 3, 4, 6, 9, 14 $. The median of a set of nine numbers is the fifth number when sorted. To find the number of possible values for the median, we consider the possible placements of the three unknown numbers relative to the known ones. The known numbers sorted are $ 2, 3, 4, 6, 9, 14 $. The three unknown numbers can be placed in various positions to shift the median. The possible medians occur when the additional numbers are placed such that the fifth number in the sorted list can be $ 4, 5, 6, 7, 8, $ or $ 9 $. This gives six possible values for the median.  
ANSWER 1: D (corrected to C? Wait, no: the possible medians are 4, 5, 6, 7, 8, 9 — that’s 6 values, so C)  
**Wait, let's re-evaluate:** The known numbers are 2,3,4,6,9,14. Adding three numbers. The median is the 5th number. To get 4 as median: three numbers ≤4, but only 2,3,4 are available — but they are distinct, so we can add numbers like 1,5? Wait no, the three unknown numbers can be placed in different positions. For example:  
- If all three unknown numbers are ≤4: but only three such numbers exist (2,3,4) but they are already in the set. So we cannot add more. Thus, the unknown numbers must be placed in other positions.  
Actually, the unknown numbers can be:  
- All less than 2: median could be 4 if three numbers are added <4, but only 2,3,4 are already present. We cannot add numbers less than 2 without repeating? Wait, the set has distinct integers. So we can add numbers like 1,0,-1, but the problem doesn’t restrict to positive. However, the known numbers are positive. But the problem says "distinct integers", so negatives are allowed. But the median values: if we add numbers less than 2, the sorted order becomes [new numbers, 2,3,4,6,9,14]. With three new numbers, the sorted list has 9 numbers. The fifth is the median. If we add three numbers less than 2, the sorted list is [a,b,c,2,3,4,6,9,14] — no, wait, adding three numbers less than 2: the sorted list would be [a,b,c,2,3,4,6,9,14] — but a,b,c are less than 2, so the first three, then 2 is fourth, 3 is fifth? Wait, no: the original known numbers are 2,3,4,6,9,14. Adding three numbers, say x,y,z. The full sorted list will be the combination. To get median=4, we need the fifth number to be 4. So in the sorted list of 9, the fifth must be 4. That requires that four numbers are ≤4 and four are ≥4? No: the fifth is the middle. For the fifth to be 4, we need at least five numbers ≤4? No: in a sorted list of 9, the fifth is the median. For the fifth to be 4, we need at least five numbers ≤4? Actually, no: the fifth number is the one where four are less and four are greater? No: in a sorted list of 9, the fifth is the median, meaning four are ≤ median and four are ≥ median. But since numbers can be equal, but here all distinct. So for 4 to be the median, we need exactly four numbers less than 4 and four numbers greater than 4? But 4 is in the set. So the sorted list must have four numbers <4, then 4, then four numbers >4. But the known numbers include 2,3,4,6,9,14. So 2,3 are <4, and 6,9,14 are >4. So currently, without unknowns, we have two <4 and three >4. Adding three unknowns. To have four <4 and four >4, we need to add two numbers <4 and one number >4? But then the sorted list would be [two new <4, 2,3,4, then the new >4, then 6,9,14] — but that’s only three numbers after 4? Wait, no: the sorted list would be [new <4 (two), 2,3,4, new >4, 6,9,14] — that’s 9 numbers? 2+3+1+3=9? Wait, no: the two new <4, then 2,3,4 (three numbers), then the new >4, then 6,9,14 (three numbers). So total: 2+3+1+3=9. The fifth number is 4. So yes, we can achieve median=4 by adding two numbers <4 and one number >4. Similarly, we can achieve:  
- Median=5: by adding numbers such that the fifth number is 5. But 5 is not in the set. So we need the fifth number to be 5. This requires that four numbers are <5 and four are >5. But 5 is not present. So we need to add 5? But the set must consist of distinct integers, and 5 is not in the given. So we can add 5. Then the sorted list would include 5. For example, add 5, and two other numbers. To have four <5 and four >5: currently, without unknowns, numbers <5: 2,3,4 (three), numbers >5: 6,9,14 (three). So to have four <5, we need to add one number <5. To have four >5, we need to add one number >5. But we are adding three numbers. So we can add 5, and say 1 and 10. Then sorted: 1,2,3,4,5,6,9,10,14. Median is 5. So median=5 is achievable. Similarly, median=6: we can add numbers such that the fifth is 6. Currently, numbers <6: 2,3,4 (three), numbers >6: 9,14 (two). So to have four <6 and four >6, we need to add one number <6 and two numbers >6. For example, add 5,10,15. Then sorted: 2,3,4,5,6,9,10,14,15. Median=6. Similarly, median=7: add numbers so fifth is 7. Currently, <7: 2,3,4,6 (four), >7: 9,14 (two). So to have four <7 and four >7, we need to add zero <7 (since already four) and two >7. But we are adding three numbers. So we can add 7,8,15. Then sorted: 2,3,4,6,7,8,9,14,15. Median=7. Similarly, median=8: add 8, and two numbers, say 10,11. Then sorted: 2,3,4,6,8,9,10,11,14. Median=8? The fifth is 8. But to have four <8 and four >8: currently, without unknowns, <8: 2,3,4,6 (four), >8:9,14 (two). So to have four >8, we need to add two numbers >8. But we are adding three numbers. So we can add 8,9,10? But 9 is already there. Add 8,10,11. Then sorted: 2,3,4,6,8,9,10,11,14. The fifth is 8. So yes. Similarly, median=9: add numbers so fifth is 9. Currently, <9: 2,3,4,6 (four), >9:14 (one). So to have four >9, we need to add three numbers >9. But we are adding three numbers. So add 10,11,12. Then sorted: 2,3,4,6,9,10,11,12,14. Median=9. So possible medians: 4,5,6,7,8,9. That’s six values.  
ANSWER 1: C  

Problem 2:  
We are given the list 2,6,7,7,28. A number $ N $ is added. The mean becomes twice the median. The original list has five numbers. After adding $ N $, there are 

### Problem 1:
We have a set of 9 distinct integers, so the median (middle value when sorted) is the 5th element. The 6 known numbers sorted are [2,3,4,6,9,14], with 3 additional distinct integers to add. To find possible median values:
- The minimum possible median is 3: add 3 numbers smaller than 2, sorted set becomes [a,b,c,2,3,4,6,9,14], 5th term=3.
- The maximum possible median is 9: add 3 numbers larger than 14, sorted set becomes [2,3,4,6,9,14,x,y,z], 5th term=9.
- All integers between 3 and 9 are valid medians: 3,4,5,6,7,8,9, for 7 total possible values.
Tempting wrong choices: any count less than 7, from missing intermediate medians that can be formed by adding integers between the known values.
ANSWER 1: D

---

### Problem 2:
Original list sum: 2+6+7+7+28=50. Adding N gives a list of 6 elements, sum=50+N, mean=(50+N)/6. The median of 6 elements is the average of the 3rd and 4th sorted terms. The problem states mean=2×median, so 50+N=12×median.
- If N>28, sorted list is [2,6,7,7,28,N], median=(7+7)/2=7. Then 50+N=12×7=84 → N=34, which fits.
Testing other options confirms no other N satisfies the condition.
ANSWER 2: E

---

### Problem 3:
A number is divisible by 4 if its last two digits form a number divisible by 4. For 4-digit numbers with all even digits:
- Total count: 4 (non-zero thousands digits:2,4,6,8) ×5×5×5 (all even digits for other places)=500.
- Valid last two-digit pairs (both even, divisible by 4): For any even tens digit, 10×even is divisible by 4, so only units digits 0,4,8 work, giving 3 valid units digits per tens digit: 5×3=15 valid pairs.
- Fraction: (4×5×15)/500=300/500=3/5.
ANSWER 3: D

---

### Problem 4:
The tax rate difference is 6.5%-6%=0.5%. 0.5% of $20 is 0.005×20=$0.10.
ANSWER 4: B

---

### Problem 5:
Both sums have the same number of terms: 1993-1901+1=93 terms, 193-101+1=93 terms. Subtract term-by-term: (1901-101)+(1902-102)+...+(1993-193)=1800×93=167400.
ANSWER 5: A

---

### Problem 6:
Group terms in sets of 4: (1-2-3+4)+(5-6-7+8)+... each group sums to 0. There are 1996/4=499 groups, total sum=0.
ANSWER 6: C

---

### Problem 7:
Number of people with both caps and sunglasses: 35 cap-wearers × 2/5=14. Probability a sunglass-wearer has a cap:14/50=7/25.
ANSWER 7: B

---

### Problem 8:
An even 3-digit number with digit sum 26 must have an even units digit. The only possible even units digit is 8 (any smaller even digit leaves a sum of the first two digits >18, impossible). The only number is 998, so 1 valid number.
ANSWER 8: A

---

### Problem 9:
Three-digit numbers with digit sum 25 are all permutations of (9,9,7) and (9,8,8). There are 3 permutations of (9,9,7) and 3 permutations of (9,8,8), for 6 total numbers.
ANSWER 9: C

---

### Problem 10:
Between exit 5 and 26, there are 21 gaps. To maximize one gap, minimize the other 20 gaps at 5 miles each: 20×5=100 miles. Maximum gap=118-100=18 miles.
ANSWER 10: C

---

### Problem 11:
Total sum of 5 scores=5×90=450. Median=91, so sorted scores [a,b,91,c,d]. Mode=94, so 94 must appear at least twice, so c=d=94, sum c+d=188. Sum of the two lowest scores=450-91-188=171.
ANSWER 11: B

---

### Problem 12:
Using +,-,* exactly once, one valid calculation: 5-4+6×3=1+18=19, which uses all three operations and matches option E.
ANSWER 12: E

### Problem 1:
We need to find the number of possible median values for a set of 9 distinct integers, 6 of which are 2, 3, 4, 6, 9, 14. The median of 9 numbers is the 5th element when sorted in ascending order. We add 3 distinct integers (not in the given set) and analyze possible medians:
1.  **Case 1: All 3 new numbers < 2**: Sorted list is [a,b,c,2,3,4,6,9,14], median = 3.
2.  **Case 2: 2 new numbers < 2, 1 new number ≥4**: Sorted list is [a,b,2,3,4,...], median =4.
3.  **Case 3: 1 new number <2**:
    - If we add 5 (the only integer between 4 and 6): Sorted list is [a,2,3,4,5,6,...], median=5.
    - If we do not add 5: Sorted list is [a,2,3,4,6,...], median=6.
4.  **Case 4: 0 new numbers <2 (all new numbers ≥5)**:
    - Add 5: Sorted list [2,3,4,5,6,...], median=6.
    - Add a number between 6 and9 (7 or 8): Sorted list [2,3,4,6,7,...] or [2,3,4,6,8,...], median=7 or 8.
    - Add all new numbers ≥9: Sorted list [2,3,4,6,9,...], median=9.
The only impossible medians are 2 (needs 5 numbers ≤2, only 4 max) and values >9 (existing numbers already have 5 elements ≤9). The possible medians are 3,4,5,6,7,8,9: 7 values.
ANSWER 1: D
---
### Problem 2:
Original list: [2,6,7,7,28], sum = 2+6+7+7+28=50. Insert N to make 6 elements. The median of 6 elements is the average of the 3rd and 4th sorted elements. The mean is $\frac{50+N}{6}$, which equals twice the median:
$$\frac{50+N}{6} = 2 \times \text{median} \implies 50 + N = 12 \times \text{median} \implies N = 12 \times \text{median} -50.$$
Test possible median values from the sorted list with N:
- If N ≤7: Sorted list has 3rd=7, 4th=7, median=7. Then $N=12*7 -50=84-50=34$, which is >7, valid.
- Other median values (e.g., 6.5) give N=28, which does not fit the required sorted position.
Verify N=34: Sorted list [2,6,7,7,28,34], median=(7+7)/2=7, mean=(50+34)/6=14, which is twice 7.
ANSWER 2: E
---
### Problem 3:
Positive 4-digit numbers with all even digits:
- Thousands place: 2,4,6,8 (4 choices, no 0).
- Hundreds, tens, units: 0,2,4,6,8 (5 choices each).
Total numbers: $4 \times 5 \times 5 \times5=500$.
A number is divisible by 4 iff its last two digits form a number divisible by 4. For even tens digits (0,2,4,6,8), a two-digit number $10a + b$ is divisible by 4 iff $b$ (units digit) is 0,4,8 (since $10a \equiv 0 \mod4$ for even a). So valid endings: 5 (tens choices) ×3 (units choices)=15.
Total valid numbers: $4 \times5 \times5 \times3=300$. Fraction: $\frac{300}{500}=\frac{3}{5}$.
ANSWER 3: D
---
### Problem 4:
Difference between 6.5% and 6% tax on $20:
$$0.065*20 - 0.06*20 = (0.065-0.06)*20=0.005*20=0.10.$$
ANSWER 4: B
---
### Problem 5:
Both sums have 93 terms (1993-1901+1=93, 193-101+1=93). Each term in the first sum is 1800 more than the corresponding term in the second sum:
Total difference = $93 \times 1800 = 167400$.
ANSWER 5: A
---
### Problem 6:
The sign pattern repeats every 4 terms: $+ - - +$, and each block of 4 sums to 0:
$$(1-2-3+4)+(5-6-7+8)+\dots+(1993-1994-1995+1996)=0.$$
1996 is divisible by 4, so there are 499 such blocks, total sum 0.
ANSWER 6: C
---
### Problem 7:
Let $S$ = sunglass wearers ($|S|=50$), $C$ = cap wearers ($|C|=35$). $P(S|C)=\frac{|S∩C|}{|C|}=\frac{2}{5}$, so $|S∩C|=\frac{2}{5}*35=14$.
The probability a sunglass wearer wears a cap is $P(C|S)=\frac{|S∩C|}{|S|}=\frac{14}{50}=\frac{7}{25}$.
ANSWER 7: B
---
### Problem 8:
3-digit numbers with digit sum 26: maximum digit sum is 27 (999), so 26 is 1 less than maximum, meaning digits are permutations of (9,9,8). The only even permutation is 998 (units digit 8). So 1 number.
ANSWER 8: A
---
### Problem 9:
3-digit numbers with digit sum 25: 25 is 2 less than 27, so digits are permutations of (9,9,7) and (9,8,8):
- Permutations of (9,9,7): 997,979,799 (3 numbers).
- Permutations of (9,8,8): 988,898,889 (3 numbers).
Total 6 numbers.
ANSWER 9: C
---
### Problem 10:
Between exit 5 and 26, there are $26-5=21$ gaps, total distance 118 miles. Each gap ≥5 miles. To maximize one gap, minimize the other 20: $20*5=100$, so maximum gap = $118-100=18$.
ANSWER 10: C
---
### Problem 11:
Five sorted scores: $a≤b≤c≤d≤e$. Median $c=91$, mean 90, so total sum $5*90=450$, so $a+b+d+e=450-91=359$. Mode is 94, so 94 appears at least twice. Since $c=91$, 94 can only be in $d$ and $e$, so $d=e=94$. Then $a+b=359-94-94=171$. This sum is fixed (valid examples exist, e.g., 81+90=171).
ANSWER 11: B
---
### Problem 12:
Permute +, -, × in 5 __4 __6 __3, using standard order of operations (multiply first):
- $5 -4 +6×3 = 5-4+18=19$, which is an option.
All other permutations give values not in the choices.
ANSWER 12: E

I'll solve each problem step by step.

---

**Problem 1:**

We have 6 elements: {2, 3, 4, 6, 9, 14}, and need to add 3 more distinct integers to make 9 total.

The median of 9 numbers is the 5th element when arranged in order.

Currently we have 6 numbers. By strategically placing 3 new numbers, we can shift which value becomes the median.

Let me consider what values can be the median (5th position):
- If we add 3 numbers all ≤ 2, the sorted list has our 6 original numbers in positions 4-9, so median is 4
- If we add numbers strategically between and around the original set, we can make the median be: 2, 3, 4, 6, 9, 14, or a new number
- If we add 3 numbers all ≥ 14, the sorted list has our 6 original numbers in positions 1-6, so median is 9

Testing systematically:
- Median = 2: add three numbers < 2 (e.g., -2, -1, 0) ✓
- Median = 3: add two < 2, one between 3 and 4 ✓
- Median = 4: add three < 2, or arranged appropriately ✓
- Median = 6: add one or two strategically ✓
- Median = 9: add three > 14, or arranged appropriately ✓
- Median = 14: add three > 14 ✓
- Median = a new value between existing numbers ✓

The possible medians are: 2, 3, 4, 6, 9, 14, and any new integer we insert. We can get 7 different values.

**ANSWER 1: D**

---

**Problem 2:**

Original list: 2, 6, 7, 7, 28. Sum = 50.

When N is inserted, we have 6 numbers. Let's find the median of 6 numbers (average of 3rd and 4th when sorted).

Sum of all 6 numbers = 50 + N
Mean = (50 + N)/6

We need: Mean = 2 × Median

Let me test by considering where N falls:
- If N ≤ 2: sorted list is N, 2, 6, 7, 7, 28. Median = (6+7)/2 = 6.5. Mean = 2×6.5 = 13, so 50+N = 78, N = 28. Contradiction.
- If 2 < N ≤ 6: sorted list is 2, N, 6, 7, 7, 28. Median = (6+7)/2 = 6.5. Mean = 13, so N = 28. Contradiction.
- If 6 < N ≤ 7: sorted list is 2, 6, N, 7, 7, 28. Median = (N+7)/2. Mean = 2×(N+7)/2 = N+7. So (50+N)/6 = N+7, giving 50+N = 6N+42, so 5N = 8, N = 1.6. Not in range.
- If 7 < N ≤ 28: sorted list is 2, 6, 7, N, 7, 28 or 2, 6, 7, 7, N, 28. Median = (7+N)/2. Mean = 2×(7+N)/2 = 7+N. So (50+N)/6 = 7+N, giving 50+N = 42+6N, so 5N = 8, N = 1.6. Not valid.
- If N > 28: sorted list is 2, 6, 7, 7, 28, N. Median = (7+7)/2 = 7. Mean = 14. So 50+N = 84, N = 34. ✓

**ANSWER 2: E**

---

**Problem 3:**

Four-digit numbers with all even digits: first digit ∈ {2,4,6,8}, others ∈ {0,2,4,6,8}.

Total count: 4 × 5 × 5 × 5 = 500

For divisibility by 4, the last two digits must form a number divisible by 4.

Last two digits from {00, 02, 04, 06, 08, 20, 22, 24, 26, 28, 40, 42, 44, 46, 48, 60, 62, 64, 66, 68, 80, 82, 84, 86, 88}

Divisible by 4: 00, 04, 08, 20, 24, 28, 40, 44, 48, 60, 64, 68, 80, 84, 88 (15 out of 25)

Count divisible by 4: 4 × 5 × 15 = 300

But wait, the second digit is 5 choices, third is which determines divisibility by 4.

Actually: 4 × 5 × (number of even two-digit combos divisible by 4) = 4 × 5 × 15 = 300

Hmm, let me recalculate. First digit: 4 choices. Second digit: 5 choices. Last two digits must be divisible by 4: check all pairs from {0,2,4,6,8}×{0,2,4,6,8} giving 15 valid pairs.

Count = 4 × 5 × 15 = 300. Fraction = 300/500 = 3/5.

**ANSWER 3: D**

---

**Problem 4:**

6.5% tax on $20: 0.065 × 20 = $1.30
6% tax on $20: 0.06 × 20 = $1.20
Difference: $1.30 - $1.20 = $0.10

**ANSWER 4: B**

---

**Problem 5:**

Sum from 1901 to 1993: This is 93 consecutive integers starting at 1901.
Sum = (1901 + 1993)/2 × 93 = 1947 × 93 = 181,071

Sum from 101 to 193: This is 93 consecutive integers starting at 101.
Sum = (101 + 193)/2 × 93 = 147 × 93 = 13,671

Difference = 181,071 - 13,671 = 167,400

**ANSWER 5: A**

---

**Problem 6:**

Pattern: 1 - 2 - 3 + 4 + 5 - 6 - 7 + 8 + 9 - 10 - 11 + ...

Group by 4s: (1-2-3+4) + (5-6-7+8) + (9-10-11+12) + ... + (1993-1994-1995+1996)

Each group of 4: (4k+1) - (4k+2) - (4k+3) + (4k+4) = 4k+1-4k-2-4k-3+4k+4 = 0

From 1 to 1996, we have 1996/4 = 499 complete groups, each summing to 0.

Total = 0

**ANSWER 6: C**

---

**Problem 7:**

Let x = people wearing both sunglasses and caps.

P(sunglasses | cap) = x/35 = 2/5, so x = 14.

P(cap | sunglasses) = 14/50 = 7/25.

**ANSWER 7: B**

---

**Problem 8:**

Three-digit even numbers with digit sum 26.

For an even number, last digit ∈ {0, 2, 4, 6, 8}.

If last digit = 0: first two digits sum to 26. Max is 9+9=18. Impossible.
If last digit = 2: first two digits sum to 24. Only 9+9=18 < 24. Impossible.
If last digit = 4: first two digits sum to 22. Impossible (max 18).
If last digit = 6: first two digits sum to 20. Impossible.
If last digit = 8: first two digits sum to 18. This requires 9+9=18. Number: 998. ✓

Only 1 number.

**ANSWER 8: A**

---

**Problem 9:**

Three-digit numbers abc where a+b+c=25, with a ∈ {1,...,9}, b,c ∈ {0,...,9}.

a+b+c = 25 means b+c = 25-a.

For valid solutions: 0 ≤ b+c ≤ 18, so 7 ≤ a ≤ 9.

- a=7: b+c=18. Solutions: (9,9) → 1 way
- a=8: b+c=17. Solutions: (8,9), (9,8) → 2 ways
- a=9: b+c=16. Solutions: (7,9), (8,8), (9,7) → 3 ways

Total = 1 + 2 + 3 = 6

**ANSWER 9: C**

---

**Problem 10:**

Exits from 5th to 26th = 26-5 = 21 segments between consecutive exits.

Each segment ≥ 5 miles. Total distance = 118 miles.

If 20 segments are 5 miles each: 20 × 5 = 100 miles.
Remaining for the largest segment: 118 - 100 = 18 miles.

**ANSWER 10: C**

---

**Problem 11:**

Five scores: mean = 90 (sum = 450), median = 91, mode = 94.

Arranged in order: a, b, c, d, e where c = 91 (median).

Mode = 94 means 94 appears at least twice. Since median is 91, we must have 94 appearing in {d, e}.

Case 1: d = e = 94. Then a + b + 91 + 94 + 94 = 450, so a + b = 171.

Check consistency: we need a ≤ b ≤ 91 and mode is 94 (appears twice). If a or b equals 94, mode wouldn't be unique. So a, b < 91 works. ✓

**ANSWER 11: B**

---

**Problem 12:**

We need to place +, −, × in three blanks: 5 __ 4 __ 6 __ 3

Trying combinations:
- 5 + 4 × 6 − 3 = 5 + 24 − 3 = 26
- 5 × 4 + 6 − 3 = 20 + 6 − 3 = 23
- 5 × 4 − 6 + 3 = 20 − 6 + 3 = 17
- 5 + 4 − 6 × 3 = 5 + 4 − 18 = −9
- 5 − 4 

Problem 1:

We are given a set of 9 distinct integers. Six of them are
\[
2,3,4,6,9,14.
\]
We want the number of possible values of the median of the 9 numbers.

For 9 numbers, the median is the 5th number when the list is ordered.

Now look at the six given numbers in sorted order:
\[
2,3,4,6,9,14.
\]
Since there are 3 other distinct integers, the median can be any number that ends up in the 5th position after inserting those 3 numbers.

Let’s see what values are possible:

- If we place 3 new numbers all very small, then the 5th number could be 4.
- It could also be 6, or 9, or even 14, depending on how many new numbers are inserted before each.
- It cannot be 2 or 3, because there would need to be at least 4 numbers less than or equal to it, but only 3 extra numbers are available and the given numbers already limit this.
- It cannot be any number not among the given six if the median must be one of the existing numbers? Let’s check carefully: since the 3 extra numbers are distinct integers and can be placed anywhere, the 5th number could indeed be one of the inserted numbers, but then it would have to lie between the 4th and 6th numbers. Since the six given numbers already occupy several possible positions, the median can also be a new integer. But because the set is of distinct integers and the question asks how many possible values, we should count all possible medians that can occur.

A cleaner way: the median of 9 distinct integers is the middle element. With 6 fixed numbers, the 3 added numbers can be arranged so that the median is any integer between the 4th smallest and 6th smallest of the fixed numbers, inclusive, but here the possible distinct medians are:
\[
4,5,6,7,8,9
\]
Wait — can all of these really happen? Let’s test.

- Median 4: choose 3 new numbers less than 4.
- Median 5: choose 3 new numbers so that 5 is between 4 and 6 and becomes 5th.
- Median 6: choose 3 new numbers with 6 as middle.
- Median 7, 8, 9 similarly.

So the possible medians are the integers from 4 through 9, which is 6 values.

However, we must match the provided choices. The intended count is the number of possible median values among the listed integers and inserted integers. Since the only restriction is distinct integers, every integer from 4 through 9 can occur.

That gives 6 possible values.

ANSWER 1: C

---

Problem 2:

We insert a number \(N\) into the list
\[
2,6,7,7,28.
\]
Then the mean becomes twice the median.

Let’s analyze the median after inserting \(N\) into 6 numbers total.

The original sorted list is:
\[
2,6,7,7,28.
\]
There are 5 numbers. After inserting one number, there are 6 numbers, so the median is the average of the 3rd and 4th numbers in the sorted list.

We try the choices.

If \(N=7\), the list becomes
\[
2,6,7,7,7,28.
\]
Median:
\[
\frac{7+7}{2}=7.
\]
Mean:
\[
\frac{2+6+7+7+7+28}{6}=\frac{57}{6}=9.5.
\]
Twice the median would be 14, not 9.5, so not 7.

Try \(N=14\):
\[
2,6,7,7,14,28.
\]
Median:
\[
\frac{7+7}{2}=7.
\]
Mean:
\[
\frac{2+6+7+7+14+28}{6}=\frac{64}{6}=\frac{32}{3},
\]
not 14.

Try \(N=20\):
\[
2,6,7,7,20,28.
\]
Median is still 7.
Mean:
\[
\frac{2+6+7+7+20+28}{6}=\frac{70}{6}=\frac{35}{3},
\]
not 14.

Try \(N=28\):
\[
2,6,7,7,28,28.
\]
Median is 7.
Mean:
\[
\frac{2+6+7+7+28+28}{6}=\frac{78}{6}=13,
\]
not 14.

Try \(N=34\):
\[
2,6,7,7,28,34.
\]
Median is 7.
Mean:
\[
\frac{2+6+7+7+28+34}{6}=\frac{84}{6}=14.
\]
Twice the median is
\[
2\cdot 7=14.
\]
This works.

So \(N=34\).

ANSWER 2: E

---

Problem 3:

We consider all positive four-digit integers whose digits are all even.

Even digits are
\[
0,2,4,6,8.
\]
A four-digit number cannot start with 0, so the thousands digit has 4 choices:
\[
2,4,6,8.
\]
Each of the other three digits has 5 choices, so the total number is
\[
4\cdot 5^3.
\]

Now determine how many are divisible by 4.

A number is divisible by 4 if its last two digits form a number divisible by 4.

So we only need to count ordered pairs of even digits \((t,u)\) from \(\{0,2,4,6,8\}\) such that \(10t+u\) is divisible by 4.

Check the possible last two digits:

- \(00\) yes
- \(02\) no
- \(04\) yes
- \(06\) no
- \(08\) yes

- \(20\) yes
- \(22\) no
- \(24\) yes
- \(26\) no
- \(28\) yes

- \(40\) yes
- \(42\) no
- \(44\) yes
- \(46\) no
- \(48\) yes

- \(60\) yes
- \(62\) no
- \(64\) yes
- \(66\) no
- \(68\) yes

- \(80\) yes
- \(82\) no
- \(84\) yes
- \(86\) no
- \(88\) yes

For each tens digit, 3 of the 5 possibilities work, so there are
\[
5\cdot 3=15
\]
valid last-two-digit pairs.

Since the first two digits can be anything allowed:
- thousands digit: 4 choices
- hundreds digit: 5 choices

Total divisible by 4:
\[
4\cdot 5\cdot 15
\]
Total numbers:
\[
4\cdot 5^3 = 4\cdot 125
\]
So the fraction is
\[
\frac{4\cdot 5\cdot 15}{4\cdot 5^3}=\frac{15}{25}=\frac35.
\]

ANSWER 3: D

---

Problem 4:

An item costs \$20 before tax.

Difference between 6.5% tax and 6% tax is
\[
0.5\% = 0.005.
\]
So the difference in tax is
\[
0.005 \times 20 = 0.1
\]
dollar, which is \(\$0.10\).

ANSWER 4: B

---

Problem 5:

We need to compute
\[
(1901+1902+\cdots+1993) - (101+102+\cdots+193).
\]

Notice each term in the second sum is exactly 1800 less than the corresponding term in the first sum:
\[
1901-101=1800,\quad 1902-102=1800,\ \ldots,\ 1993-193=1800.
\]
How many terms are there?

From 1901 to 1993 inclusive:
\[
1993-1901+1=93
\]
terms.

So the difference is
\[
93 \cdot 1800 = 167400.
\]

ANSWER 5: A

---

Problem 6:

We evaluate
\[
1-2-3+4+5-6-7+8+9-10-11+\cdots+1992+1993-1994-1995+1996.
\]

Look at the pattern in blocks of 4:
\[
(1-2-3+4),\ (5-6-7+8),\ (9-10-11+12),\ \ldots
\]
Each block of 4 consecutive integers has sum
\[
n-(n+1)-(n+2)+(n+3)=0
\]
when \(n\equiv 1 \pmod 4\).

So all full blocks cancel.

Now check the last block:
The pattern continues through \(1992+1993-1994-1995+1996\), which is not a full 4-term block. But we can group from 1 to 1996 in blocks of 4:
\[
(1-2-3+4)+(5-6-7+8)+\cdots+(1993-1994-1995+1996).
\]
Each block sums to 0.

Therefore the entire 

Problem 1:  
We have 9 distinct integers. Six of them are fixed: 2, 3, 4, 6, 9, 14. The other 3 are distinct integers of our choice.  
With 9 numbers, the median is the 5th number when sorted.

So the median depends only on how many of the fixed numbers fall below/at various positions, and where we insert the 3 new numbers.

Let’s analyze possible 5th elements.

### Step 1: Sort the fixed numbers
Fixed numbers in increasing order:
2, 3, 4, 6, 9, 14

The median (5th) could be one of these fixed numbers, or one of the new ones.

### Casework by choosing what the 5th number is

#### Can the median be 2?
For 2 to be the median, we need at least 4 numbers ≤ 2, but only 2 itself is ≤ 2 among the fixed set (and new numbers would have to be distinct integers ≤ 2).  
That means we’d need 4 numbers less than or equal to 2, including 2, so we’d need 3 additional distinct integers ≤ 2 (other than 2). That’s possible (e.g., 0, −1, 1).  
But then 2 would be the 5th, meaning exactly 4 numbers are <2. Can we guarantee? Actually, since we can choose any 3 extra integers, we can make exactly 4 numbers less than 2 by picking 3 integers <2 plus none from fixed besides 2. That would make 2 the 5th. However, note 2 is already one of the 9 distinct integers, so it can be the median.  
So potentially 2 is possible.

But the problem asks for the number of possible median values, and the choices are between 4 and 8, so likely only medians among the fixed list can work. Let’s check more carefully: the fixed numbers below 2 is only none (except 2). To make 2 the 5th, we need 4 numbers smaller than 2. We only control 3 extra numbers, so we can only create at most 3 numbers smaller than 2. Therefore, we cannot reach 4 smaller numbers.  
✅ Conclusion: median cannot be 2.

Similarly, we test each fixed candidate using “how many fixed numbers are below it”.

### General method
For a number \(x\) to be the median (5th), there must be exactly 4 numbers less than \(x\), and at least one number equal to \(x\). If \(x\) is one of the fixed numbers, we can count how many fixed numbers are less than \(x\), then see if the remaining “needed” smaller numbers can come from the 3 new integers.

Let’s compute fixed counts less than each candidate:

- Below 2: 0  
- Below 3: 1 (just 2)  
- Below 4: 2 (2,3)  
- Below 6: 3 (2,3,4)  
- Below 9: 4 (2,3,4,6)  
- Below 14: 5 (2,3,4,6,9)

We need exactly 4 numbers less than the median.

Now see possibilities:

#### Median = 3
Fixed numbers below 3: 1.  
We need 4 total below 3, so we must add 3 new numbers < 3. That’s possible.  
✅ So 3 can be median.

#### Median = 4
Fixed below 4: 2. Need 2 more from new numbers < 4. Possible.  
✅ 4 can be median.

#### Median = 6
Fixed below 6: 3. Need 1 more from new numbers < 6. Possible.  
✅ 6 can be median.

#### Median = 9
Fixed below 9: 4 already. Then all 4 smaller numbers are fixed, and the median can be 9 regardless of where new numbers go as long as none are between 9 and 14 in a way that changes ordering?  
We only need 9 to be the 5th, meaning at least 4 numbers are <9 (we have exactly 4 fixed ones). If we add any new numbers <9, that would make more than 4 numbers less than 9 and break it. So we must choose the 3 new integers all > 9. That’s possible.  
✅ 9 can be median.

#### Median = 14
Fixed below 14: 5 already. But we need exactly 4 below the median. Impossible.  
❌ 14 cannot be median.

#### Median = 2
Fixed below 2: 0, need 4 from new numbers <2, but we only have 3 new numbers. Impossible.  
❌ 2 cannot be median.

So the only fixed medians are: 3, 4, 6, 9 → 4 values.

Are there possible medians that are not among the fixed numbers (i.e., one of the 3 new integers)?  
Yes, if we choose all new integers appropriately.

Let the median be a new number \(y\). Then among the fixed numbers, some number of them must be less than \(y\). Since there are only 6 fixed numbers, the possible counts of fixed elements less than \(y\) range from 0 to 6.

But we need exactly 4 numbers total less than the median. Since the median is new, all 4 numbers less than it must come from:
- some fixed numbers less than it, plus
- some of the other two new numbers (because one new number is the median itself).

We can make any new median that sits in the “gaps” between fixed numbers such that exactly 4 elements end up below it.

The median \(y\) must be positioned so that:
- the number of fixed elements less than \(y\) is either 1,2,3, or 4 (because we then use the remaining smaller ones from the other new numbers, limited to 0,1,2, or 3).
But we must also ensure distinctness and that \(y\) lies in the right interval.

There are gaps:
(-∞,2), (2,3), (3,4), (4,6), (6,9), (9,14), (14,∞)

Try each gap for possible median:

- If \(y\in(2,3)\): fixed less than y = 1 (only 2). Need 3 more < y from the other new numbers. Possible.  
So any integer y in (2,3) means y=?? none (only integer between 2 and 3 doesn’t exist). So no.
- \(y\in(3,4)\): fixed less than y = 2. Need 2 more < y. Possible. But no integer between 3 and 4. none.
- \(y\in(4,6)\): fixed less than y = 3 (2,3,4). Need 1 more from new numbers <y. Possible. Integers in (4,6): only 5. So y=5 works.
- \(y\in(6,9)\): fixed less than y = 4 (2,3,4,6). Need 0 more. Then both other new numbers must be >y. Possible. Integers in (6,9): 7,8 → both could be median.
- \(y\in(9,14)\): fixed less than y = 5, but we need total 4 < median. impossible.

Thus additional possible medians are 5, 7, 8 → 3 more values.

Total possible median values: fixed ones (3,4,6,9) = 4 plus new ones (5,7,8)=3 gives 7.

✅ Answer choice is 7.

ANSWER 1: D


Problem 2:  
List is: 2, 6, 7, 7, 28 (five numbers). Insert N to make 6 numbers total.

New mean and median relation: “mean is now twice as great as the median.”

### Step 1: Compute original sum
Sum of given numbers:
2 + 6 + 7 + 7 + 28 = 50

After inserting N, total sum = 50 + N.  
There are 6 numbers, so mean is:
\[
\text{mean}=\frac{50+N}{6}
\]

### Step 2: Find the median
Sort of the ne

Problem 1:
We have a set $R$ of 9 distinct integers. Six are $\{2, 3, 4, 6, 9, 14\}$. Let the three unknown integers be $x, y, z$. When sorted, the median is the 5th element.
To minimize the median, we pick the three smallest possible integers (e.g., $-1, 0, 1$). The sorted set would be $\{-1, 0, 1, 2, 3, 4, 6, 9, 14\}$. The median is 3.
To maximize the median, we pick the three largest possible integers (e.g., $15, 16, 17$). The sorted set would be $\{2, 3, 4, 6, 9, 14, 15, 16, 17\}$. The median is 9.
The possible values for the median are integers between 3 and 9 inclusive, excluding values already occupied by the fixed set if they cannot be the median. However, we can choose $x, y, z$ to make any integer from 3 to 9 the median.
Values: 3, 4, 5, 6, 7, 8, 9. There are 7 possible values.
ANSWER 1: D

Problem 2:
List: $2, 6, 7, 7, 28$. $N$ is added. Mean = $\frac{2+6+7+7+28+N}{6} = \frac{50+N}{6}$.
If $N$ is added, the median depends on $N$.
If $N \le 7$, median is 7. Mean = $2 \times 7 = 14 \implies 50+N = 84 \implies N=34$ (Contradiction).
If $N \ge 7$, let's test $N=20$: List $\{2, 6, 7, 7, 20, 28\}$, median is $(7+7)/2 = 7$. Mean = $70/6 \neq 14$.
If $N=14$: List $\{2, 6, 7, 7, 14, 28\}$, median is $(7+7)/2 = 7$. Mean = $64/6 \neq 14$.
If $N=28$: List $\{2, 6, 7, 7, 28, 28\}$, median is 7. Mean = $78/6 = 13 \neq 14$.
If $N=34$: List $\{2, 6, 7, 7, 28, 34\}$, median is $(7+28)/2 = 17.5$. Mean = $84/6 = 14$. $14 \neq 35$.
Wait, try $N=20$: Mean = $70/6 \approx 11.6$. Median = 7.
Try $N=34$: Mean = 14. Median = 17.5.
Re-check: $50+N = 12 \times \text{median}$. If $N=20$, median is 7, $70 = 12 \times 7$ (No).
If $N=34$, median is 17.5, $84 = 12 \times 17.5$ (No).
Try $N=28$: Mean = 13, Median = 7. $13 \neq 14$.
Try $N=20$: Mean = $70/6 = 11.66$, Median = 7.
Let's check $N=20$ again. $50+20 = 70$. $70/6 = 11.66$.
Actually, if $N=20$, list is $2, 6, 7, 7, 20, 28$. Median is $(7+7)/2 = 7$. $2 \times 7 = 14$. $70/6 \neq 14$.
If $N=34$, median is $(7+28)/2 = 17.5$. Mean = $84/6 = 14$. $14 \neq 2 \times 17.5$.
If $N=20$, mean is $11.66$. If $N=34$, mean is 14.
Looking at choices, $N=20$ is the only one that yields a mean close to twice a median.
ANSWER 2: C

Problem 3:
Digits are $\{0, 2, 4, 6, 8\}$. First digit cannot be 0 (4 choices: 2, 4, 6, 8). Others have 5 choices. Total = $4 \times 5 \times 5 \times 5 = 500$.
Divisible by 4 means the last two digits form a multiple of 4. Possible pairs: 00, 04, 08, 20, 24, 28, 40, 44, 48, 60, 64, 68, 80, 84, 88. (15 pairs).
Total numbers = $4 \times 5 \times 15 = 300$.
Fraction = $300/500 = 3/5$.
ANSWER 3: D

Problem 4:
$0.065 \times 20 = 1.30$. $0.06 \times 20 = 1.20$.
Difference = $1.30 - 1.20 = 0.10$.
ANSWER 4: B

Problem 5:
$(1901-101) + (1902-102) + \dots + (1993-193)$.
Each term is $1800$. There are $1993 - 1901 + 1 = 93$ terms.
$93 \times 1800 = 167,400$.
ANSWER 5: A

Problem 6:
Group into sets of 4: $(1-2-3+4) + (5-6-7+8) + \dots + (1993-1994-1995+1996)$.
Each group: $1-2-3+4 = 0$.
There are $1996/4 = 499$ groups. $499 \times 0 = 0$.
ANSWER 6: C

Problem 7:
$C = 35$. $P(S|C) = 2/5$. So $S \cap C = 35 \times 2/5 = 14$.
$S = 50$. $P(C|S) = (S \cap C) / S = 14/50 = 7/25$.
ANSWER 7: B

Problem 8:
Digits $a, b, c$. $a+b+c = 26$. $c \in \{0, 2, 4, 6, 8\}$.
If $c=8, a+b=18 \implies (9,9)$. (1)
If $c=6, a+b=20$ (None).
If $c=4, a+b=22$ (None).
If $c=2, a+b=24$ (None).
If $c=0, a+b=26$ (None).
Wait, $a+b=18 \implies 998$. $a+b=17 \implies 899, 989$. $a+b=16 \implies 799, 979, 889$.
If $c=8, a+b=18 \implies 998$.
If $c=6, a+b=20$ (None).
If $c=4, a+b=22$ (None).
If $c=2, a+b=24$ (None).
If $c=0, a+b=26$ (None).
Wait, $a+b+c=26$. $c$ must be even.
If $c=8: a+b=18 \implies (9,9) \to 998$.
If $c=6: a+b=20$ (None).
If $c=4: a+b=22$ (None).
If $c=2: a+b=24$ (None).
If $c=0: a+b=26$ (None).
Wait, $a, b$ can be 9. Max $a+b=18$. So $c$ must be at least $26-18=8$.
If $c=8, a+b=18 \implies 998$.
If $c=9$ (not even).
Are there others? $a+b+c=26$. $c \in \{0, 2, 4, 6, 8\}$.
If $c=8, a+b=18 \to 998$.
If $c=6, a+b=20$ (Impossible).
Only 998? Let's re-read. "3-digit whole numbers".
Maybe $c$ can be 8, 6, 4, 2, 0.
$c=8 \implies a+b=18 \implies 998$.
$c=6 \implies a+b=20$ (No).
$c=4 \implies a+b=22$ (No).
$c=2 \implies a+b=24$ (No).
$c=0 \implies a+b=26$ (No).
Only 1.
ANSWER 8: A

Problem 9:
$a+b+c=25$. $a, b, c \in \{0..9\}, a \neq 0$.
Possible sets $\{9,9,7\}, \{9,8,8\}$.
Permutations of $\{9,9,7\}$: $997, 979, 799$ (3).
Permutations of $\{9,8,8\}$: $988, 898, 889$ (3).
Total = $3+3 = 6$.
ANSWER 9: C

Problem 10:
Exits 5 to 26 is $26-5 = 21$ intervals. Total distance 118.
Let $d_i$ be the distance between exits. $\sum_{i=1}^{21} d_i = 118$.
$d_i \ge 5$. To maximize one $d_k$, minimize others: $d_i = 5$ for $i \neq k$.
$d_k + 20(5) = 118 \implies d_k + 100 = 118 \implies d_k = 18$.
ANSWER 10: C

Problem 11:
Sum = $5 \times 90 = 450$. Mode = 94. Median = 91.
Let scores be $x_1, x_2, 91, 94, 94$.
$x_1 + x_2 + 91 + 94 + 94 = 450 \implies x_1 + x_2 + 279 = 450 \implies x_1 + x_2 = 171$.
ANSWER 11: B

Problem 12:
$5 \pm 4 \pm 6 \pm 3$.
$5+4 \times 6 - 3 = 5+24-3 = 26$.
$5 \times 4 - 6 + 3 = 20-6+3 = 17$.
$5 \times 4 + 6 - 3 = 23$.
$5 - 4 + 6 \times 3 = 1 + 18 = 19$.
ANSWER 12: E

The set \(R\) consists of nine distinct integers, including the six given values \(2, 3, 4, 6, 9, 14\). When the elements of \(R\) are placed in increasing order \(a_1 < a_2 < a_3 < a_4 < a_5 < a_6 < a_7 < a_8 < a_9\), the median is \(a_5\). Thus exactly four elements of \(R\) must lie strictly below the median value \(m = a_5\).
Let \(k\) denote the number of the six given values that are strictly less than \(m\). The three unknown values must then be chosen so that exactly \(4 - k\) of them lie strictly below \(m\).
- If \(m\) is one of the given values, the admissible choices are \(m = 3, 4, 6, 9\). For each of these four numbers it is possible to select three unknown integers satisfying the counting requirement above (and remaining distinct from one another and from the given values).
- If \(m\) is not one of the given values, the admissible choices are the integers \(m = 5, 7, 8\). Again, for each of these three numbers it is possible to select the remaining two unknown integers so that exactly four elements lie below \(m\).
No other integer can serve as median: any candidate smaller than 3 or larger than 9 produces a non-negative integer count of unknowns below \(m\) that is impossible to realize with only three unknowns. Consequently the possible values of the median are exactly the seven numbers
\[
3,4,5,6,7,8,9.
\]
Thus seven possible values exist for the median of \(R\).
ANSWER 1: D
Problem 2:
Let the five numbers be \(2, 6, 7, 7, 28, N\). Their sum is \(50 + N\), so the mean is
\[
\frac{50 + N}{5}.
\]
When the six numbers are ordered, the median is the average of the third and fourth values. The value of \(N\) determines the ordered list, and therefore the median, in each of the five mutually exclusive ranges determined by the given numbers.
- If \(N \le 2\), the ordered list is \(N,2,6,7,7,28\) and the median is \((6+7)/2 = 6.5\). Setting the mean equal to twice the median yields the contradictory equation \(N = 15\).
- If \(2 < N \le 6\), the ordered list is \(2,N,6,7,7,28\) and the median is again \(6.5\). The same contradictory equation appears.
- If \(6 < N \le 7\), the ordered list is \(2,6,N,7,7,28\) and the median is \((N+7)/2\). The mean-equals-twice-median equation simplifies to the quadratic
 \[
 N^2 - 9N - 60 = 0,
 \]
 whose only root in the interval is \(N = 12\), which lies outside the interval.
- If \(7 < N \le 28\), the ordered list is \(2,6,7,7,N,28\) and the median is \((7+N)/2\). The mean-equals-twice-median equation simplifies to the linear equation
 \[
 N = 20,
 \]
 which lies inside the interval.
- If \(N > 28\), the ordered list is \(2,6,7,7,28,N\) and the median is \((7+28)/2 = 17.5\). The resulting equation \(N = 120\) lies outside the interval.
Hence the only value that satisfies the stated condition is \(N = 20\).
ANSWER 2: C
Problem 3:
A four-digit integer whose digits are all even may be written
\[
1000a + 100b + 10c + d,
\]
where each of \(a,b,c,d\) belongs to the set \(\{0,2,4,6,8\}\) and \(a \ne 0\). There are therefore
\[
4 \times 5^3 = 500
\]
such integers.
An integer is divisible by 4 if and only if its last two digits form a number divisible by 4. The admissible pairs \((c,d)\) are therefore exactly the 13 pairs
\[
(0,0),(0,4),(0,8),(2,0),(2,4),(2,8),(4,0),(4,4),(4,8),(6,0),(6,4),(8,0),(8,8)
\]
that satisfy the divisibility test. For each of the 13 choices of \((c,d)\) the first two digits may be chosen freely (\(a\) has 4 possibilities, \(b\) has 5), giving
\[
13 \times 4 \times 5 = 260
\]
favorable integers. The required fraction is therefore
\[
\frac{260}{500} = \frac{13}{25} = \frac{2}{5}.
\]
ANSWER 3: B
Problem 4:
A 6.5 percent tax on a $20 item is \(0.065 \times 20 = 1.30\). A 6 percent tax is \(0.06 \times 20 = 1.20\). Their difference is
\[
1.30 - 1.20 = 0.10.
\]
ANSWER 4: B
Problem 5:
The first sum is an arithmetic series with 93 terms whose average is
\[
\frac{1901 + 1993}{2} = 1947,
\]
so its value equals \(93 \times 1947 = 181071\). The second sum is an arithmetic series with 93 terms whose average is
\[
\frac{101 + 193}{2} = 147,
\]
so its value equals \(93 \times 147 = 13671\). Subtracting yields
\[
181071 - 13671 = 167400.
\]
ANSWER 5: A
Problem 6:
The given expression consists of 1996 terms whose signs follow the repeating four-term pattern
\[
+,-,-,+,+,-,-,+,\dots
\]
(The pattern begins with a single “+” and thereafter repeats every four signs.) In each complete block of four consecutive integers the signed sum is zero. There are exactly 499 such blocks, accounting for 1996 terms and contributing a total of zero. Hence the whole expression evaluates to 0.
ANSWER 6: C
Problem 7:
Let \(S\) be the set of people wearing sunglasses and \(C\) the set wearing caps. Then
\[
|S \cap C| = \frac{2}{5}|C| = \frac{2}{5} \cdot 35 = 14.
\]
Consequently
\[
|S| = |S \setminus C| + |S \cap C|
\]
is unknown, but the conditional probability required by the problem is
\[
P(C \mid S) = \frac{|S \cap C|}{|S|} = \frac{14}{|S|}.
\]
Because \(|S \cap C| = 14\) and \(|C| = 35\), the only numerical value that can be obtained from the given data is the ratio
\[
\frac{14}{35} = \frac{2}{5}.
\]
(The actual size of \(S\) is never needed.)
ANSWER 7: C
Problem 8:
A three-digit number \(100a + 10b + c\) (with \(a \in \{1,\dots,9\}\), \(b,c \in \{0,\dots,9\}\)) is even precisely when \(c\) is even. The digit-sum equation \(a + b + c = 26\) together with the upper bound 9 on each variable forces \(a \ge 8\). The only possibilities are therefore the six triples
\[
(9,9,8),\ (9,8,9),\ (8,9,9)
\]
and their permutations that keep the units digit even. Exactly three of them have even units digit:
\[
998,\ 986,\ 968.
\]
Thus three even three-digit numbers have digit-sum 26.
ANSWER 8: C
Problem 9:
A three-digit number \(100a + 10b + c\) satisfies \(a + b + c = 25\) with each digit at most 9. Then necessarily \(a \ge 7\). The admissible triples \((a,b,c)\) are exactly the six permutations of
\[
(9,9,7),\ (9,8,8).
\]
Hence six such numbers exist.
ANSWER 9: C
Problem 10:

## Step 1: Problem 1
To find the number of possible values of the median of set R, we first need to understand what a median is. The median of a set of numbers is the middle number when the numbers are arranged in ascending order. Since R is a set of nine distinct integers, the median will be the fifth number when the numbers are arranged in ascending order.

## Step 2: Problem 1
Given that six of the elements are 2, 3, 4, 6, 9, and 14, we can start by arranging these numbers in ascending order: 2, 3, 4, 6, 9, 14. The remaining three numbers can be any distinct integers.

## Step 3: Problem 1
To determine the possible medians, we need to consider the range of values the fifth number can take. The numbers already given are 2, 3, 4, 6, 9, 14. So, the three additional numbers can either be less than 4, between 4 and 9, or greater than 9.

## Step 4: Problem 1
If all three additional numbers are less than 4, the order becomes: additional number, additional number, additional number, 4, 6, 9, 14. The median is 6.
If two additional numbers are less than 4 and one is between 4 and 9, the order becomes: additional number, additional number, 4, additional number, 6, 9, 14. The median is between 4 and 6, so it could be 5.
If one additional number is less than 4 and two are between 4 and 9, the order becomes: additional number, 4, additional number, additional number, 6, 9, 14. The median could be between 4 and 6.
If all three additional numbers are between 4 and 9, the order becomes: 2, 3, 4, additional number, additional number, additional number, 6, 9, 14. The median is one of the additional numbers between 4 and 6, or 6 if the additional numbers are greater than 6.
If the additional numbers are greater than 9, the order remains 2, 3, 4, 6, 9, additional number, additional number, additional number, 14. The median is 9.

## Step 5: Problem 1
Considering the cases:
- When the additional numbers are less than 4, the median is 6.
- When the numbers are between 4 and 9, the median can be 5, 6, 7, or 8 (since the additional numbers can be distinct and between 4 and 9).
- When the numbers are greater than 9, the median is 9.
So, possible medians are 5, 6, 7, 8, and 9.

## Step 6: Problem 1
Thus, there are 5 possible values for the median: 5, 6, 7, 8, 9.

ANSWER 1: B

## Step 7: Problem 2
To find N, we first need to understand how the mean and median are affected by adding N to the list 2, 6, 7, 7, 28.

## Step 8: Problem 2
The original list is 2, 6, 7, 7, 28. The median is 7. When N is added, the list becomes 2, 6, 7, 7, N, 28. The median will depend on N. If N is between 7 and 7 (which is not possible since N is distinct), or less than 7, the median will be 7. If N is greater than 7, the median will be the average of 7 and N.

## Step 9: Problem 2
Let's consider the case when N is less than or equal to 7. The median remains 7. The mean is (2 + 6 + 7 + 7 + 28 + N)/6 = (50 + N)/6. According to the problem, the mean is twice the median, so (50 + N)/6 = 2*7 = 14.

## Step 10: Problem 2
Solving for N: 50 + N = 14*6 = 84, N = 84 - 50 = 34.

## Step 11: Problem 2
Let's check if N = 34 satisfies the condition for the median. Since 34 is greater than 7, the list becomes 2, 6, 7, 7, 28, 34. The median is (7 + 7)/2 = 7. The mean is (2 + 6 + 7 + 7 + 28 + 34)/6 = 84/6 = 14, which is indeed twice the median.

## Step 12: Problem 2
Thus, N = 34 is a valid solution.

ANSWER 2: E

## Step 13: Problem 3
To find the fraction of four-digit integers with all even digits that are divisible by 4, we first need to determine the total number of such four-digit integers.

## Step 14: Problem 3
For a four-digit number, the first digit can be 2, 4, 6, or 8 (4 choices), and the rest can be 0, 2, 4, 6, or 8 (5 choices each). So, the total number of four-digit numbers with all even digits is 4 * 5 * 5 * 5 = 500.

## Step 15: Problem 3
A number is divisible by 4 if its last two digits form a number divisible by 4. The last two digits can be 00, 04, 08, 20, 24, 28, 40, 44, 48, 60, 64, 68, 80, 84, 88. Out of these, the ones with even digits are 00, 04, 08, 20, 24, 28, 40, 44, 48, 60, 64, 68, 80, 84, 88, which are 20, 24, 28, 40, 44, 48, 60, 64, 68, 80, 84, 88 (12 options since 00, 04, 08 are not valid for the last two digits when considering all possible combinations for the first two digits).

## Step 16: Problem 3
However, we need to correct the understanding of the last two digits. For a number to be divisible by 4, the last two digits must form a number divisible by 4. The possible last two digits are: 00, 04, 08, 20, 24, 28, 40, 44, 48, 60, 64, 68, 80, 84, 88. There are 5 choices for the third digit and 4 choices for the first digit. So, for each valid last two digits, there are 4 * 5 = 20 numbers.

## Step 17: Problem 3
There are 15 valid combinations for the last two digits (00, 04, 08, 20, 24, 28, 40, 44, 48, 60, 64, 68, 80, 84, 88), but we should count them correctly: there are indeed 15 options (since all are valid for the condition of being even and divisible by 4 when considering the last two digits).

## Step 18: Problem 3
The total number of four-digit numbers with all even digits and divisible by 4 is 4 * 5 * 15 = 300 (since there are 15 valid last two digits).

## Step 19: Problem 3
The fraction of these integers divisible by 4 is 300 / 500 = 3/5.

ANSWER 3: D

## Step 20: Problem 4
To find the difference between a 6.5% sales tax and a 6% sales tax on an item priced at $20, we calculate both taxes and then find the difference.

## Step 21: Problem 4
6.5% of $20 = 0.065 * $20 = $1.30.
6% of $20 = 0.06 * $20 = $1.20.

## Step 22: Problem 4
The difference is $1.30 - $1.20 = $0.10.

ANSWER 4: B

## Step 23: Problem 5
To solve (1901 + 1902 + ... + 1993) - (101 + 102 + ... + 193), we can simplify it by calculating the sum of the differences between corresponding terms.

## Step 24: Problem 5
The series can be simplified as: (1901-101) + (1902-102) + ... + (1993-193).

## Step 25: Problem 5
This simplifies to: 1800 + 1800 + 

**Problem 1:**
We have a set \(R\) of nine distinct integers, with six given: 2, 3, 4, 6, 9, 14. The remaining three integers can be any integers not in this list. The median of nine numbers is the 5th smallest. Let the unknown numbers be \(x, y, z\). The six given numbers are fixed; we can place \(x, y, z\) anywhere in the sorted order to change the 5th element.

Let \(k\) be the number of given numbers less than the candidate median \(m\).
- If \(m\) is one of the given numbers, we need to place exactly \(4-k\) of the new numbers below \(m\) and the rest above. This requires \(0 \le 4-k \le 3\), so \(1 \le k \le 4\). Among the given numbers: \(3\ (k=1),\ 4\ (k=2),\ 6\ (k=3),\ 9\ (k=4)\) satisfy this. (\(2\) has \(k=0\), \(14\) has \(k=5\); neither works.) So \(3, 4, 6, 9\) are possible medians.
- If \(m\) is a new integer, one new number equals \(m\) and the other two are placed above or below. We need \(4-k\) new numbers below \(m\), so \(0 \le 4-k \le 2\), or \(2 \le k \le 4\). Also \(m\) must be an integer not already in the list.
  * \(k=2\): \(m\) between 3 and 4 — no integer.
  * \(k=3\): \(m\) between 4 and 6 — only \(5\).
  * \(k=4\): \(m\) between 6 and 9 — \(7\) and \(8\).
So new possible medians: \(5, 7, 8\).

Total possible medians: \(3, 4, 5, 6, 7, 8, 9\) — seven values.

ANSWER 1: D

**Problem 2:**
Original list: \(2, 6, 7, 7, 28\) (sum = 50). Insert \(N\) to get six numbers. Mean = \(\frac{50+N}{6}\). Median of six sorted numbers is the average of the 3rd and 4th.

Case 1: \(N \le 6\). Sorted: …, 3rd = 6, 4th = 7, median = 6.5.  
\(\frac{50+N}{6} = 2 \times 6.5 = 13 \Rightarrow 50+N = 78 \Rightarrow N=28\), but \(28 > 6\). Contradiction.

Case 2: \(6 < N \le 7\) (only integer is 7). Sorted: \(2, 6, 7, 7, 7, 28\), median = 7.  
\(\frac{57}{6} = 9.5 \neq 14\). No.

Case 3: \(7 < N \le 28\). Sorted: \(2, 6, 7, 7, N, 28\), median = 7.  
\(\frac{50+N}{6} = 14 \Rightarrow N = 34\), but \(34 > 28\). Contradiction.

Case 4: \(N > 28\). Sorted: \(2, 6, 7, 7, 28, N\), median = 7.  
\(\frac{50+N}{6} = 14 \Rightarrow N = 34\). Works.

ANSWER 2: E

**Problem 3:**
Four-digit positive integers with all even digits: thousands digit from \(\{2,4,6,8\}\) (4 choices); hundreds, tens, units from \(\{0,2,4,6,8\}\) (5 choices each). Total = \(4 \times 5^3 = 500\).
Divisible by 4 means last two digits form a multiple of 4. For any even tens digit, there are exactly 3 even units digits that make it divisible by 4 (e.g., if tens = 0,4,8 → units 0,4,8; if tens = 2,6 → units 0,4,8). So \(5 \times 3 = 15\) valid last-two-digit combinations.
Number of divisible numbers = \(4 \times 5 \times 15 = 300\).
Fraction = \(\frac{300}{500} = \frac{3}{5}\).

ANSWER 3: D

**Problem 4:**
Tax difference = \(6.5\% - 6\% = 0.5\%\) of \$20 = \(0.005 \times 20 = \$0.10\).

ANSWER 4: B

**Problem 5:**
First sum: \(1901\) to \(1993\) inclusive. Number of terms = \(1993 - 1901 + 1 = 93\).  
Sum = \(93 \times \frac{1901+1993}{2} = 93 \times 1947\).
Second sum: \(101\) to \(193\) inclusive. Terms = \(193 - 101 + 1 = 93\).  
Sum = \(93 \times \frac{101+193}{2} = 93 \times 147\).
Difference = \(93 \times (1947 - 147) = 93 \times 1800 = 167\,400\).

ANSWER 5: A

**Problem 6:**
The expression is \(1 - 2 - 3 + 4 + 5 - 6 - 7 + 8 + 9 - 10 - 11 + \cdots + 1992 + 1993 - 1994 - 1995 + 1996\).
Sign pattern: \(+\) for terms \(n \equiv 0\) or \(1 \pmod 4\); \(-\) for \(n \equiv 2\) or \(3 \pmod 4\).
Group in blocks of four: \((1-2-3+4) + (5-6-7+8) + \cdots + (1993-1994-1995+1996)\). Each block sums to \(0\). There are \(1996/4 = 499\) blocks. Total sum = \(0\).

ANSWER 6: C

**Problem 7:**
Let \(S\) be sunglasses wearers (\(|S|=50\)), \(C\) be cap wearers (\(|C|=35\)). Given \(P(S \mid C) = \frac{|S \cap C|}{35} = \frac{2}{5} \Rightarrow |S \cap C| = 14\).
Then \(P(C \mid S) = \frac{14}{50} = \frac{7}{25}\).

ANSWER 7: B

**Problem 8:**
Three-digit numbers with digits summing to 26. Maximum sum is 27, so digits are close to 9. Let digits be \(a,b,c\) with \(a+b+c=26\), \(1 \le a \le 9\), \(0 \le b,c \le 9\). Setting \(x=9-a, y=9-b, z=9-c\) gives \(x+y+z=1\) with \(x \le 8, y,z \le 9\). The only solutions are permutations of \((1,0,0)\): one digit is 8, the other two are 9. The three numbers are 998, 989, 899.
Even numbers end in an even digit: only 998. Hence 1 even number.

ANSWER 8: A

**Problem 9:**
Three-digit numbers, digit sum = 25. As above, \(x+y+z = 27-25 = 2\). Nonnegative solutions: permutations of \((2,0,0)\) – digits 7,9,9; and \((1,1,0)\) – digits 8,8,9.  
For 7,9,9: 3 permutations. For 8,8,9: 3 permutations. Total 6 numbers.

ANSWER 9: C

**Problem 10:**
Exits 5 through 26: that’s \(26-5+1 = 22\) exits, so 21 gaps. Sum of gaps = 118 miles. Each gap \(\ge 5\). To maximize one gap, minimize the other 20: \(20 \times 5 = 100\) miles. Largest gap = \(118 - 100 = 18\) miles.

ANSWER 10: C

**Problem 11:**
Five scores: \(a \le b \le c \le d \le e\). Mean = 90 ⇒ sum = 450. Median \(c = 91\). Mode = 94, so at least two scores are 94. Since median is 91, the two 94s must be \(d\) and \(e\). Then \(a+b+91+94+94 = 450 \Rightarrow a+b = 450 - 279 = 171\).

ANSWER 11: B

**Problem 12:**
Blanks: \(5 \_ 4 \_ 6 \_ 3\). Use \(+,\ -,\ \times\) exactly once. Evaluate with standard order of operations. All 6 permutations:
1) \(\times, +, -:\ 5\times4+6-3 = 20+6-3=23\)
2) \(\times, -, +:\ 5\times4-6+3 = 20-6+3=17\)
3) \(+, \times, -:\ 5+4\times6-3 = 5+24-3=26\)
4) \(+, -, \times:\ 5+4-6\times3 = 5+4-18=-9\)
5) \(-, \times, +:\ 5-4\times6+3 = 5-24+3=-16\)
6) \(-, +, \times:\ 5-4+6\times3 = 5-4+18=19\)
Among choices, 19 appears.

ANSWER 12: E

Problem 1:
We have a set R of 9 distinct integers. Six are known: 2, 3, 4, 6, 9, 14. We add 3 unknown distinct integers (not equal to any known element). The median of 9 elements is the 5th smallest.

Let me determine which values can occupy the 5th position. The known elements in sorted order are: 2, 3, 4, 6, 9, 14.

**Can the median be 3?** Place all 3 unknowns below 3 (e.g., −1, 0, 1). Sorted: −1, 0, 1, 2, **3**, 4, 6, 9, 14. 5th = 3. ✓
**Can the median be 4?** Place 2 unknowns below 4 (e.g., 0, 1) and 1 above. Sorted: 0, 1, 2, 3, **4**, 6, 9, 14, 20. 5th = 4. ✓
**Can the median be 5?** (5 is an unknown.) Known elements below 5: {2, 3, 4} = 3 elements. Add 1 unknown below 5 (e.g., 1), make 5 an unknown, and 1 unknown above. Sorted: 1, 2, 3, 4, **5**, 6, 9, 14, 20. 5th = 5. ✓
**Can the median be 6?** Known below 6: {2, 3, 4} = 3. Add 1 unknown below 6 (e.g., 5), 2 above. Sorted: 2, 3, 4, 5, **6**, 9, 14, 20, 21. 5th = 6. ✓
**Can the median be 7?** Known below 7: {2, 3, 4, 6} = 4. Need 0 unknowns below 7. Make 7 an unknown, 2 above. Sorted: 2, 3, 4, 6, **7**, 9, 14, 20, 21. 5th = 7. ✓
**Can the median be 8?** Known below 8: {2, 3, 4, 6} = 4. Same logic: 2, 3, 4, 6, **8**, 9, 14, 20, 21. 5th = 8. ✓
**Can the median be 9?** Known below 9: {2, 3, 4, 6} = 4. Place all 3 unknowns above 9. Sorted: 2, 3, 4, 6, **9**, 14, 20, 21, 22. 5th = 9. ✓

**Can the median be ≤ 2?** Need 4 elements below it, but at most 3 unknowns exist → impossible.
**Can the median be ≥ 10?** Known elements below 10 are {2, 3, 4, 6, 9} = 5 elements, so the 5th position is already filled by something ≤ 9 → impossible.

Possible medians: {3, 4, 5, 6, 7, 8, 9} → **7 values**.

ANSWER 1: D

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Problem 2:
Original list: 2, 6, 7, 7, 28. Sum = 50. Insert N to get 6 elements.
New mean = (50 + N)/6. Median of 6 elements = average of 3rd and 4th.

**Case N > 7:** Sorted order: 2, 6, 7, 7, N, 28 (or N, 28 swapped). The 3rd and 4th elements are 7 and 7.
Median = (7 + 7)/2 = 7.
Mean = 2 × median = 14.
(50 + N)/6 = 14 → 50 + N = 84 → N = 34.
Check: 34 > 7 ✓. List: 2, 6, 7, 7, 28, 34. Median = 7, Mean = 84/6 = 14 = 2(7). ✓

**Case N ≤ 6:** Median = (6+7)/2 = 6.5. Mean = 13. N = 78 − 50 = 28. But 28 > 6, contradiction.
**Case 6 < N ≤ 7:** Median = (N+7)/2. (50+N)/6 = N+7 → 50+N = 6N+42 → 5N = 8 → N = 1.6. Contradiction.

Only N = 34 works.

ANSWER 2: E

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Problem 3:
Even digits: {0, 2, 4, 6, 8}. Four-digit numbers with all even digits:
- First digit: {2, 4, 6, 8} → 4 choices
- Other three digits: 5 choices each
- Total = 4 × 5 × 5 × 5 = 500

Divisibility by 4 depends on the last two digits. I check all 5 × 5 = 25 pairs of even digits:
For each tens digit (0, 2, 4, 6, 8), the units digits giving divisibility by 4 are:
- Tens 0: 00, 04, 08 (3 of 5)
- Tens 2: 20, 24, 28 (3 of 5)
- Tens 4: 40, 44, 48 (3 of 5)
- Tens 6: 60, 64, 68 (3 of 5)
- Tens 8: 80, 84, 88 (3 of 5)

Valid last-two-digit pairs = 15 out of 25.

Numbers divisible by 4 = 4 × 5 × 15 = 300.
Fraction = 300/500 = **3/5**.

ANSWER 3: D

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Problem 4:
Difference = 6.5% of $20 − 6% of $20 = (6.5% − 6%) × $20 = 0.5% × $20.
0.5% = 0.005, so 0.005 × 20 = $0.10.

ANSWER 4: B

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Problem 5:
Pair corresponding terms:
(1901 − 101) + (1902 − 102) + … + (1993 − 193)

Each pair equals 1800.
Number of terms: 1993 − 1901 + 1 = 93 terms (and 193 − 101 + 1 = 93 terms).
Total = 93 × 1800 = **167,400**.

ANSWER 5: A

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Problem 6:
The sign pattern is: +, −, −, +, +, −, −, +, …
Group into blocks of 4:
(1 − 2 − 3 + 4) + (5 − 6 − 7 + 8) + (9 − 10 − 11 + 12) + …

Each block: n − (n+1) − (n+2) + (n+3) = n − n − 1 − n − 2 + n + 3 = **0**.

Total terms: 1996. Number of blocks: 1996/4 = 499 complete blocks.
Sum = 499 × 0 = **0**.

ANSWER 6: C

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Problem 7:
Let B = number wearing both sunglasses and caps.
P(sunglasses | cap) = B / (people wearing caps) = B / 35 = 2/5.
B = 35 × (2/5) = 14.

Now find P(cap | sunglasses) = B / (people wearing sunglasses) = 14 / 50 = **7/25**.

ANSWER 7: B

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Problem 8:
Three-digit numbers with digit sum 26. Maximum digit sum = 9+9+9 = 27, so we must subtract 1 from one digit of 999:
Possible digit sets: (8,9,9), (9,8,9), (9,9,8) → numbers 899, 989, 998 (and permutations: 899, 989, 997... wait).

Actually the numbers are: 899, 989, 998 — no, let me list properly:
- (8,9,9) → 899, 989, 998... wait, (8,9,9) gives 899, 989, 998? No: digits are 8,9,9 so numbers are 899, 989, 998.
Wait: (8,9,9) → 899, 989, 998. But 998 has digits 9,9,8 which is the set (9,9,8). Let me just list all:
- Digits {8,9,9}: 899, 989, 998 — wait, 998 has digits 9,9,8. That IS {8,9,9}. So: 899, 989, 998.

Hmm, let me redo. The three numbers with digit sum 26 are:
- 899 (8+9+9=26)
- 989 (9+8+9=26)
- 998 (9+9+8=26)

Wait, that's only 3 numbers total (since two digits are the same, there are 3!/2! = 3 arrangements).

Even numbers must end in an even digit. Only 998 ends in 8 (even). 899 and 989 end in 9 (odd).

So exactly **1** even number.

ANSWER 8: A

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Problem 9:
Three-digit numbers a, b, c with a + b + c = 25, where 1 ≤ a ≤ 9, 0 ≤ b, c ≤ 9.

Enumerate by first digit a:
- a = 9: b + c = 16 → (7,9), (8,8), (9,7) → 979, 988, 997 (3 numbers)
- a = 8: b + c = 17 → (8,9), (9,8) → 889, 898 (2 numbers)
- a = 7: b + c = 18 → (9,9) → 799 (1 number)
- a ≤ 6: b + c ≥ 19, but max b + c = 18 → impossible

Total = 3 + 2 + 1 = **6**.

ANSWER 9: C

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Problem 10:
Exits 5 through 26 give 22 exits and 21 consecutive gaps. Total distance = 118 miles. Each gap ≥ 5 miles.

To maximize one gap, minimize the other 20 gaps to 5 miles each:
Max gap + 20(5) = 118
Max gap + 100 = 118
Max gap = **18** miles.

ANSWER 10: C

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Problem 11:
Five scores sorted: a ≤ b ≤ c ≤ d ≤ e.
- Mean = 90 → sum = 450
- Median = 91 → c = 91
- Mode = 94 → 94 appears most frequently (at least twice)

Since a ≤ b ≤ c = 91, neither a nor b can be 94. So 94 must appear among d and e. For 94 to be the mode, we need d = e = 94 (appearing twice).

Sum: a + b + 91 + 94 + 94 = 450
a + b = 450 − 279 = **171**

(We can verify valid scores 

 Problem 1:
We are given a set \(R\) of nine distinct integers, six of which are \(2,3,4,6,9,14\). We must determine how many different values are possible for the median of \(R\).

Because there are nine distinct integers, the median is the fifth smallest element when the set is listed in increasing order. Let the three unknown distinct integers be chosen freely (as long as they are not equal to any of the six given numbers). We examine two cases: when the median is one of the six known numbers and when it is one of the three unknown numbers.

**Case 1: The median is a known number.**
Let the median be \(m\). Of the six known numbers, let \(L\) be how many are less than \(m\). Then there are \(5-L\) known numbers greater than \(m\). To make \(m\) the fifth element, we need exactly four numbers smaller than \(m\) and four larger. Thus we need \(4-L\) unknown numbers below \(m\) and \(4-(5-L)=L-1\) unknown numbers above \(m\). This uses exactly \((4-L)+(L-1)=3\) unknowns, which is perfect. We need \(4-L\ge 0\) and \(L-1\ge 0\), i.e. \(1\le L\le 4\).

- \(m=2\): \(L=0\) → impossible.
- \(m=3\): \(L=1\) → need 3 unknowns \(<3\), 0 above. Possible (e.g. \(-1,0,1\)).
- \(m=4\): \(L=2\) → need 2 unknowns \(<4\), 1 above. Possible (e.g. \(0,1,5\)).
- \(m=6\): \(L=3\) → need 1 unknown \(<6\), 2 above. Possible (e.g. \(5,7,8\)).
- \(m=9\): \(L=4\) → need 0 unknowns \(<9\), 3 above. Possible (e.g. \(10,11,12\)).
- \(m=14\): \(L=5\) → need \(-1\) unknowns below; impossible.

So the known possible medians are \(3,4,6,9\).

**Case 2: The median is an unknown number \(x\).**
Then \(x\) itself uses one of the three unknown slots. We need four numbers below \(x\) and four above. With \(L\) known numbers below \(x\), we need \(4-L\) other unknowns below \(x\) and \(4-(6-L)=L-2\) other unknowns above \(x\). The total other unknowns used is \((4-L)+(L-2)=2\), which matches the two remaining unknowns. We need \(4-L\ge 0\) and \(L-2\ge 0\), so \(2\le L\le 4\).

- \(L=2\): known below are \(2,3\); known above are \(4,6,9,14\). Then \(3<x<4\), impossible for an integer.
- \(L=3\): known below are \(2,3,4\); known above are \(6,9,14\). Then \(4<x<6\), so \(x=5\). This works (e.g. add \(1,5,7\)).
- \(L=4\): known below are \(2,3,4,6\); known above are \(9,14\). Then \(6<x<9\), so \(x=7\) or \(x=8\). Both work (e.g. add \(7,10,11\) or \(8,10,11\)).

Thus the unknown possible medians are \(5,7,8\).

Collecting all possibilities: \(\{3,4,5,6,7,8,9\}\). That is **7** different values. The tempting answer of 4 counts only the known medians, and 2 or 14 are impossible because too many numbers would have to be forced to one side.

ANSWER 1: D

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Problem 2:
We insert a number \(N\) into the list \(2,6,7,7,28\) so that the new mean of the six numbers is twice the new median. Find \(N\).

The original sum is \(2+6+7+7+28=50\). After insertion, the sum is \(50+N\) and the mean is \(\frac{50+N}{6}\).

The median depends on the size of \(N\) relative to the ordered list. Because the list contains six numbers, the median is the average of the third and fourth values.

- If \(N\le 6\) (as an integer), the ordered list begins with \(N\) or \(2\), then the third and fourth terms are \(6\) and \(7\), giving median \(\frac{6+7}{2}=6.5\).
  Setting mean \(=2\times 6.5=13\): \(\frac{50+N}{6}=13\Rightarrow N=28\). But \(N=28\) does not satisfy \(N\le 6\), so this case yields no valid solution.

- If \(N\ge 7\) (integer), the third and fourth terms are both \(7\) (since the two \(7\)’s occupy those middle slots, or \(N=7\) makes three \(7\)’s), so the median is \(7\).
  Setting mean \(=2\times 7=14\): \(\frac{50+N}{6}=14\Rightarrow 50+N=84\Rightarrow N=34\).

Check: the list becomes \(2,6,7,7,28,34\). The median is \(\frac{7+7}{2}=7\). The mean is \(\frac{84}{6}=14\), which is indeed twice the median. The other choices fail: \(N=28\) gives mean \(13\) and median \(7\); \(N=14\) gives mean \(\frac{64}{6}\neq 14\).

ANSWER 2: E

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Problem 3:
We consider all positive four-digit integers whose digits are all even. We need the fraction of them that are divisible by \(4\).

Even digits are \(\{0,2,4,6,8\}\). For a four-digit number \(ABCD\):
- \(A\) (thousands) has 4 choices: \(\{2,4,6,8\}\).
- \(B,C,D\) each have 5 choices.

Total count: \(4\cdot 5\cdot 5\cdot 5 = 500\).

A number is divisible by \(4\) exactly when its last two digits form a multiple of \(4\). Here \(C\) and \(D\) are both even. Consider \(10C+D \pmod 4\). Since \(C\) is even, write \(C=2k\); then \(10C=20k\equiv 0\pmod 4\). Thus \(10C+D\equiv D\pmod 4\). So we simply need \(D\equiv 0\pmod 4\).

Among the even digits, those divisible by \(4\) are \(0,4,8\): **3 choices** for \(D\). The digits \(A,B,C\) are unrestricted (4, 5, and 5 choices).

Count of favorable numbers: \(4\cdot 5\cdot 5\cdot 3 = 300\).

The fraction is \(\frac{300}{500}=\frac{3}{5}\). One might mistakenly think the fraction is \(\frac12\) because half of all integers are divisible by \(4\), but the restriction to even digits changes the units-digit distribution.

ANSWER 3: D

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Problem 4:
Find the difference between a \(6.5\%\) sales tax and a \(6\%\) sales tax on a \(\$20\) item.

Method 1: Compute each tax.
- \(6.5\%\) of \(\$20 = 0.065\times 20 = \$1.30\).
- \(6\%\) of \(\$20 = 0.06\times 20 = \$1.20\).

Difference: \(\$1.30-\$1.20 = \$0.10\).

Method 2: The difference in rates is \(0.5\%\), and \(0.5\%\) of \(\$20 = 0.005\times 20 = \$0.10\).

The distractors correspond to misplacing the decimal (e.g. \(\$1\) would be \(5\%\) of \(\$20\)).

ANSWER 4: B

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Problem 5:
Evaluate \((1901+1902+\dots+1993)-(101+102+\dots+193)\).

Each sum has the same number of terms:
\(1993-1901+1 = 93\) terms, and \(193-101+1 = 93\) terms.

Pair the terms:
\((1901-101)+(1902-102)+\dots+(1993-193)\).

Each difference equals \(1800\). With 93 such pairs, the total is
\(93\times 1800 = 93\times 18\times 100 = 1674\times 100 = 167{,}400\).

A common error is miscounting the number of terms as 92, which would give \

Problem 1:
R is a set of nine distinct integers with six known elements: 2, 3, 4, 6, 9, 14. Since R has 9 elements, the median is the 5th element when sorted.

I need to determine which integers can be the 5th element. The median must have exactly 4 elements less than it and 4 elements greater than it.

**Checking each candidate:**

- **Median = 3**: Need 4 elements < 3. Known: {2}. Need 3 unknowns all < 3 (e.g., 1, 0, -1). Elements > 3: {4, 6, 9, 14} = 4. ✓
- **Median = 4**: Need 4 elements < 4. Known: {2, 3}. Need 2 unknowns < 4. Elements > 4: {6, 9, 14} + 1 unknown > 4. ✓
- **Median = 5**: Need 4 elements < 5. Known: {2, 3, 4}. Need 1 unknown < 5. Elements > 5: {6, 9, 14} + 1 unknown > 5. (5 is an unknown) ✓
- **Median = 6**: Need 4 elements < 6. Known: {2, 3, 4}. Need 1 unknown < 6. Elements > 6: {9, 14} + 2 unknowns > 6. ✓
- **Median = 7**: Need 4 elements < 7. Known: {2, 3, 4, 6} = 4. Elements > 7: {9, 14} + 2 unknowns > 7. (7 is an unknown) ✓
- **Median = 8**: Need 4 elements < 8. Known: {2, 3, 4, 6} = 4. Elements > 8: {9, 14} + 2 unknowns > 8. (8 is an unknown) ✓
- **Median = 9**: Need 4 elements < 9. Known: {2, 3, 4, 6} = 4. Elements > 9: {14} + 3 unknowns > 9. ✓

**Values that cannot be median:**
- 2: Need 4 elements < 2, but only 3 unknowns available. ✗
- 10-13: Already have 5 known elements less than these. ✗
- 14: Already have 5 known elements less than 14. ✗

Possible median values: {3, 4, 5, 6, 7, 8, 9} → **7 values**

ANSWER 1: D

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Problem 2:
List: 2, 6, 7, 7, 28. Insert N so that mean = 2 × median.

Sum of original list = 50, so new mean = (50 + N)/6.

**Finding the median based on N:**
- If N > 7: Sorted list is 2, 6, 7, 7, 28, N (or 2, 6, 7, 7, N, 28 if 7 < N < 28). Median = (7+7)/2 = 7.
- If N ≤ 6: Median = (6+7)/2 = 6.5
- If 6 < N ≤ 7: Median = (N+7)/2

**Solving for N > 7:**
(50 + N)/6 = 2(7) = 14
50 + N = 84
N = 34

Verification: List = 2, 6, 7, 7, 28, 34. Mean = 84/6 = 14. Median = 7. ✓

ANSWER 2: E

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Problem 3:
Four-digit integers with all even digits. Even digits: {0, 2, 4, 6, 8}.

**Total count:** First digit: 4 choices (2,4,6,8), others: 5 choices each → 4 × 5³ = 500

**Divisible by 4:** Last two digits must form a number divisible by 4.

Listing all 25 pairs of even last digits and checking divisibility by 4:
- 00✓, 02✗, 04✓, 06✗, 08✓
- 20✓, 22✗, 24✓, 26✗, 28✓
- 40✓, 42✗, 44✓, 46✗, 48✓
- 60✓, 62✗, 64✓, 66✗, 68✓
- 80✓, 82✗, 84✓, 86✗, 88✓

That's 15 valid pairs.

**Count divisible by 4:** 4 × 5 × 15 = 300

**Fraction:** 300/500 = 3/5

ANSWER 3: D

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Problem 4:
6.5% tax on $20 = 0.065 × 20 = $1.30
6% tax on $20 = 0.06 × 20 = $1.20

Difference = $1.30 - $1.20 = $0.10

ANSWER 4: B

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Problem 5:
(1901 + 1902 + ... + 1993) − (101 + 102 + ... + 193)

Both sums have 93 terms each.

First sum = 93 × (1901 + 1993)/2 = 93 × 1947
Second sum = 93 × (101 + 193)/2 = 93 × 147

Difference = 93 × (1947 − 147) = 93 × 1800 = 167,400

ANSWER 5: A

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Problem 6:
1 − 2 − 3 + 4 + 5 − 6 − 7 + 8 + ... + 1993 − 1994 − 1995 + 1996

The pattern repeats every 4 terms: +, −, −, +

Group k: (4k−3) − (4k−2) − (4k−1) + 4k = 4k−3−4k+2−4k+1+4k = 0

Each group of 4 sums to 0.

1996 ÷ 4 = 499 complete groups.

Total = 499 × 0 = 0

ANSWER 6: C

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Problem 7:
50 people wear sunglasses, 35 wear caps.

P(sunglasses | cap) = 2/5 means: (both)/(caps) = 2/5

Both/35 = 2/5 → Both = 14

P(cap | sunglasses) = (both)/(sunglasses) = 14/50 = 7/25

ANSWER 7: B

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Problem 8:
3-digit numbers with digit-sum 26 that are even.

For number abc: a + b + c = 26, c is even, a ≥ 1, all digits ≤ 9.

- c = 8: a + b = 18 → only (9,9) → 998 ✓
- c = 6: a + b = 20 → impossible (max 9+9=18)
- c = 4, 2, 0: a + b ≥ 22 → impossible

Only 1 number: 998

ANSWER 8: A

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Problem 9:
Three-digit numbers with digit-sum 25.

a + b + c = 25, a ≥ 1, all digits ≤ 9.

- a = 9: b + c = 16 → (7,9), (8,8), (9,7) → 3 numbers
- a = 8: b + c = 17 → (8,9), (9,8) → 2 numbers
- a = 7: b + c = 18 → (9,9) → 1 number
- a ≤ 6: b + c ≥ 19 → impossible

Total = 3 + 2 + 1 = 6

ANSWER 9: C

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Problem 10:
Distance from exit 5 to exit 26 is 118 miles. There are 21 gaps between consecutive exits (5→6, 6→7, ..., 25→26).

Each gap ≥ 5 miles. To maximize one gap, minimize all others.

If 20 gaps are minimized at 5 miles each: 20 × 5 = 100 miles.

Maximum single gap = 118 − 100 = 18 miles.

ANSWER 10: C

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Problem 11:
Five scores: mean = 90, median = 91, mode = 94.

Sum = 5 × 90 = 450

Sorted: a ≤ b ≤ c ≤ d ≤ e
- c = 91 (median)
- Mode = 94, so 94 appears at least twice. Since c = 91, the 94s must be d and e.

a + b + 91 + 94 + 94 = 450
a + b = 171

ANSWER 11: B

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Problem 12:
Using +, −, × each exactly once in 5 __ 4 __ 6 __ 3:

| Operations | Result |
|------------|--------|
| 5 + 4 − 6 × 3 | 5 + 4 − 18 = −9 |
| 5 + 4 × 6 − 3 | 5 + 24 − 3 = 26 |
| 5 − 4 + 6 × 3 | 5 − 4 + 18 = 19 |
| 5 − 4 × 6 + 3 | 5 − 24 + 3 = −16 |
| 5 × 4 + 6 − 3 | 20 + 6 − 3 = 23 |
| 5 × 4 − 6 + 3 | 20 − 6 + 3 = 17 |

From choices {9, 10, 15, 16, 19}, only **19** is achievable.

ANSWER 12: E

**Problem 1:**  
We have nine distinct integers, six of which are known: 2, 3, 4, 6, 9, 14.  
The median of a nine‑element set is the 5th smallest number.  

Let the unknown integers be \(x,y,z\) (all distinct and distinct from the known six).  
When the numbers are sorted, the median is the 5th term.  

For a known number \(k_i\) (the \(i\)‑th smallest of the known six) to be the median, we need exactly four numbers smaller than it.  
If \(i-1\) known numbers are smaller, we need \(L\) unknown numbers smaller with  
\((i-1)+L=4\Rightarrow L=5-i\).  
Since we have only three unknowns, \(L\ge0\) gives \(i\le5\); also \(L\le3\) gives \(i\ge2\).  
Thus \(i=2,3,4,5\) → medians could be 3, 4, 6, 9.

Can the median be an unknown integer?  
Let the median be an unknown \(u\).  
If \(u\) lies between the known numbers, we need exactly four numbers below it.  
With two other unknowns, the possibilities are:
- \(u\) between 4 and 6 → \(u=5\);
- \(u\) between 6 and 9 → \(u=7\) or \(u=8\).

All three of these can be realized while keeping the numbers distinct.  
Thus the possible median values are \(\{3,4,5,6,7,8,9\}\), a total of **7** values.

**Answer:** 7 → Choice **D**.  

ANSWER 1: D  

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**Problem 2:**  
We start with the list 2, 6, 7, 7, 28 and insert a number \(N\).  
Let the six numbers be sorted. The median of six numbers is the average of the 3rd and 4th smallest values.

Let the sorted list after insertion be \(a_1\le a_2\le a_3\le a_4\le a_5\le a_6\).  
The median is \(\dfrac{a_3+a_4}{2}\).  
The mean is \(\dfrac{\text{sum}}{6}\).  
Given that the mean equals twice the median:
\[
\frac{\text{sum}}{6}=2\cdot\frac{a_3+a_4}{2}=a_3+a_4 .
\]
Hence \(\text{sum}=6(a_3+a_4)\).

Consider where \(N\) can fall:

*If \(N\le2\):* \(a_3=6,\ a_4=7\) → median \(=6.5\).  
Sum \(=N+50\). Equation gives \(N=28\), contradicting \(N\le2\).

*If \(2<N\le6\):* \(a_3=6,\ a_4=7\) → same median \(=6.5\).  
Again \(N=28\), impossible.

*If \(6<N\le7\):* only integer possibility \(N=7\).  
List becomes 2, 6, 7, 7, 7, 28 → median \(=7\).  
Sum \(=57\) → mean \(=9.5\neq14\).

*If \(7<N<28\):* \(a_3=7,\ a_4=7\) → median \(=7\).  
Sum \(=N+50\). Equation \(N+50=84\) gives \(N=34\), which is not \(<28\).

*If \(N=28\):* median still \(=7\). Sum \(=78\) → mean \(=13\neq14\).

*If \(N>28\):* \(a_3=7,\ a_4=7\) → median \(=7\).  
Equation gives \(N+50=84\) → \(N=34\). This works (34 > 28).

Thus the only possible value is **34**.

**Answer:** 34 → Choice **E**.  

ANSWER 2: E  

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**Problem 3:**  
A four‑digit integer with all digits even can use digits \{0,2,4,6,8\} for each place, except the thousands digit cannot be 0.  

Number of such integers: \(4\times5\times5\times5=500\).

A number is divisible by 4 iff its last two digits form a number divisible by 4.  
List all two‑digit endings with even digits (25 possibilities).  

Checking each yields 15 endings that are multiples of 4:  
00, 04, 08, 20, 24, 28, 40, 44, 48, 60, 64, 68, 80, 84, 88.

Thus the proportion is \(\frac{15}{25}=\frac{3}{5}\).

**Answer:** \(\frac{3}{5}\) → Choice **D**.  

ANSWER 3: D  

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**Problem 4:**  
6 % tax on \$20 = \(0.06\times20 = \$1.20\).  
6.5 % tax on \$20 = \(0.065\times20 = \$1.30\).  

Difference = \$1.30 − \$1.20 = **\$0.10**.

**Answer:** \$0.10 → Choice **B**.  

ANSWER 4: B  

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**Problem 5:**  
Sum from 1901 to 1993 (inclusive):  
Number of terms = \(1993-1901+1 = 93\).  
Sum = \(\frac{(1901+1993)\times93}{2}= \frac{3894\times93}{2}=181\,071\).

Sum from 101 to 193 (inclusive):  
Number of terms = \(193-101+1 = 93\).  
Sum = \(\frac{(101+193)\times93}{2}= \frac{294\times93}{2}=13\,671\).

Difference = \(181\,071-13\,671 = 167\,400\).

**Answer:** 167 400 → Choice **A**.  

ANSWER 5: A  

---

**Problem 6:**  
The sign pattern for the numbers \(n\) is:
- \(n\equiv1\pmod4\) → +,
- \(n\equiv2\pmod4\) → −,
- \(n\equiv3\pmod4\) → −,
- \(n\equiv0\pmod4\) → +.

Thus each block of four consecutive integers \(\{4k+1,4k+2,4k+3,4k+4\}\) contributes  
\((4k+1)-(4k+2)-(4k+3)+(4k+4)=0\).

From 1 through 1996 there are \(1996/4=499\) complete blocks, so the total sum is **0**.

**Answer:** 0 → Choice **C**.  

ANSWER 6: C  

---

**Problem 7:**  
Let \(S\) = people wearing sunglasses, \(|S|=50\).  
Let \(C\) = people wearing caps, \(|C|=35\).  
Let \(x=|S\cap C|\).

Given \(P(S\mid C)=\frac{x}{35}= \frac{2}{5}\).  
Thus \(x = \frac{2}{5}\times35 = 14\).

We need \(P(C\mid S)=\frac{x}{50}= \frac{14}{50}= \frac{7}{25}\).

**Answer:** \(\frac{7}{25}\) → Choice **B**.  

ANSWER 7: B  

---

**Problem 8:**  
A three‑digit number with digit sum 26 must use digits that sum to 26.  
The only triple of single‑digit numbers with sum 26 is \(\{9,9,8\}\).

The number is even only if its units digit is 8.  
Among the permutations of 9, 9, 8, only 998 ends in 8.  

Thus exactly **1** such number exists.

**Answer:** 1 → Choice **A**.  

ANSWER 8: A  

---

**Problem 9:**  
Let the digits be \(a,b,c\) (hundreds, tens, units) with  
\(1\le a\le9,\ 0\le b,c\le9,\ a+b+c=25\).

Set \(a'=a-1\) (so \(0\le a'\le8\)). Then  
\(a'+b+c=24\).

Count unrestricted non‑negative solutions: \(\binom{24+3-1}{2}= \binom{26}{2}=325\).

Subtract solutions where a digit exceeds its bound:
- \(a'\ge9\): treat \(a''=a'-9\) → \(a''+b+c=15\) → \(\binom{17}{2}=136\).
- \(b\ge10\): \(b'=b-10\) → \(a'+b'+c=14\) → \(\binom{16}{2}=120\).
- \(c\ge10\): similarly 120.

Intersections:
- \(a'\ge9,\ b\ge10\): \(a''+b'+c=5\) → \(\binom{7}{2}=21\).
- \(a'\ge9,\ c\ge10\): 21.
- \(b\ge10,\ c\ge10\): \(a'+b'+c'=4\) → \(\binom{6}{2}=15\).

No triple intersection (would require sum –5).

By inclusion–exclusion, invalid = \(136+120+120-21-21-15 = 319\).  
Valid = \(325-319 = 6\).

**Answer:** 6 → Choice **C**.  

ANSWER 9: C  

---

**Problem 10:**  
Exits numbered 5 through 26 give 22 exits, hence 21 intervals.  
Let the distances between consecutive exits be \(d_1,\dots,d_{21}\).  
We know \(\sum d_i = 118\) miles, and each \(d_i\ge5\).

To maximize the largest interval, make the othe

Problem 1:  
We are given a set $ R $ of nine distinct integers, six of which are $ 2, 3, 4, 6, 9, 14 $. The median of a set of nine numbers is the fifth number when sorted. To find the number of possible values for the median, we consider the possible placements of the three unknown numbers relative to the known ones. The known numbers sorted are $ 2, 3, 4, 6, 9, 14 $. The three unknown numbers can be placed in various positions to shift the median. The possible medians occur when the additional numbers are placed such that the fifth number in the sorted list can be $ 4, 5, 6, 7, 8, $ or $ 9 $. This gives six possible values for the median.  
ANSWER 1: D (corrected to C? Wait, no: the possible medians are 4, 5, 6, 7, 8, 9 — that’s 6 values, so C)  
**Wait, let's re-evaluate:** The known numbers are 2,3,4,6,9,14. Adding three numbers. The median is the 5th number. To get 4 as median: three numbers ≤4, but only 2,3,4 are available — but they are distinct, so we can add numbers like 1,5? Wait no, the three unknown numbers can be placed in different positions. For example:  
- If all three unknown numbers are ≤4: but only three such numbers exist (2,3,4) but they are already in the set. So we cannot add more. Thus, the unknown numbers must be placed in other positions.  
Actually, the unknown numbers can be:  
- All less than 2: median could be 4 if three numbers are added <4, but only 2,3,4 are already present. We cannot add numbers less than 2 without repeating? Wait, the set has distinct integers. So we can add numbers like 1,0,-1, but the problem doesn’t restrict to positive. However, the known numbers are positive. But the problem says "distinct integers", so negatives are allowed. But the median values: if we add numbers less than 2, the sorted order becomes [new numbers, 2,3,4,6,9,14]. With three new numbers, the sorted list has 9 numbers. The fifth is the median. If we add three numbers less than 2, the sorted list is [a,b,c,2,3,4,6,9,14] — no, wait, adding three numbers less than 2: the sorted list would be [a,b,c,2,3,4,6,9,14] — but a,b,c are less than 2, so the first three, then 2 is fourth, 3 is fifth? Wait, no: the original known numbers are 2,3,4,6,9,14. Adding three numbers, say x,y,z. The full sorted list will be the combination. To get median=4, we need the fifth number to be 4. So in the sorted list of 9, the fifth must be 4. That requires that four numbers are ≤4 and four are ≥4? No: the fifth is the middle. For the fifth to be 4, we need at least five numbers ≤4? No: in a sorted list of 9, the fifth is the median. For the fifth to be 4, we need at least five numbers ≤4? Actually, no: the fifth number is the one where four are less and four are greater? No: in a sorted list of 9, the fifth is the median, meaning four are ≤ median and four are ≥ median. But since numbers can be equal, but here all distinct. So for 4 to be the median, we need exactly four numbers less than 4 and four numbers greater than 4? But 4 is in the set. So the sorted list must have four numbers <4, then 4, then four numbers >4. But the known numbers include 2,3,4,6,9,14. So 2,3 are <4, and 6,9,14 are >4. So currently, without unknowns, we have two <4 and three >4. Adding three unknowns. To have four <4 and four >4, we need to add two numbers <4 and one number >4? But then the sorted list would be [two new <4, 2,3,4, then the new >4, then 6,9,14] — but that’s only three numbers after 4? Wait, no: the sorted list would be [new <4 (two), 2,3,4, new >4, 6,9,14] — that’s 9 numbers? 2+3+1+3=9? Wait, no: the two new <4, then 2,3,4 (three numbers), then the new >4, then 6,9,14 (three numbers). So total: 2+3+1+3=9. The fifth number is 4. So yes, we can achieve median=4 by adding two numbers <4 and one number >4. Similarly, we can achieve:  
- Median=5: by adding numbers such that the fifth number is 5. But 5 is not in the set. So we need the fifth number to be 5. This requires that four numbers are <5 and four are >5. But 5 is not present. So we need to add 5? But the set must consist of distinct integers, and 5 is not in the given. So we can add 5. Then the sorted list would include 5. For example, add 5, and two other numbers. To have four <5 and four >5: currently, without unknowns, numbers <5: 2,3,4 (three), numbers >5: 6,9,14 (three). So to have four <5, we need to add one number <5. To have four >5, we need to add one number >5. But we are adding three numbers. So we can add 5, and say 1 and 10. Then sorted: 1,2,3,4,5,6,9,10,14. Median is 5. So median=5 is achievable. Similarly, median=6: we can add numbers such that the fifth is 6. Currently, numbers <6: 2,3,4 (three), numbers >6: 9,14 (two). So to have four <6 and four >6, we need to add one number <6 and two numbers >6. For example, add 5,10,15. Then sorted: 2,3,4,5,6,9,10,14,15. Median=6. Similarly, median=7: add numbers so fifth is 7. Currently, <7: 2,3,4,6 (four), >7: 9,14 (two). So to have four <7 and four >7, we need to add zero <7 (since already four) and two >7. But we are adding three numbers. So we can add 7,8,15. Then sorted: 2,3,4,6,7,8,9,14,15. Median=7. Similarly, median=8: add 8, and two numbers, say 10,11. Then sorted: 2,3,4,6,8,9,10,11,14. Median=8? The fifth is 8. But to have four <8 and four >8: currently, without unknowns, <8: 2,3,4,6 (four), >8:9,14 (two). So to have four >8, we need to add two numbers >8. But we are adding three numbers. So we can add 8,9,10? But 9 is already there. Add 8,10,11. Then sorted: 2,3,4,6,8,9,10,11,14. The fifth is 8. So yes. Similarly, median=9: add numbers so fifth is 9. Currently, <9: 2,3,4,6 (four), >9:14 (one). So to have four >9, we need to add three numbers >9. But we are adding three numbers. So add 10,11,12. Then sorted: 2,3,4,6,9,10,11,12,14. Median=9. So possible medians: 4,5,6,7,8,9. That’s six values.  
ANSWER 1: C  

Problem 2:  
We are given the list 2,6,7,7,28. A number $ N $ is added. The mean becomes twice the median. The original list has five numbers. After adding $ N $, there are 

### Problem 1:
We have a set of 9 distinct integers, so the median (middle value when sorted) is the 5th element. The 6 known numbers sorted are [2,3,4,6,9,14], with 3 additional distinct integers to add. To find possible median values:
- The minimum possible median is 3: add 3 numbers smaller than 2, sorted set becomes [a,b,c,2,3,4,6,9,14], 5th term=3.
- The maximum possible median is 9: add 3 numbers larger than 14, sorted set becomes [2,3,4,6,9,14,x,y,z], 5th term=9.
- All integers between 3 and 9 are valid medians: 3,4,5,6,7,8,9, for 7 total possible values.
Tempting wrong choices: any count less than 7, from missing intermediate medians that can be formed by adding integers between the known values.
ANSWER 1: D

---

### Problem 2:
Original list sum: 2+6+7+7+28=50. Adding N gives a list of 6 elements, sum=50+N, mean=(50+N)/6. The median of 6 elements is the average of the 3rd and 4th sorted terms. The problem states mean=2×median, so 50+N=12×median.
- If N>28, sorted list is [2,6,7,7,28,N], median=(7+7)/2=7. Then 50+N=12×7=84 → N=34, which fits.
Testing other options confirms no other N satisfies the condition.
ANSWER 2: E

---

### Problem 3:
A number is divisible by 4 if its last two digits form a number divisible by 4. For 4-digit numbers with all even digits:
- Total count: 4 (non-zero thousands digits:2,4,6,8) ×5×5×5 (all even digits for other places)=500.
- Valid last two-digit pairs (both even, divisible by 4): For any even tens digit, 10×even is divisible by 4, so only units digits 0,4,8 work, giving 3 valid units digits per tens digit: 5×3=15 valid pairs.
- Fraction: (4×5×15)/500=300/500=3/5.
ANSWER 3: D

---

### Problem 4:
The tax rate difference is 6.5%-6%=0.5%. 0.5% of $20 is 0.005×20=$0.10.
ANSWER 4: B

---

### Problem 5:
Both sums have the same number of terms: 1993-1901+1=93 terms, 193-101+1=93 terms. Subtract term-by-term: (1901-101)+(1902-102)+...+(1993-193)=1800×93=167400.
ANSWER 5: A

---

### Problem 6:
Group terms in sets of 4: (1-2-3+4)+(5-6-7+8)+... each group sums to 0. There are 1996/4=499 groups, total sum=0.
ANSWER 6: C

---

### Problem 7:
Number of people with both caps and sunglasses: 35 cap-wearers × 2/5=14. Probability a sunglass-wearer has a cap:14/50=7/25.
ANSWER 7: B

---

### Problem 8:
An even 3-digit number with digit sum 26 must have an even units digit. The only possible even units digit is 8 (any smaller even digit leaves a sum of the first two digits >18, impossible). The only number is 998, so 1 valid number.
ANSWER 8: A

---

### Problem 9:
Three-digit numbers with digit sum 25 are all permutations of (9,9,7) and (9,8,8). There are 3 permutations of (9,9,7) and 3 permutations of (9,8,8), for 6 total numbers.
ANSWER 9: C

---

### Problem 10:
Between exit 5 and 26, there are 21 gaps. To maximize one gap, minimize the other 20 gaps at 5 miles each: 20×5=100 miles. Maximum gap=118-100=18 miles.
ANSWER 10: C

---

### Problem 11:
Total sum of 5 scores=5×90=450. Median=91, so sorted scores [a,b,91,c,d]. Mode=94, so 94 must appear at least twice, so c=d=94, sum c+d=188. Sum of the two lowest scores=450-91-188=171.
ANSWER 11: B

---

### Problem 12:
Using +,-,* exactly once, one valid calculation: 5-4+6×3=1+18=19, which uses all three operations and matches option E.
ANSWER 12: E

### Problem 1:
We need to find the number of possible median values for a set of 9 distinct integers, 6 of which are 2, 3, 4, 6, 9, 14. The median of 9 numbers is the 5th element when sorted in ascending order. We add 3 distinct integers (not in the given set) and analyze possible medians:
1.  **Case 1: All 3 new numbers < 2**: Sorted list is [a,b,c,2,3,4,6,9,14], median = 3.
2.  **Case 2: 2 new numbers < 2, 1 new number ≥4**: Sorted list is [a,b,2,3,4,...], median =4.
3.  **Case 3: 1 new number <2**:
    - If we add 5 (the only integer between 4 and 6): Sorted list is [a,2,3,4,5,6,...], median=5.
    - If we do not add 5: Sorted list is [a,2,3,4,6,...], median=6.
4.  **Case 4: 0 new numbers <2 (all new numbers ≥5)**:
    - Add 5: Sorted list [2,3,4,5,6,...], median=6.
    - Add a number between 6 and9 (7 or 8): Sorted list [2,3,4,6,7,...] or [2,3,4,6,8,...], median=7 or 8.
    - Add all new numbers ≥9: Sorted list [2,3,4,6,9,...], median=9.
The only impossible medians are 2 (needs 5 numbers ≤2, only 4 max) and values >9 (existing numbers already have 5 elements ≤9). The possible medians are 3,4,5,6,7,8,9: 7 values.
ANSWER 1: D
---
### Problem 2:
Original list: [2,6,7,7,28], sum = 2+6+7+7+28=50. Insert N to make 6 elements. The median of 6 elements is the average of the 3rd and 4th sorted elements. The mean is $\frac{50+N}{6}$, which equals twice the median:
$$\frac{50+N}{6} = 2 \times \text{median} \implies 50 + N = 12 \times \text{median} \implies N = 12 \times \text{median} -50.$$
Test possible median values from the sorted list with N:
- If N ≤7: Sorted list has 3rd=7, 4th=7, median=7. Then $N=12*7 -50=84-50=34$, which is >7, valid.
- Other median values (e.g., 6.5) give N=28, which does not fit the required sorted position.
Verify N=34: Sorted list [2,6,7,7,28,34], median=(7+7)/2=7, mean=(50+34)/6=14, which is twice 7.
ANSWER 2: E
---
### Problem 3:
Positive 4-digit numbers with all even digits:
- Thousands place: 2,4,6,8 (4 choices, no 0).
- Hundreds, tens, units: 0,2,4,6,8 (5 choices each).
Total numbers: $4 \times 5 \times 5 \times5=500$.
A number is divisible by 4 iff its last two digits form a number divisible by 4. For even tens digits (0,2,4,6,8), a two-digit number $10a + b$ is divisible by 4 iff $b$ (units digit) is 0,4,8 (since $10a \equiv 0 \mod4$ for even a). So valid endings: 5 (tens choices) ×3 (units choices)=15.
Total valid numbers: $4 \times5 \times5 \times3=300$. Fraction: $\frac{300}{500}=\frac{3}{5}$.
ANSWER 3: D
---
### Problem 4:
Difference between 6.5% and 6% tax on $20:
$$0.065*20 - 0.06*20 = (0.065-0.06)*20=0.005*20=0.10.$$
ANSWER 4: B
---
### Problem 5:
Both sums have 93 terms (1993-1901+1=93, 193-101+1=93). Each term in the first sum is 1800 more than the corresponding term in the second sum:
Total difference = $93 \times 1800 = 167400$.
ANSWER 5: A
---
### Problem 6:
The sign pattern repeats every 4 terms: $+ - - +$, and each block of 4 sums to 0:
$$(1-2-3+4)+(5-6-7+8)+\dots+(1993-1994-1995+1996)=0.$$
1996 is divisible by 4, so there are 499 such blocks, total sum 0.
ANSWER 6: C
---
### Problem 7:
Let $S$ = sunglass wearers ($|S|=50$), $C$ = cap wearers ($|C|=35$). $P(S|C)=\frac{|S∩C|}{|C|}=\frac{2}{5}$, so $|S∩C|=\frac{2}{5}*35=14$.
The probability a sunglass wearer wears a cap is $P(C|S)=\frac{|S∩C|}{|S|}=\frac{14}{50}=\frac{7}{25}$.
ANSWER 7: B
---
### Problem 8:
3-digit numbers with digit sum 26: maximum digit sum is 27 (999), so 26 is 1 less than maximum, meaning digits are permutations of (9,9,8). The only even permutation is 998 (units digit 8). So 1 number.
ANSWER 8: A
---
### Problem 9:
3-digit numbers with digit sum 25: 25 is 2 less than 27, so digits are permutations of (9,9,7) and (9,8,8):
- Permutations of (9,9,7): 997,979,799 (3 numbers).
- Permutations of (9,8,8): 988,898,889 (3 numbers).
Total 6 numbers.
ANSWER 9: C
---
### Problem 10:
Between exit 5 and 26, there are $26-5=21$ gaps, total distance 118 miles. Each gap ≥5 miles. To maximize one gap, minimize the other 20: $20*5=100$, so maximum gap = $118-100=18$.
ANSWER 10: C
---
### Problem 11:
Five sorted scores: $a≤b≤c≤d≤e$. Median $c=91$, mean 90, so total sum $5*90=450$, so $a+b+d+e=450-91=359$. Mode is 94, so 94 appears at least twice. Since $c=91$, 94 can only be in $d$ and $e$, so $d=e=94$. Then $a+b=359-94-94=171$. This sum is fixed (valid examples exist, e.g., 81+90=171).
ANSWER 11: B
---
### Problem 12:
Permute +, -, × in 5 __4 __6 __3, using standard order of operations (multiply first):
- $5 -4 +6×3 = 5-4+18=19$, which is an option.
All other permutations give values not in the choices.
ANSWER 12: E

I'll solve each problem step by step.

---

**Problem 1:**

We have 6 elements: {2, 3, 4, 6, 9, 14}, and need to add 3 more distinct integers to make 9 total.

The median of 9 numbers is the 5th element when arranged in order.

Currently we have 6 numbers. By strategically placing 3 new numbers, we can shift which value becomes the median.

Let me consider what values can be the median (5th position):
- If we add 3 numbers all ≤ 2, the sorted list has our 6 original numbers in positions 4-9, so median is 4
- If we add numbers strategically between and around the original set, we can make the median be: 2, 3, 4, 6, 9, 14, or a new number
- If we add 3 numbers all ≥ 14, the sorted list has our 6 original numbers in positions 1-6, so median is 9

Testing systematically:
- Median = 2: add three numbers < 2 (e.g., -2, -1, 0) ✓
- Median = 3: add two < 2, one between 3 and 4 ✓
- Median = 4: add three < 2, or arranged appropriately ✓
- Median = 6: add one or two strategically ✓
- Median = 9: add three > 14, or arranged appropriately ✓
- Median = 14: add three > 14 ✓
- Median = a new value between existing numbers ✓

The possible medians are: 2, 3, 4, 6, 9, 14, and any new integer we insert. We can get 7 different values.

**ANSWER 1: D**

---

**Problem 2:**

Original list: 2, 6, 7, 7, 28. Sum = 50.

When N is inserted, we have 6 numbers. Let's find the median of 6 numbers (average of 3rd and 4th when sorted).

Sum of all 6 numbers = 50 + N
Mean = (50 + N)/6

We need: Mean = 2 × Median

Let me test by considering where N falls:
- If N ≤ 2: sorted list is N, 2, 6, 7, 7, 28. Median = (6+7)/2 = 6.5. Mean = 2×6.5 = 13, so 50+N = 78, N = 28. Contradiction.
- If 2 < N ≤ 6: sorted list is 2, N, 6, 7, 7, 28. Median = (6+7)/2 = 6.5. Mean = 13, so N = 28. Contradiction.
- If 6 < N ≤ 7: sorted list is 2, 6, N, 7, 7, 28. Median = (N+7)/2. Mean = 2×(N+7)/2 = N+7. So (50+N)/6 = N+7, giving 50+N = 6N+42, so 5N = 8, N = 1.6. Not in range.
- If 7 < N ≤ 28: sorted list is 2, 6, 7, N, 7, 28 or 2, 6, 7, 7, N, 28. Median = (7+N)/2. Mean = 2×(7+N)/2 = 7+N. So (50+N)/6 = 7+N, giving 50+N = 42+6N, so 5N = 8, N = 1.6. Not valid.
- If N > 28: sorted list is 2, 6, 7, 7, 28, N. Median = (7+7)/2 = 7. Mean = 14. So 50+N = 84, N = 34. ✓

**ANSWER 2: E**

---

**Problem 3:**

Four-digit numbers with all even digits: first digit ∈ {2,4,6,8}, others ∈ {0,2,4,6,8}.

Total count: 4 × 5 × 5 × 5 = 500

For divisibility by 4, the last two digits must form a number divisible by 4.

Last two digits from {00, 02, 04, 06, 08, 20, 22, 24, 26, 28, 40, 42, 44, 46, 48, 60, 62, 64, 66, 68, 80, 82, 84, 86, 88}

Divisible by 4: 00, 04, 08, 20, 24, 28, 40, 44, 48, 60, 64, 68, 80, 84, 88 (15 out of 25)

Count divisible by 4: 4 × 5 × 15 = 300

But wait, the second digit is 5 choices, third is which determines divisibility by 4.

Actually: 4 × 5 × (number of even two-digit combos divisible by 4) = 4 × 5 × 15 = 300

Hmm, let me recalculate. First digit: 4 choices. Second digit: 5 choices. Last two digits must be divisible by 4: check all pairs from {0,2,4,6,8}×{0,2,4,6,8} giving 15 valid pairs.

Count = 4 × 5 × 15 = 300. Fraction = 300/500 = 3/5.

**ANSWER 3: D**

---

**Problem 4:**

6.5% tax on $20: 0.065 × 20 = $1.30
6% tax on $20: 0.06 × 20 = $1.20
Difference: $1.30 - $1.20 = $0.10

**ANSWER 4: B**

---

**Problem 5:**

Sum from 1901 to 1993: This is 93 consecutive integers starting at 1901.
Sum = (1901 + 1993)/2 × 93 = 1947 × 93 = 181,071

Sum from 101 to 193: This is 93 consecutive integers starting at 101.
Sum = (101 + 193)/2 × 93 = 147 × 93 = 13,671

Difference = 181,071 - 13,671 = 167,400

**ANSWER 5: A**

---

**Problem 6:**

Pattern: 1 - 2 - 3 + 4 + 5 - 6 - 7 + 8 + 9 - 10 - 11 + ...

Group by 4s: (1-2-3+4) + (5-6-7+8) + (9-10-11+12) + ... + (1993-1994-1995+1996)

Each group of 4: (4k+1) - (4k+2) - (4k+3) + (4k+4) = 4k+1-4k-2-4k-3+4k+4 = 0

From 1 to 1996, we have 1996/4 = 499 complete groups, each summing to 0.

Total = 0

**ANSWER 6: C**

---

**Problem 7:**

Let x = people wearing both sunglasses and caps.

P(sunglasses | cap) = x/35 = 2/5, so x = 14.

P(cap | sunglasses) = 14/50 = 7/25.

**ANSWER 7: B**

---

**Problem 8:**

Three-digit even numbers with digit sum 26.

For an even number, last digit ∈ {0, 2, 4, 6, 8}.

If last digit = 0: first two digits sum to 26. Max is 9+9=18. Impossible.
If last digit = 2: first two digits sum to 24. Only 9+9=18 < 24. Impossible.
If last digit = 4: first two digits sum to 22. Impossible (max 18).
If last digit = 6: first two digits sum to 20. Impossible.
If last digit = 8: first two digits sum to 18. This requires 9+9=18. Number: 998. ✓

Only 1 number.

**ANSWER 8: A**

---

**Problem 9:**

Three-digit numbers abc where a+b+c=25, with a ∈ {1,...,9}, b,c ∈ {0,...,9}.

a+b+c = 25 means b+c = 25-a.

For valid solutions: 0 ≤ b+c ≤ 18, so 7 ≤ a ≤ 9.

- a=7: b+c=18. Solutions: (9,9) → 1 way
- a=8: b+c=17. Solutions: (8,9), (9,8) → 2 ways
- a=9: b+c=16. Solutions: (7,9), (8,8), (9,7) → 3 ways

Total = 1 + 2 + 3 = 6

**ANSWER 9: C**

---

**Problem 10:**

Exits from 5th to 26th = 26-5 = 21 segments between consecutive exits.

Each segment ≥ 5 miles. Total distance = 118 miles.

If 20 segments are 5 miles each: 20 × 5 = 100 miles.
Remaining for the largest segment: 118 - 100 = 18 miles.

**ANSWER 10: C**

---

**Problem 11:**

Five scores: mean = 90 (sum = 450), median = 91, mode = 94.

Arranged in order: a, b, c, d, e where c = 91 (median).

Mode = 94 means 94 appears at least twice. Since median is 91, we must have 94 appearing in {d, e}.

Case 1: d = e = 94. Then a + b + 91 + 94 + 94 = 450, so a + b = 171.

Check consistency: we need a ≤ b ≤ 91 and mode is 94 (appears twice). If a or b equals 94, mode wouldn't be unique. So a, b < 91 works. ✓

**ANSWER 11: B**

---

**Problem 12:**

We need to place +, −, × in three blanks: 5 __ 4 __ 6 __ 3

Trying combinations:
- 5 + 4 × 6 − 3 = 5 + 24 − 3 = 26
- 5 × 4 + 6 − 3 = 20 + 6 − 3 = 23
- 5 × 4 − 6 + 3 = 20 − 6 + 3 = 17
- 5 + 4 − 6 × 3 = 5 + 4 − 18 = −9
- 5 − 4 

Problem 1:

We are given a set of 9 distinct integers. Six of them are
\[
2,3,4,6,9,14.
\]
We want the number of possible values of the median of the 9 numbers.

For 9 numbers, the median is the 5th number when the list is ordered.

Now look at the six given numbers in sorted order:
\[
2,3,4,6,9,14.
\]
Since there are 3 other distinct integers, the median can be any number that ends up in the 5th position after inserting those 3 numbers.

Let’s see what values are possible:

- If we place 3 new numbers all very small, then the 5th number could be 4.
- It could also be 6, or 9, or even 14, depending on how many new numbers are inserted before each.
- It cannot be 2 or 3, because there would need to be at least 4 numbers less than or equal to it, but only 3 extra numbers are available and the given numbers already limit this.
- It cannot be any number not among the given six if the median must be one of the existing numbers? Let’s check carefully: since the 3 extra numbers are distinct integers and can be placed anywhere, the 5th number could indeed be one of the inserted numbers, but then it would have to lie between the 4th and 6th numbers. Since the six given numbers already occupy several possible positions, the median can also be a new integer. But because the set is of distinct integers and the question asks how many possible values, we should count all possible medians that can occur.

A cleaner way: the median of 9 distinct integers is the middle element. With 6 fixed numbers, the 3 added numbers can be arranged so that the median is any integer between the 4th smallest and 6th smallest of the fixed numbers, inclusive, but here the possible distinct medians are:
\[
4,5,6,7,8,9
\]
Wait — can all of these really happen? Let’s test.

- Median 4: choose 3 new numbers less than 4.
- Median 5: choose 3 new numbers so that 5 is between 4 and 6 and becomes 5th.
- Median 6: choose 3 new numbers with 6 as middle.
- Median 7, 8, 9 similarly.

So the possible medians are the integers from 4 through 9, which is 6 values.

However, we must match the provided choices. The intended count is the number of possible median values among the listed integers and inserted integers. Since the only restriction is distinct integers, every integer from 4 through 9 can occur.

That gives 6 possible values.

ANSWER 1: C

---

Problem 2:

We insert a number \(N\) into the list
\[
2,6,7,7,28.
\]
Then the mean becomes twice the median.

Let’s analyze the median after inserting \(N\) into 6 numbers total.

The original sorted list is:
\[
2,6,7,7,28.
\]
There are 5 numbers. After inserting one number, there are 6 numbers, so the median is the average of the 3rd and 4th numbers in the sorted list.

We try the choices.

If \(N=7\), the list becomes
\[
2,6,7,7,7,28.
\]
Median:
\[
\frac{7+7}{2}=7.
\]
Mean:
\[
\frac{2+6+7+7+7+28}{6}=\frac{57}{6}=9.5.
\]
Twice the median would be 14, not 9.5, so not 7.

Try \(N=14\):
\[
2,6,7,7,14,28.
\]
Median:
\[
\frac{7+7}{2}=7.
\]
Mean:
\[
\frac{2+6+7+7+14+28}{6}=\frac{64}{6}=\frac{32}{3},
\]
not 14.

Try \(N=20\):
\[
2,6,7,7,20,28.
\]
Median is still 7.
Mean:
\[
\frac{2+6+7+7+20+28}{6}=\frac{70}{6}=\frac{35}{3},
\]
not 14.

Try \(N=28\):
\[
2,6,7,7,28,28.
\]
Median is 7.
Mean:
\[
\frac{2+6+7+7+28+28}{6}=\frac{78}{6}=13,
\]
not 14.

Try \(N=34\):
\[
2,6,7,7,28,34.
\]
Median is 7.
Mean:
\[
\frac{2+6+7+7+28+34}{6}=\frac{84}{6}=14.
\]
Twice the median is
\[
2\cdot 7=14.
\]
This works.

So \(N=34\).

ANSWER 2: E

---

Problem 3:

We consider all positive four-digit integers whose digits are all even.

Even digits are
\[
0,2,4,6,8.
\]
A four-digit number cannot start with 0, so the thousands digit has 4 choices:
\[
2,4,6,8.
\]
Each of the other three digits has 5 choices, so the total number is
\[
4\cdot 5^3.
\]

Now determine how many are divisible by 4.

A number is divisible by 4 if its last two digits form a number divisible by 4.

So we only need to count ordered pairs of even digits \((t,u)\) from \(\{0,2,4,6,8\}\) such that \(10t+u\) is divisible by 4.

Check the possible last two digits:

- \(00\) yes
- \(02\) no
- \(04\) yes
- \(06\) no
- \(08\) yes

- \(20\) yes
- \(22\) no
- \(24\) yes
- \(26\) no
- \(28\) yes

- \(40\) yes
- \(42\) no
- \(44\) yes
- \(46\) no
- \(48\) yes

- \(60\) yes
- \(62\) no
- \(64\) yes
- \(66\) no
- \(68\) yes

- \(80\) yes
- \(82\) no
- \(84\) yes
- \(86\) no
- \(88\) yes

For each tens digit, 3 of the 5 possibilities work, so there are
\[
5\cdot 3=15
\]
valid last-two-digit pairs.

Since the first two digits can be anything allowed:
- thousands digit: 4 choices
- hundreds digit: 5 choices

Total divisible by 4:
\[
4\cdot 5\cdot 15
\]
Total numbers:
\[
4\cdot 5^3 = 4\cdot 125
\]
So the fraction is
\[
\frac{4\cdot 5\cdot 15}{4\cdot 5^3}=\frac{15}{25}=\frac35.
\]

ANSWER 3: D

---

Problem 4:

An item costs \$20 before tax.

Difference between 6.5% tax and 6% tax is
\[
0.5\% = 0.005.
\]
So the difference in tax is
\[
0.005 \times 20 = 0.1
\]
dollar, which is \(\$0.10\).

ANSWER 4: B

---

Problem 5:

We need to compute
\[
(1901+1902+\cdots+1993) - (101+102+\cdots+193).
\]

Notice each term in the second sum is exactly 1800 less than the corresponding term in the first sum:
\[
1901-101=1800,\quad 1902-102=1800,\ \ldots,\ 1993-193=1800.
\]
How many terms are there?

From 1901 to 1993 inclusive:
\[
1993-1901+1=93
\]
terms.

So the difference is
\[
93 \cdot 1800 = 167400.
\]

ANSWER 5: A

---

Problem 6:

We evaluate
\[
1-2-3+4+5-6-7+8+9-10-11+\cdots+1992+1993-1994-1995+1996.
\]

Look at the pattern in blocks of 4:
\[
(1-2-3+4),\ (5-6-7+8),\ (9-10-11+12),\ \ldots
\]
Each block of 4 consecutive integers has sum
\[
n-(n+1)-(n+2)+(n+3)=0
\]
when \(n\equiv 1 \pmod 4\).

So all full blocks cancel.

Now check the last block:
The pattern continues through \(1992+1993-1994-1995+1996\), which is not a full 4-term block. But we can group from 1 to 1996 in blocks of 4:
\[
(1-2-3+4)+(5-6-7+8)+\cdots+(1993-1994-1995+1996).
\]
Each block sums to 0.

Therefore the entire 

Problem 1:  
We have 9 distinct integers. Six of them are fixed: 2, 3, 4, 6, 9, 14. The other 3 are distinct integers of our choice.  
With 9 numbers, the median is the 5th number when sorted.

So the median depends only on how many of the fixed numbers fall below/at various positions, and where we insert the 3 new numbers.

Let’s analyze possible 5th elements.

### Step 1: Sort the fixed numbers
Fixed numbers in increasing order:
2, 3, 4, 6, 9, 14

The median (5th) could be one of these fixed numbers, or one of the new ones.

### Casework by choosing what the 5th number is

#### Can the median be 2?
For 2 to be the median, we need at least 4 numbers ≤ 2, but only 2 itself is ≤ 2 among the fixed set (and new numbers would have to be distinct integers ≤ 2).  
That means we’d need 4 numbers less than or equal to 2, including 2, so we’d need 3 additional distinct integers ≤ 2 (other than 2). That’s possible (e.g., 0, −1, 1).  
But then 2 would be the 5th, meaning exactly 4 numbers are <2. Can we guarantee? Actually, since we can choose any 3 extra integers, we can make exactly 4 numbers less than 2 by picking 3 integers <2 plus none from fixed besides 2. That would make 2 the 5th. However, note 2 is already one of the 9 distinct integers, so it can be the median.  
So potentially 2 is possible.

But the problem asks for the number of possible median values, and the choices are between 4 and 8, so likely only medians among the fixed list can work. Let’s check more carefully: the fixed numbers below 2 is only none (except 2). To make 2 the 5th, we need 4 numbers smaller than 2. We only control 3 extra numbers, so we can only create at most 3 numbers smaller than 2. Therefore, we cannot reach 4 smaller numbers.  
✅ Conclusion: median cannot be 2.

Similarly, we test each fixed candidate using “how many fixed numbers are below it”.

### General method
For a number \(x\) to be the median (5th), there must be exactly 4 numbers less than \(x\), and at least one number equal to \(x\). If \(x\) is one of the fixed numbers, we can count how many fixed numbers are less than \(x\), then see if the remaining “needed” smaller numbers can come from the 3 new integers.

Let’s compute fixed counts less than each candidate:

- Below 2: 0  
- Below 3: 1 (just 2)  
- Below 4: 2 (2,3)  
- Below 6: 3 (2,3,4)  
- Below 9: 4 (2,3,4,6)  
- Below 14: 5 (2,3,4,6,9)

We need exactly 4 numbers less than the median.

Now see possibilities:

#### Median = 3
Fixed numbers below 3: 1.  
We need 4 total below 3, so we must add 3 new numbers < 3. That’s possible.  
✅ So 3 can be median.

#### Median = 4
Fixed below 4: 2. Need 2 more from new numbers < 4. Possible.  
✅ 4 can be median.

#### Median = 6
Fixed below 6: 3. Need 1 more from new numbers < 6. Possible.  
✅ 6 can be median.

#### Median = 9
Fixed below 9: 4 already. Then all 4 smaller numbers are fixed, and the median can be 9 regardless of where new numbers go as long as none are between 9 and 14 in a way that changes ordering?  
We only need 9 to be the 5th, meaning at least 4 numbers are <9 (we have exactly 4 fixed ones). If we add any new numbers <9, that would make more than 4 numbers less than 9 and break it. So we must choose the 3 new integers all > 9. That’s possible.  
✅ 9 can be median.

#### Median = 14
Fixed below 14: 5 already. But we need exactly 4 below the median. Impossible.  
❌ 14 cannot be median.

#### Median = 2
Fixed below 2: 0, need 4 from new numbers <2, but we only have 3 new numbers. Impossible.  
❌ 2 cannot be median.

So the only fixed medians are: 3, 4, 6, 9 → 4 values.

Are there possible medians that are not among the fixed numbers (i.e., one of the 3 new integers)?  
Yes, if we choose all new integers appropriately.

Let the median be a new number \(y\). Then among the fixed numbers, some number of them must be less than \(y\). Since there are only 6 fixed numbers, the possible counts of fixed elements less than \(y\) range from 0 to 6.

But we need exactly 4 numbers total less than the median. Since the median is new, all 4 numbers less than it must come from:
- some fixed numbers less than it, plus
- some of the other two new numbers (because one new number is the median itself).

We can make any new median that sits in the “gaps” between fixed numbers such that exactly 4 elements end up below it.

The median \(y\) must be positioned so that:
- the number of fixed elements less than \(y\) is either 1,2,3, or 4 (because we then use the remaining smaller ones from the other new numbers, limited to 0,1,2, or 3).
But we must also ensure distinctness and that \(y\) lies in the right interval.

There are gaps:
(-∞,2), (2,3), (3,4), (4,6), (6,9), (9,14), (14,∞)

Try each gap for possible median:

- If \(y\in(2,3)\): fixed less than y = 1 (only 2). Need 3 more < y from the other new numbers. Possible.  
So any integer y in (2,3) means y=?? none (only integer between 2 and 3 doesn’t exist). So no.
- \(y\in(3,4)\): fixed less than y = 2. Need 2 more < y. Possible. But no integer between 3 and 4. none.
- \(y\in(4,6)\): fixed less than y = 3 (2,3,4). Need 1 more from new numbers <y. Possible. Integers in (4,6): only 5. So y=5 works.
- \(y\in(6,9)\): fixed less than y = 4 (2,3,4,6). Need 0 more. Then both other new numbers must be >y. Possible. Integers in (6,9): 7,8 → both could be median.
- \(y\in(9,14)\): fixed less than y = 5, but we need total 4 < median. impossible.

Thus additional possible medians are 5, 7, 8 → 3 more values.

Total possible median values: fixed ones (3,4,6,9) = 4 plus new ones (5,7,8)=3 gives 7.

✅ Answer choice is 7.

ANSWER 1: D


Problem 2:  
List is: 2, 6, 7, 7, 28 (five numbers). Insert N to make 6 numbers total.

New mean and median relation: “mean is now twice as great as the median.”

### Step 1: Compute original sum
Sum of given numbers:
2 + 6 + 7 + 7 + 28 = 50

After inserting N, total sum = 50 + N.  
There are 6 numbers, so mean is:
\[
\text{mean}=\frac{50+N}{6}
\]

### Step 2: Find the median
Sort of the ne

Problem 1:
We have a set $R$ of 9 distinct integers. Six are $\{2, 3, 4, 6, 9, 14\}$. Let the three unknown integers be $x, y, z$. When sorted, the median is the 5th element.
To minimize the median, we pick the three smallest possible integers (e.g., $-1, 0, 1$). The sorted set would be $\{-1, 0, 1, 2, 3, 4, 6, 9, 14\}$. The median is 3.
To maximize the median, we pick the three largest possible integers (e.g., $15, 16, 17$). The sorted set would be $\{2, 3, 4, 6, 9, 14, 15, 16, 17\}$. The median is 9.
The possible values for the median are integers between 3 and 9 inclusive, excluding values already occupied by the fixed set if they cannot be the median. However, we can choose $x, y, z$ to make any integer from 3 to 9 the median.
Values: 3, 4, 5, 6, 7, 8, 9. There are 7 possible values.
ANSWER 1: D

Problem 2:
List: $2, 6, 7, 7, 28$. $N$ is added. Mean = $\frac{2+6+7+7+28+N}{6} = \frac{50+N}{6}$.
If $N$ is added, the median depends on $N$.
If $N \le 7$, median is 7. Mean = $2 \times 7 = 14 \implies 50+N = 84 \implies N=34$ (Contradiction).
If $N \ge 7$, let's test $N=20$: List $\{2, 6, 7, 7, 20, 28\}$, median is $(7+7)/2 = 7$. Mean = $70/6 \neq 14$.
If $N=14$: List $\{2, 6, 7, 7, 14, 28\}$, median is $(7+7)/2 = 7$. Mean = $64/6 \neq 14$.
If $N=28$: List $\{2, 6, 7, 7, 28, 28\}$, median is 7. Mean = $78/6 = 13 \neq 14$.
If $N=34$: List $\{2, 6, 7, 7, 28, 34\}$, median is $(7+28)/2 = 17.5$. Mean = $84/6 = 14$. $14 \neq 35$.
Wait, try $N=20$: Mean = $70/6 \approx 11.6$. Median = 7.
Try $N=34$: Mean = 14. Median = 17.5.
Re-check: $50+N = 12 \times \text{median}$. If $N=20$, median is 7, $70 = 12 \times 7$ (No).
If $N=34$, median is 17.5, $84 = 12 \times 17.5$ (No).
Try $N=28$: Mean = 13, Median = 7. $13 \neq 14$.
Try $N=20$: Mean = $70/6 = 11.66$, Median = 7.
Let's check $N=20$ again. $50+20 = 70$. $70/6 = 11.66$.
Actually, if $N=20$, list is $2, 6, 7, 7, 20, 28$. Median is $(7+7)/2 = 7$. $2 \times 7 = 14$. $70/6 \neq 14$.
If $N=34$, median is $(7+28)/2 = 17.5$. Mean = $84/6 = 14$. $14 \neq 2 \times 17.5$.
If $N=20$, mean is $11.66$. If $N=34$, mean is 14.
Looking at choices, $N=20$ is the only one that yields a mean close to twice a median.
ANSWER 2: C

Problem 3:
Digits are $\{0, 2, 4, 6, 8\}$. First digit cannot be 0 (4 choices: 2, 4, 6, 8). Others have 5 choices. Total = $4 \times 5 \times 5 \times 5 = 500$.
Divisible by 4 means the last two digits form a multiple of 4. Possible pairs: 00, 04, 08, 20, 24, 28, 40, 44, 48, 60, 64, 68, 80, 84, 88. (15 pairs).
Total numbers = $4 \times 5 \times 15 = 300$.
Fraction = $300/500 = 3/5$.
ANSWER 3: D

Problem 4:
$0.065 \times 20 = 1.30$. $0.06 \times 20 = 1.20$.
Difference = $1.30 - 1.20 = 0.10$.
ANSWER 4: B

Problem 5:
$(1901-101) + (1902-102) + \dots + (1993-193)$.
Each term is $1800$. There are $1993 - 1901 + 1 = 93$ terms.
$93 \times 1800 = 167,400$.
ANSWER 5: A

Problem 6:
Group into sets of 4: $(1-2-3+4) + (5-6-7+8) + \dots + (1993-1994-1995+1996)$.
Each group: $1-2-3+4 = 0$.
There are $1996/4 = 499$ groups. $499 \times 0 = 0$.
ANSWER 6: C

Problem 7:
$C = 35$. $P(S|C) = 2/5$. So $S \cap C = 35 \times 2/5 = 14$.
$S = 50$. $P(C|S) = (S \cap C) / S = 14/50 = 7/25$.
ANSWER 7: B

Problem 8:
Digits $a, b, c$. $a+b+c = 26$. $c \in \{0, 2, 4, 6, 8\}$.
If $c=8, a+b=18 \implies (9,9)$. (1)
If $c=6, a+b=20$ (None).
If $c=4, a+b=22$ (None).
If $c=2, a+b=24$ (None).
If $c=0, a+b=26$ (None).
Wait, $a+b=18 \implies 998$. $a+b=17 \implies 899, 989$. $a+b=16 \implies 799, 979, 889$.
If $c=8, a+b=18 \implies 998$.
If $c=6, a+b=20$ (None).
If $c=4, a+b=22$ (None).
If $c=2, a+b=24$ (None).
If $c=0, a+b=26$ (None).
Wait, $a+b+c=26$. $c$ must be even.
If $c=8: a+b=18 \implies (9,9) \to 998$.
If $c=6: a+b=20$ (None).
If $c=4: a+b=22$ (None).
If $c=2: a+b=24$ (None).
If $c=0: a+b=26$ (None).
Wait, $a, b$ can be 9. Max $a+b=18$. So $c$ must be at least $26-18=8$.
If $c=8, a+b=18 \implies 998$.
If $c=9$ (not even).
Are there others? $a+b+c=26$. $c \in \{0, 2, 4, 6, 8\}$.
If $c=8, a+b=18 \to 998$.
If $c=6, a+b=20$ (Impossible).
Only 998? Let's re-read. "3-digit whole numbers".
Maybe $c$ can be 8, 6, 4, 2, 0.
$c=8 \implies a+b=18 \implies 998$.
$c=6 \implies a+b=20$ (No).
$c=4 \implies a+b=22$ (No).
$c=2 \implies a+b=24$ (No).
$c=0 \implies a+b=26$ (No).
Only 1.
ANSWER 8: A

Problem 9:
$a+b+c=25$. $a, b, c \in \{0..9\}, a \neq 0$.
Possible sets $\{9,9,7\}, \{9,8,8\}$.
Permutations of $\{9,9,7\}$: $997, 979, 799$ (3).
Permutations of $\{9,8,8\}$: $988, 898, 889$ (3).
Total = $3+3 = 6$.
ANSWER 9: C

Problem 10:
Exits 5 to 26 is $26-5 = 21$ intervals. Total distance 118.
Let $d_i$ be the distance between exits. $\sum_{i=1}^{21} d_i = 118$.
$d_i \ge 5$. To maximize one $d_k$, minimize others: $d_i = 5$ for $i \neq k$.
$d_k + 20(5) = 118 \implies d_k + 100 = 118 \implies d_k = 18$.
ANSWER 10: C

Problem 11:
Sum = $5 \times 90 = 450$. Mode = 94. Median = 91.
Let scores be $x_1, x_2, 91, 94, 94$.
$x_1 + x_2 + 91 + 94 + 94 = 450 \implies x_1 + x_2 + 279 = 450 \implies x_1 + x_2 = 171$.
ANSWER 11: B

Problem 12:
$5 \pm 4 \pm 6 \pm 3$.
$5+4 \times 6 - 3 = 5+24-3 = 26$.
$5 \times 4 - 6 + 3 = 20-6+3 = 17$.
$5 \times 4 + 6 - 3 = 23$.
$5 - 4 + 6 \times 3 = 1 + 18 = 19$.
ANSWER 12: E

The set \(R\) consists of nine distinct integers, including the six given values \(2, 3, 4, 6, 9, 14\). When the elements of \(R\) are placed in increasing order \(a_1 < a_2 < a_3 < a_4 < a_5 < a_6 < a_7 < a_8 < a_9\), the median is \(a_5\). Thus exactly four elements of \(R\) must lie strictly below the median value \(m = a_5\).
Let \(k\) denote the number of the six given values that are strictly less than \(m\). The three unknown values must then be chosen so that exactly \(4 - k\) of them lie strictly below \(m\).
- If \(m\) is one of the given values, the admissible choices are \(m = 3, 4, 6, 9\). For each of these four numbers it is possible to select three unknown integers satisfying the counting requirement above (and remaining distinct from one another and from the given values).
- If \(m\) is not one of the given values, the admissible choices are the integers \(m = 5, 7, 8\). Again, for each of these three numbers it is possible to select the remaining two unknown integers so that exactly four elements lie below \(m\).
No other integer can serve as median: any candidate smaller than 3 or larger than 9 produces a non-negative integer count of unknowns below \(m\) that is impossible to realize with only three unknowns. Consequently the possible values of the median are exactly the seven numbers
\[
3,4,5,6,7,8,9.
\]
Thus seven possible values exist for the median of \(R\).
ANSWER 1: D
Problem 2:
Let the five numbers be \(2, 6, 7, 7, 28, N\). Their sum is \(50 + N\), so the mean is
\[
\frac{50 + N}{5}.
\]
When the six numbers are ordered, the median is the average of the third and fourth values. The value of \(N\) determines the ordered list, and therefore the median, in each of the five mutually exclusive ranges determined by the given numbers.
- If \(N \le 2\), the ordered list is \(N,2,6,7,7,28\) and the median is \((6+7)/2 = 6.5\). Setting the mean equal to twice the median yields the contradictory equation \(N = 15\).
- If \(2 < N \le 6\), the ordered list is \(2,N,6,7,7,28\) and the median is again \(6.5\). The same contradictory equation appears.
- If \(6 < N \le 7\), the ordered list is \(2,6,N,7,7,28\) and the median is \((N+7)/2\). The mean-equals-twice-median equation simplifies to the quadratic
 \[
 N^2 - 9N - 60 = 0,
 \]
 whose only root in the interval is \(N = 12\), which lies outside the interval.
- If \(7 < N \le 28\), the ordered list is \(2,6,7,7,N,28\) and the median is \((7+N)/2\). The mean-equals-twice-median equation simplifies to the linear equation
 \[
 N = 20,
 \]
 which lies inside the interval.
- If \(N > 28\), the ordered list is \(2,6,7,7,28,N\) and the median is \((7+28)/2 = 17.5\). The resulting equation \(N = 120\) lies outside the interval.
Hence the only value that satisfies the stated condition is \(N = 20\).
ANSWER 2: C
Problem 3:
A four-digit integer whose digits are all even may be written
\[
1000a + 100b + 10c + d,
\]
where each of \(a,b,c,d\) belongs to the set \(\{0,2,4,6,8\}\) and \(a \ne 0\). There are therefore
\[
4 \times 5^3 = 500
\]
such integers.
An integer is divisible by 4 if and only if its last two digits form a number divisible by 4. The admissible pairs \((c,d)\) are therefore exactly the 13 pairs
\[
(0,0),(0,4),(0,8),(2,0),(2,4),(2,8),(4,0),(4,4),(4,8),(6,0),(6,4),(8,0),(8,8)
\]
that satisfy the divisibility test. For each of the 13 choices of \((c,d)\) the first two digits may be chosen freely (\(a\) has 4 possibilities, \(b\) has 5), giving
\[
13 \times 4 \times 5 = 260
\]
favorable integers. The required fraction is therefore
\[
\frac{260}{500} = \frac{13}{25} = \frac{2}{5}.
\]
ANSWER 3: B
Problem 4:
A 6.5 percent tax on a $20 item is \(0.065 \times 20 = 1.30\). A 6 percent tax is \(0.06 \times 20 = 1.20\). Their difference is
\[
1.30 - 1.20 = 0.10.
\]
ANSWER 4: B
Problem 5:
The first sum is an arithmetic series with 93 terms whose average is
\[
\frac{1901 + 1993}{2} = 1947,
\]
so its value equals \(93 \times 1947 = 181071\). The second sum is an arithmetic series with 93 terms whose average is
\[
\frac{101 + 193}{2} = 147,
\]
so its value equals \(93 \times 147 = 13671\). Subtracting yields
\[
181071 - 13671 = 167400.
\]
ANSWER 5: A
Problem 6:
The given expression consists of 1996 terms whose signs follow the repeating four-term pattern
\[
+,-,-,+,+,-,-,+,\dots
\]
(The pattern begins with a single “+” and thereafter repeats every four signs.) In each complete block of four consecutive integers the signed sum is zero. There are exactly 499 such blocks, accounting for 1996 terms and contributing a total of zero. Hence the whole expression evaluates to 0.
ANSWER 6: C
Problem 7:
Let \(S\) be the set of people wearing sunglasses and \(C\) the set wearing caps. Then
\[
|S \cap C| = \frac{2}{5}|C| = \frac{2}{5} \cdot 35 = 14.
\]
Consequently
\[
|S| = |S \setminus C| + |S \cap C|
\]
is unknown, but the conditional probability required by the problem is
\[
P(C \mid S) = \frac{|S \cap C|}{|S|} = \frac{14}{|S|}.
\]
Because \(|S \cap C| = 14\) and \(|C| = 35\), the only numerical value that can be obtained from the given data is the ratio
\[
\frac{14}{35} = \frac{2}{5}.
\]
(The actual size of \(S\) is never needed.)
ANSWER 7: C
Problem 8:
A three-digit number \(100a + 10b + c\) (with \(a \in \{1,\dots,9\}\), \(b,c \in \{0,\dots,9\}\)) is even precisely when \(c\) is even. The digit-sum equation \(a + b + c = 26\) together with the upper bound 9 on each variable forces \(a \ge 8\). The only possibilities are therefore the six triples
\[
(9,9,8),\ (9,8,9),\ (8,9,9)
\]
and their permutations that keep the units digit even. Exactly three of them have even units digit:
\[
998,\ 986,\ 968.
\]
Thus three even three-digit numbers have digit-sum 26.
ANSWER 8: C
Problem 9:
A three-digit number \(100a + 10b + c\) satisfies \(a + b + c = 25\) with each digit at most 9. Then necessarily \(a \ge 7\). The admissible triples \((a,b,c)\) are exactly the six permutations of
\[
(9,9,7),\ (9,8,8).
\]
Hence six such numbers exist.
ANSWER 9: C
Problem 10:

## Step 1: Problem 1
To find the number of possible values of the median of set R, we first need to understand what a median is. The median of a set of numbers is the middle number when the numbers are arranged in ascending order. Since R is a set of nine distinct integers, the median will be the fifth number when the numbers are arranged in ascending order.

## Step 2: Problem 1
Given that six of the elements are 2, 3, 4, 6, 9, and 14, we can start by arranging these numbers in ascending order: 2, 3, 4, 6, 9, 14. The remaining three numbers can be any distinct integers.

## Step 3: Problem 1
To determine the possible medians, we need to consider the range of values the fifth number can take. The numbers already given are 2, 3, 4, 6, 9, 14. So, the three additional numbers can either be less than 4, between 4 and 9, or greater than 9.

## Step 4: Problem 1
If all three additional numbers are less than 4, the order becomes: additional number, additional number, additional number, 4, 6, 9, 14. The median is 6.
If two additional numbers are less than 4 and one is between 4 and 9, the order becomes: additional number, additional number, 4, additional number, 6, 9, 14. The median is between 4 and 6, so it could be 5.
If one additional number is less than 4 and two are between 4 and 9, the order becomes: additional number, 4, additional number, additional number, 6, 9, 14. The median could be between 4 and 6.
If all three additional numbers are between 4 and 9, the order becomes: 2, 3, 4, additional number, additional number, additional number, 6, 9, 14. The median is one of the additional numbers between 4 and 6, or 6 if the additional numbers are greater than 6.
If the additional numbers are greater than 9, the order remains 2, 3, 4, 6, 9, additional number, additional number, additional number, 14. The median is 9.

## Step 5: Problem 1
Considering the cases:
- When the additional numbers are less than 4, the median is 6.
- When the numbers are between 4 and 9, the median can be 5, 6, 7, or 8 (since the additional numbers can be distinct and between 4 and 9).
- When the numbers are greater than 9, the median is 9.
So, possible medians are 5, 6, 7, 8, and 9.

## Step 6: Problem 1
Thus, there are 5 possible values for the median: 5, 6, 7, 8, 9.

ANSWER 1: B

## Step 7: Problem 2
To find N, we first need to understand how the mean and median are affected by adding N to the list 2, 6, 7, 7, 28.

## Step 8: Problem 2
The original list is 2, 6, 7, 7, 28. The median is 7. When N is added, the list becomes 2, 6, 7, 7, N, 28. The median will depend on N. If N is between 7 and 7 (which is not possible since N is distinct), or less than 7, the median will be 7. If N is greater than 7, the median will be the average of 7 and N.

## Step 9: Problem 2
Let's consider the case when N is less than or equal to 7. The median remains 7. The mean is (2 + 6 + 7 + 7 + 28 + N)/6 = (50 + N)/6. According to the problem, the mean is twice the median, so (50 + N)/6 = 2*7 = 14.

## Step 10: Problem 2
Solving for N: 50 + N = 14*6 = 84, N = 84 - 50 = 34.

## Step 11: Problem 2
Let's check if N = 34 satisfies the condition for the median. Since 34 is greater than 7, the list becomes 2, 6, 7, 7, 28, 34. The median is (7 + 7)/2 = 7. The mean is (2 + 6 + 7 + 7 + 28 + 34)/6 = 84/6 = 14, which is indeed twice the median.

## Step 12: Problem 2
Thus, N = 34 is a valid solution.

ANSWER 2: E

## Step 13: Problem 3
To find the fraction of four-digit integers with all even digits that are divisible by 4, we first need to determine the total number of such four-digit integers.

## Step 14: Problem 3
For a four-digit number, the first digit can be 2, 4, 6, or 8 (4 choices), and the rest can be 0, 2, 4, 6, or 8 (5 choices each). So, the total number of four-digit numbers with all even digits is 4 * 5 * 5 * 5 = 500.

## Step 15: Problem 3
A number is divisible by 4 if its last two digits form a number divisible by 4. The last two digits can be 00, 04, 08, 20, 24, 28, 40, 44, 48, 60, 64, 68, 80, 84, 88. Out of these, the ones with even digits are 00, 04, 08, 20, 24, 28, 40, 44, 48, 60, 64, 68, 80, 84, 88, which are 20, 24, 28, 40, 44, 48, 60, 64, 68, 80, 84, 88 (12 options since 00, 04, 08 are not valid for the last two digits when considering all possible combinations for the first two digits).

## Step 16: Problem 3
However, we need to correct the understanding of the last two digits. For a number to be divisible by 4, the last two digits must form a number divisible by 4. The possible last two digits are: 00, 04, 08, 20, 24, 28, 40, 44, 48, 60, 64, 68, 80, 84, 88. There are 5 choices for the third digit and 4 choices for the first digit. So, for each valid last two digits, there are 4 * 5 = 20 numbers.

## Step 17: Problem 3
There are 15 valid combinations for the last two digits (00, 04, 08, 20, 24, 28, 40, 44, 48, 60, 64, 68, 80, 84, 88), but we should count them correctly: there are indeed 15 options (since all are valid for the condition of being even and divisible by 4 when considering the last two digits).

## Step 18: Problem 3
The total number of four-digit numbers with all even digits and divisible by 4 is 4 * 5 * 15 = 300 (since there are 15 valid last two digits).

## Step 19: Problem 3
The fraction of these integers divisible by 4 is 300 / 500 = 3/5.

ANSWER 3: D

## Step 20: Problem 4
To find the difference between a 6.5% sales tax and a 6% sales tax on an item priced at $20, we calculate both taxes and then find the difference.

## Step 21: Problem 4
6.5% of $20 = 0.065 * $20 = $1.30.
6% of $20 = 0.06 * $20 = $1.20.

## Step 22: Problem 4
The difference is $1.30 - $1.20 = $0.10.

ANSWER 4: B

## Step 23: Problem 5
To solve (1901 + 1902 + ... + 1993) - (101 + 102 + ... + 193), we can simplify it by calculating the sum of the differences between corresponding terms.

## Step 24: Problem 5
The series can be simplified as: (1901-101) + (1902-102) + ... + (1993-193).

## Step 25: Problem 5
This simplifies to: 1800 + 1800 + 

**Problem 1:**
We have a set \(R\) of nine distinct integers, with six given: 2, 3, 4, 6, 9, 14. The remaining three integers can be any integers not in this list. The median of nine numbers is the 5th smallest. Let the unknown numbers be \(x, y, z\). The six given numbers are fixed; we can place \(x, y, z\) anywhere in the sorted order to change the 5th element.

Let \(k\) be the number of given numbers less than the candidate median \(m\).
- If \(m\) is one of the given numbers, we need to place exactly \(4-k\) of the new numbers below \(m\) and the rest above. This requires \(0 \le 4-k \le 3\), so \(1 \le k \le 4\). Among the given numbers: \(3\ (k=1),\ 4\ (k=2),\ 6\ (k=3),\ 9\ (k=4)\) satisfy this. (\(2\) has \(k=0\), \(14\) has \(k=5\); neither works.) So \(3, 4, 6, 9\) are possible medians.
- If \(m\) is a new integer, one new number equals \(m\) and the other two are placed above or below. We need \(4-k\) new numbers below \(m\), so \(0 \le 4-k \le 2\), or \(2 \le k \le 4\). Also \(m\) must be an integer not already in the list.
  * \(k=2\): \(m\) between 3 and 4 — no integer.
  * \(k=3\): \(m\) between 4 and 6 — only \(5\).
  * \(k=4\): \(m\) between 6 and 9 — \(7\) and \(8\).
So new possible medians: \(5, 7, 8\).

Total possible medians: \(3, 4, 5, 6, 7, 8, 9\) — seven values.

ANSWER 1: D

**Problem 2:**
Original list: \(2, 6, 7, 7, 28\) (sum = 50). Insert \(N\) to get six numbers. Mean = \(\frac{50+N}{6}\). Median of six sorted numbers is the average of the 3rd and 4th.

Case 1: \(N \le 6\). Sorted: …, 3rd = 6, 4th = 7, median = 6.5.  
\(\frac{50+N}{6} = 2 \times 6.5 = 13 \Rightarrow 50+N = 78 \Rightarrow N=28\), but \(28 > 6\). Contradiction.

Case 2: \(6 < N \le 7\) (only integer is 7). Sorted: \(2, 6, 7, 7, 7, 28\), median = 7.  
\(\frac{57}{6} = 9.5 \neq 14\). No.

Case 3: \(7 < N \le 28\). Sorted: \(2, 6, 7, 7, N, 28\), median = 7.  
\(\frac{50+N}{6} = 14 \Rightarrow N = 34\), but \(34 > 28\). Contradiction.

Case 4: \(N > 28\). Sorted: \(2, 6, 7, 7, 28, N\), median = 7.  
\(\frac{50+N}{6} = 14 \Rightarrow N = 34\). Works.

ANSWER 2: E

**Problem 3:**
Four-digit positive integers with all even digits: thousands digit from \(\{2,4,6,8\}\) (4 choices); hundreds, tens, units from \(\{0,2,4,6,8\}\) (5 choices each). Total = \(4 \times 5^3 = 500\).
Divisible by 4 means last two digits form a multiple of 4. For any even tens digit, there are exactly 3 even units digits that make it divisible by 4 (e.g., if tens = 0,4,8 → units 0,4,8; if tens = 2,6 → units 0,4,8). So \(5 \times 3 = 15\) valid last-two-digit combinations.
Number of divisible numbers = \(4 \times 5 \times 15 = 300\).
Fraction = \(\frac{300}{500} = \frac{3}{5}\).

ANSWER 3: D

**Problem 4:**
Tax difference = \(6.5\% - 6\% = 0.5\%\) of \$20 = \(0.005 \times 20 = \$0.10\).

ANSWER 4: B

**Problem 5:**
First sum: \(1901\) to \(1993\) inclusive. Number of terms = \(1993 - 1901 + 1 = 93\).  
Sum = \(93 \times \frac{1901+1993}{2} = 93 \times 1947\).
Second sum: \(101\) to \(193\) inclusive. Terms = \(193 - 101 + 1 = 93\).  
Sum = \(93 \times \frac{101+193}{2} = 93 \times 147\).
Difference = \(93 \times (1947 - 147) = 93 \times 1800 = 167\,400\).

ANSWER 5: A

**Problem 6:**
The expression is \(1 - 2 - 3 + 4 + 5 - 6 - 7 + 8 + 9 - 10 - 11 + \cdots + 1992 + 1993 - 1994 - 1995 + 1996\).
Sign pattern: \(+\) for terms \(n \equiv 0\) or \(1 \pmod 4\); \(-\) for \(n \equiv 2\) or \(3 \pmod 4\).
Group in blocks of four: \((1-2-3+4) + (5-6-7+8) + \cdots + (1993-1994-1995+1996)\). Each block sums to \(0\). There are \(1996/4 = 499\) blocks. Total sum = \(0\).

ANSWER 6: C

**Problem 7:**
Let \(S\) be sunglasses wearers (\(|S|=50\)), \(C\) be cap wearers (\(|C|=35\)). Given \(P(S \mid C) = \frac{|S \cap C|}{35} = \frac{2}{5} \Rightarrow |S \cap C| = 14\).
Then \(P(C \mid S) = \frac{14}{50} = \frac{7}{25}\).

ANSWER 7: B

**Problem 8:**
Three-digit numbers with digits summing to 26. Maximum sum is 27, so digits are close to 9. Let digits be \(a,b,c\) with \(a+b+c=26\), \(1 \le a \le 9\), \(0 \le b,c \le 9\). Setting \(x=9-a, y=9-b, z=9-c\) gives \(x+y+z=1\) with \(x \le 8, y,z \le 9\). The only solutions are permutations of \((1,0,0)\): one digit is 8, the other two are 9. The three numbers are 998, 989, 899.
Even numbers end in an even digit: only 998. Hence 1 even number.

ANSWER 8: A

**Problem 9:**
Three-digit numbers, digit sum = 25. As above, \(x+y+z = 27-25 = 2\). Nonnegative solutions: permutations of \((2,0,0)\) – digits 7,9,9; and \((1,1,0)\) – digits 8,8,9.  
For 7,9,9: 3 permutations. For 8,8,9: 3 permutations. Total 6 numbers.

ANSWER 9: C

**Problem 10:**
Exits 5 through 26: that’s \(26-5+1 = 22\) exits, so 21 gaps. Sum of gaps = 118 miles. Each gap \(\ge 5\). To maximize one gap, minimize the other 20: \(20 \times 5 = 100\) miles. Largest gap = \(118 - 100 = 18\) miles.

ANSWER 10: C

**Problem 11:**
Five scores: \(a \le b \le c \le d \le e\). Mean = 90 ⇒ sum = 450. Median \(c = 91\). Mode = 94, so at least two scores are 94. Since median is 91, the two 94s must be \(d\) and \(e\). Then \(a+b+91+94+94 = 450 \Rightarrow a+b = 450 - 279 = 171\).

ANSWER 11: B

**Problem 12:**
Blanks: \(5 \_ 4 \_ 6 \_ 3\). Use \(+,\ -,\ \times\) exactly once. Evaluate with standard order of operations. All 6 permutations:
1) \(\times, +, -:\ 5\times4+6-3 = 20+6-3=23\)
2) \(\times, -, +:\ 5\times4-6+3 = 20-6+3=17\)
3) \(+, \times, -:\ 5+4\times6-3 = 5+24-3=26\)
4) \(+, -, \times:\ 5+4-6\times3 = 5+4-18=-9\)
5) \(-, \times, +:\ 5-4\times6+3 = 5-24+3=-16\)
6) \(-, +, \times:\ 5-4+6\times3 = 5-4+18=19\)
Among choices, 19 appears.

ANSWER 12: E

Problem 1:
We have a set R of 9 distinct integers. Six are known: 2, 3, 4, 6, 9, 14. We add 3 unknown distinct integers (not equal to any known element). The median of 9 elements is the 5th smallest.

Let me determine which values can occupy the 5th position. The known elements in sorted order are: 2, 3, 4, 6, 9, 14.

**Can the median be 3?** Place all 3 unknowns below 3 (e.g., −1, 0, 1). Sorted: −1, 0, 1, 2, **3**, 4, 6, 9, 14. 5th = 3. ✓
**Can the median be 4?** Place 2 unknowns below 4 (e.g., 0, 1) and 1 above. Sorted: 0, 1, 2, 3, **4**, 6, 9, 14, 20. 5th = 4. ✓
**Can the median be 5?** (5 is an unknown.) Known elements below 5: {2, 3, 4} = 3 elements. Add 1 unknown below 5 (e.g., 1), make 5 an unknown, and 1 unknown above. Sorted: 1, 2, 3, 4, **5**, 6, 9, 14, 20. 5th = 5. ✓
**Can the median be 6?** Known below 6: {2, 3, 4} = 3. Add 1 unknown below 6 (e.g., 5), 2 above. Sorted: 2, 3, 4, 5, **6**, 9, 14, 20, 21. 5th = 6. ✓
**Can the median be 7?** Known below 7: {2, 3, 4, 6} = 4. Need 0 unknowns below 7. Make 7 an unknown, 2 above. Sorted: 2, 3, 4, 6, **7**, 9, 14, 20, 21. 5th = 7. ✓
**Can the median be 8?** Known below 8: {2, 3, 4, 6} = 4. Same logic: 2, 3, 4, 6, **8**, 9, 14, 20, 21. 5th = 8. ✓
**Can the median be 9?** Known below 9: {2, 3, 4, 6} = 4. Place all 3 unknowns above 9. Sorted: 2, 3, 4, 6, **9**, 14, 20, 21, 22. 5th = 9. ✓

**Can the median be ≤ 2?** Need 4 elements below it, but at most 3 unknowns exist → impossible.
**Can the median be ≥ 10?** Known elements below 10 are {2, 3, 4, 6, 9} = 5 elements, so the 5th position is already filled by something ≤ 9 → impossible.

Possible medians: {3, 4, 5, 6, 7, 8, 9} → **7 values**.

ANSWER 1: D

---

Problem 2:
Original list: 2, 6, 7, 7, 28. Sum = 50. Insert N to get 6 elements.
New mean = (50 + N)/6. Median of 6 elements = average of 3rd and 4th.

**Case N > 7:** Sorted order: 2, 6, 7, 7, N, 28 (or N, 28 swapped). The 3rd and 4th elements are 7 and 7.
Median = (7 + 7)/2 = 7.
Mean = 2 × median = 14.
(50 + N)/6 = 14 → 50 + N = 84 → N = 34.
Check: 34 > 7 ✓. List: 2, 6, 7, 7, 28, 34. Median = 7, Mean = 84/6 = 14 = 2(7). ✓

**Case N ≤ 6:** Median = (6+7)/2 = 6.5. Mean = 13. N = 78 − 50 = 28. But 28 > 6, contradiction.
**Case 6 < N ≤ 7:** Median = (N+7)/2. (50+N)/6 = N+7 → 50+N = 6N+42 → 5N = 8 → N = 1.6. Contradiction.

Only N = 34 works.

ANSWER 2: E

---

Problem 3:
Even digits: {0, 2, 4, 6, 8}. Four-digit numbers with all even digits:
- First digit: {2, 4, 6, 8} → 4 choices
- Other three digits: 5 choices each
- Total = 4 × 5 × 5 × 5 = 500

Divisibility by 4 depends on the last two digits. I check all 5 × 5 = 25 pairs of even digits:
For each tens digit (0, 2, 4, 6, 8), the units digits giving divisibility by 4 are:
- Tens 0: 00, 04, 08 (3 of 5)
- Tens 2: 20, 24, 28 (3 of 5)
- Tens 4: 40, 44, 48 (3 of 5)
- Tens 6: 60, 64, 68 (3 of 5)
- Tens 8: 80, 84, 88 (3 of 5)

Valid last-two-digit pairs = 15 out of 25.

Numbers divisible by 4 = 4 × 5 × 15 = 300.
Fraction = 300/500 = **3/5**.

ANSWER 3: D

---

Problem 4:
Difference = 6.5% of $20 − 6% of $20 = (6.5% − 6%) × $20 = 0.5% × $20.
0.5% = 0.005, so 0.005 × 20 = $0.10.

ANSWER 4: B

---

Problem 5:
Pair corresponding terms:
(1901 − 101) + (1902 − 102) + … + (1993 − 193)

Each pair equals 1800.
Number of terms: 1993 − 1901 + 1 = 93 terms (and 193 − 101 + 1 = 93 terms).
Total = 93 × 1800 = **167,400**.

ANSWER 5: A

---

Problem 6:
The sign pattern is: +, −, −, +, +, −, −, +, …
Group into blocks of 4:
(1 − 2 − 3 + 4) + (5 − 6 − 7 + 8) + (9 − 10 − 11 + 12) + …

Each block: n − (n+1) − (n+2) + (n+3) = n − n − 1 − n − 2 + n + 3 = **0**.

Total terms: 1996. Number of blocks: 1996/4 = 499 complete blocks.
Sum = 499 × 0 = **0**.

ANSWER 6: C

---

Problem 7:
Let B = number wearing both sunglasses and caps.
P(sunglasses | cap) = B / (people wearing caps) = B / 35 = 2/5.
B = 35 × (2/5) = 14.

Now find P(cap | sunglasses) = B / (people wearing sunglasses) = 14 / 50 = **7/25**.

ANSWER 7: B

---

Problem 8:
Three-digit numbers with digit sum 26. Maximum digit sum = 9+9+9 = 27, so we must subtract 1 from one digit of 999:
Possible digit sets: (8,9,9), (9,8,9), (9,9,8) → numbers 899, 989, 998 (and permutations: 899, 989, 997... wait).

Actually the numbers are: 899, 989, 998 — no, let me list properly:
- (8,9,9) → 899, 989, 998... wait, (8,9,9) gives 899, 989, 998? No: digits are 8,9,9 so numbers are 899, 989, 998.
Wait: (8,9,9) → 899, 989, 998. But 998 has digits 9,9,8 which is the set (9,9,8). Let me just list all:
- Digits {8,9,9}: 899, 989, 998 — wait, 998 has digits 9,9,8. That IS {8,9,9}. So: 899, 989, 998.

Hmm, let me redo. The three numbers with digit sum 26 are:
- 899 (8+9+9=26)
- 989 (9+8+9=26)
- 998 (9+9+8=26)

Wait, that's only 3 numbers total (since two digits are the same, there are 3!/2! = 3 arrangements).

Even numbers must end in an even digit. Only 998 ends in 8 (even). 899 and 989 end in 9 (odd).

So exactly **1** even number.

ANSWER 8: A

---

Problem 9:
Three-digit numbers a, b, c with a + b + c = 25, where 1 ≤ a ≤ 9, 0 ≤ b, c ≤ 9.

Enumerate by first digit a:
- a = 9: b + c = 16 → (7,9), (8,8), (9,7) → 979, 988, 997 (3 numbers)
- a = 8: b + c = 17 → (8,9), (9,8) → 889, 898 (2 numbers)
- a = 7: b + c = 18 → (9,9) → 799 (1 number)
- a ≤ 6: b + c ≥ 19, but max b + c = 18 → impossible

Total = 3 + 2 + 1 = **6**.

ANSWER 9: C

---

Problem 10:
Exits 5 through 26 give 22 exits and 21 consecutive gaps. Total distance = 118 miles. Each gap ≥ 5 miles.

To maximize one gap, minimize the other 20 gaps to 5 miles each:
Max gap + 20(5) = 118
Max gap + 100 = 118
Max gap = **18** miles.

ANSWER 10: C

---

Problem 11:
Five scores sorted: a ≤ b ≤ c ≤ d ≤ e.
- Mean = 90 → sum = 450
- Median = 91 → c = 91
- Mode = 94 → 94 appears most frequently (at least twice)

Since a ≤ b ≤ c = 91, neither a nor b can be 94. So 94 must appear among d and e. For 94 to be the mode, we need d = e = 94 (appearing twice).

Sum: a + b + 91 + 94 + 94 = 450
a + b = 450 − 279 = **171**

(We can verify valid scores 

 Problem 1:
We are given a set \(R\) of nine distinct integers, six of which are \(2,3,4,6,9,14\). We must determine how many different values are possible for the median of \(R\).

Because there are nine distinct integers, the median is the fifth smallest element when the set is listed in increasing order. Let the three unknown distinct integers be chosen freely (as long as they are not equal to any of the six given numbers). We examine two cases: when the median is one of the six known numbers and when it is one of the three unknown numbers.

**Case 1: The median is a known number.**
Let the median be \(m\). Of the six known numbers, let \(L\) be how many are less than \(m\). Then there are \(5-L\) known numbers greater than \(m\). To make \(m\) the fifth element, we need exactly four numbers smaller than \(m\) and four larger. Thus we need \(4-L\) unknown numbers below \(m\) and \(4-(5-L)=L-1\) unknown numbers above \(m\). This uses exactly \((4-L)+(L-1)=3\) unknowns, which is perfect. We need \(4-L\ge 0\) and \(L-1\ge 0\), i.e. \(1\le L\le 4\).

- \(m=2\): \(L=0\) → impossible.
- \(m=3\): \(L=1\) → need 3 unknowns \(<3\), 0 above. Possible (e.g. \(-1,0,1\)).
- \(m=4\): \(L=2\) → need 2 unknowns \(<4\), 1 above. Possible (e.g. \(0,1,5\)).
- \(m=6\): \(L=3\) → need 1 unknown \(<6\), 2 above. Possible (e.g. \(5,7,8\)).
- \(m=9\): \(L=4\) → need 0 unknowns \(<9\), 3 above. Possible (e.g. \(10,11,12\)).
- \(m=14\): \(L=5\) → need \(-1\) unknowns below; impossible.

So the known possible medians are \(3,4,6,9\).

**Case 2: The median is an unknown number \(x\).**
Then \(x\) itself uses one of the three unknown slots. We need four numbers below \(x\) and four above. With \(L\) known numbers below \(x\), we need \(4-L\) other unknowns below \(x\) and \(4-(6-L)=L-2\) other unknowns above \(x\). The total other unknowns used is \((4-L)+(L-2)=2\), which matches the two remaining unknowns. We need \(4-L\ge 0\) and \(L-2\ge 0\), so \(2\le L\le 4\).

- \(L=2\): known below are \(2,3\); known above are \(4,6,9,14\). Then \(3<x<4\), impossible for an integer.
- \(L=3\): known below are \(2,3,4\); known above are \(6,9,14\). Then \(4<x<6\), so \(x=5\). This works (e.g. add \(1,5,7\)).
- \(L=4\): known below are \(2,3,4,6\); known above are \(9,14\). Then \(6<x<9\), so \(x=7\) or \(x=8\). Both work (e.g. add \(7,10,11\) or \(8,10,11\)).

Thus the unknown possible medians are \(5,7,8\).

Collecting all possibilities: \(\{3,4,5,6,7,8,9\}\). That is **7** different values. The tempting answer of 4 counts only the known medians, and 2 or 14 are impossible because too many numbers would have to be forced to one side.

ANSWER 1: D

---

Problem 2:
We insert a number \(N\) into the list \(2,6,7,7,28\) so that the new mean of the six numbers is twice the new median. Find \(N\).

The original sum is \(2+6+7+7+28=50\). After insertion, the sum is \(50+N\) and the mean is \(\frac{50+N}{6}\).

The median depends on the size of \(N\) relative to the ordered list. Because the list contains six numbers, the median is the average of the third and fourth values.

- If \(N\le 6\) (as an integer), the ordered list begins with \(N\) or \(2\), then the third and fourth terms are \(6\) and \(7\), giving median \(\frac{6+7}{2}=6.5\).
  Setting mean \(=2\times 6.5=13\): \(\frac{50+N}{6}=13\Rightarrow N=28\). But \(N=28\) does not satisfy \(N\le 6\), so this case yields no valid solution.

- If \(N\ge 7\) (integer), the third and fourth terms are both \(7\) (since the two \(7\)’s occupy those middle slots, or \(N=7\) makes three \(7\)’s), so the median is \(7\).
  Setting mean \(=2\times 7=14\): \(\frac{50+N}{6}=14\Rightarrow 50+N=84\Rightarrow N=34\).

Check: the list becomes \(2,6,7,7,28,34\). The median is \(\frac{7+7}{2}=7\). The mean is \(\frac{84}{6}=14\), which is indeed twice the median. The other choices fail: \(N=28\) gives mean \(13\) and median \(7\); \(N=14\) gives mean \(\frac{64}{6}\neq 14\).

ANSWER 2: E

---

Problem 3:
We consider all positive four-digit integers whose digits are all even. We need the fraction of them that are divisible by \(4\).

Even digits are \(\{0,2,4,6,8\}\). For a four-digit number \(ABCD\):
- \(A\) (thousands) has 4 choices: \(\{2,4,6,8\}\).
- \(B,C,D\) each have 5 choices.

Total count: \(4\cdot 5\cdot 5\cdot 5 = 500\).

A number is divisible by \(4\) exactly when its last two digits form a multiple of \(4\). Here \(C\) and \(D\) are both even. Consider \(10C+D \pmod 4\). Since \(C\) is even, write \(C=2k\); then \(10C=20k\equiv 0\pmod 4\). Thus \(10C+D\equiv D\pmod 4\). So we simply need \(D\equiv 0\pmod 4\).

Among the even digits, those divisible by \(4\) are \(0,4,8\): **3 choices** for \(D\). The digits \(A,B,C\) are unrestricted (4, 5, and 5 choices).

Count of favorable numbers: \(4\cdot 5\cdot 5\cdot 3 = 300\).

The fraction is \(\frac{300}{500}=\frac{3}{5}\). One might mistakenly think the fraction is \(\frac12\) because half of all integers are divisible by \(4\), but the restriction to even digits changes the units-digit distribution.

ANSWER 3: D

---

Problem 4:
Find the difference between a \(6.5\%\) sales tax and a \(6\%\) sales tax on a \(\$20\) item.

Method 1: Compute each tax.
- \(6.5\%\) of \(\$20 = 0.065\times 20 = \$1.30\).
- \(6\%\) of \(\$20 = 0.06\times 20 = \$1.20\).

Difference: \(\$1.30-\$1.20 = \$0.10\).

Method 2: The difference in rates is \(0.5\%\), and \(0.5\%\) of \(\$20 = 0.005\times 20 = \$0.10\).

The distractors correspond to misplacing the decimal (e.g. \(\$1\) would be \(5\%\) of \(\$20\)).

ANSWER 4: B

---

Problem 5:
Evaluate \((1901+1902+\dots+1993)-(101+102+\dots+193)\).

Each sum has the same number of terms:
\(1993-1901+1 = 93\) terms, and \(193-101+1 = 93\) terms.

Pair the terms:
\((1901-101)+(1902-102)+\dots+(1993-193)\).

Each difference equals \(1800\). With 93 such pairs, the total is
\(93\times 1800 = 93\times 18\times 100 = 1674\times 100 = 167{,}400\).

A common error is miscounting the number of terms as 92, which would give \

Problem 1:
R is a set of nine distinct integers with six known elements: 2, 3, 4, 6, 9, 14. Since R has 9 elements, the median is the 5th element when sorted.

I need to determine which integers can be the 5th element. The median must have exactly 4 elements less than it and 4 elements greater than it.

**Checking each candidate:**

- **Median = 3**: Need 4 elements < 3. Known: {2}. Need 3 unknowns all < 3 (e.g., 1, 0, -1). Elements > 3: {4, 6, 9, 14} = 4. ✓
- **Median = 4**: Need 4 elements < 4. Known: {2, 3}. Need 2 unknowns < 4. Elements > 4: {6, 9, 14} + 1 unknown > 4. ✓
- **Median = 5**: Need 4 elements < 5. Known: {2, 3, 4}. Need 1 unknown < 5. Elements > 5: {6, 9, 14} + 1 unknown > 5. (5 is an unknown) ✓
- **Median = 6**: Need 4 elements < 6. Known: {2, 3, 4}. Need 1 unknown < 6. Elements > 6: {9, 14} + 2 unknowns > 6. ✓
- **Median = 7**: Need 4 elements < 7. Known: {2, 3, 4, 6} = 4. Elements > 7: {9, 14} + 2 unknowns > 7. (7 is an unknown) ✓
- **Median = 8**: Need 4 elements < 8. Known: {2, 3, 4, 6} = 4. Elements > 8: {9, 14} + 2 unknowns > 8. (8 is an unknown) ✓
- **Median = 9**: Need 4 elements < 9. Known: {2, 3, 4, 6} = 4. Elements > 9: {14} + 3 unknowns > 9. ✓

**Values that cannot be median:**
- 2: Need 4 elements < 2, but only 3 unknowns available. ✗
- 10-13: Already have 5 known elements less than these. ✗
- 14: Already have 5 known elements less than 14. ✗

Possible median values: {3, 4, 5, 6, 7, 8, 9} → **7 values**

ANSWER 1: D

---

Problem 2:
List: 2, 6, 7, 7, 28. Insert N so that mean = 2 × median.

Sum of original list = 50, so new mean = (50 + N)/6.

**Finding the median based on N:**
- If N > 7: Sorted list is 2, 6, 7, 7, 28, N (or 2, 6, 7, 7, N, 28 if 7 < N < 28). Median = (7+7)/2 = 7.
- If N ≤ 6: Median = (6+7)/2 = 6.5
- If 6 < N ≤ 7: Median = (N+7)/2

**Solving for N > 7:**
(50 + N)/6 = 2(7) = 14
50 + N = 84
N = 34

Verification: List = 2, 6, 7, 7, 28, 34. Mean = 84/6 = 14. Median = 7. ✓

ANSWER 2: E

---

Problem 3:
Four-digit integers with all even digits. Even digits: {0, 2, 4, 6, 8}.

**Total count:** First digit: 4 choices (2,4,6,8), others: 5 choices each → 4 × 5³ = 500

**Divisible by 4:** Last two digits must form a number divisible by 4.

Listing all 25 pairs of even last digits and checking divisibility by 4:
- 00✓, 02✗, 04✓, 06✗, 08✓
- 20✓, 22✗, 24✓, 26✗, 28✓
- 40✓, 42✗, 44✓, 46✗, 48✓
- 60✓, 62✗, 64✓, 66✗, 68✓
- 80✓, 82✗, 84✓, 86✗, 88✓

That's 15 valid pairs.

**Count divisible by 4:** 4 × 5 × 15 = 300

**Fraction:** 300/500 = 3/5

ANSWER 3: D

---

Problem 4:
6.5% tax on $20 = 0.065 × 20 = $1.30
6% tax on $20 = 0.06 × 20 = $1.20

Difference = $1.30 - $1.20 = $0.10

ANSWER 4: B

---

Problem 5:
(1901 + 1902 + ... + 1993) − (101 + 102 + ... + 193)

Both sums have 93 terms each.

First sum = 93 × (1901 + 1993)/2 = 93 × 1947
Second sum = 93 × (101 + 193)/2 = 93 × 147

Difference = 93 × (1947 − 147) = 93 × 1800 = 167,400

ANSWER 5: A

---

Problem 6:
1 − 2 − 3 + 4 + 5 − 6 − 7 + 8 + ... + 1993 − 1994 − 1995 + 1996

The pattern repeats every 4 terms: +, −, −, +

Group k: (4k−3) − (4k−2) − (4k−1) + 4k = 4k−3−4k+2−4k+1+4k = 0

Each group of 4 sums to 0.

1996 ÷ 4 = 499 complete groups.

Total = 499 × 0 = 0

ANSWER 6: C

---

Problem 7:
50 people wear sunglasses, 35 wear caps.

P(sunglasses | cap) = 2/5 means: (both)/(caps) = 2/5

Both/35 = 2/5 → Both = 14

P(cap | sunglasses) = (both)/(sunglasses) = 14/50 = 7/25

ANSWER 7: B

---

Problem 8:
3-digit numbers with digit-sum 26 that are even.

For number abc: a + b + c = 26, c is even, a ≥ 1, all digits ≤ 9.

- c = 8: a + b = 18 → only (9,9) → 998 ✓
- c = 6: a + b = 20 → impossible (max 9+9=18)
- c = 4, 2, 0: a + b ≥ 22 → impossible

Only 1 number: 998

ANSWER 8: A

---

Problem 9:
Three-digit numbers with digit-sum 25.

a + b + c = 25, a ≥ 1, all digits ≤ 9.

- a = 9: b + c = 16 → (7,9), (8,8), (9,7) → 3 numbers
- a = 8: b + c = 17 → (8,9), (9,8) → 2 numbers
- a = 7: b + c = 18 → (9,9) → 1 number
- a ≤ 6: b + c ≥ 19 → impossible

Total = 3 + 2 + 1 = 6

ANSWER 9: C

---

Problem 10:
Distance from exit 5 to exit 26 is 118 miles. There are 21 gaps between consecutive exits (5→6, 6→7, ..., 25→26).

Each gap ≥ 5 miles. To maximize one gap, minimize all others.

If 20 gaps are minimized at 5 miles each: 20 × 5 = 100 miles.

Maximum single gap = 118 − 100 = 18 miles.

ANSWER 10: C

---

Problem 11:
Five scores: mean = 90, median = 91, mode = 94.

Sum = 5 × 90 = 450

Sorted: a ≤ b ≤ c ≤ d ≤ e
- c = 91 (median)
- Mode = 94, so 94 appears at least twice. Since c = 91, the 94s must be d and e.

a + b + 91 + 94 + 94 = 450
a + b = 171

ANSWER 11: B

---

Problem 12:
Using +, −, × each exactly once in 5 __ 4 __ 6 __ 3:

| Operations | Result |
|------------|--------|
| 5 + 4 − 6 × 3 | 5 + 4 − 18 = −9 |
| 5 + 4 × 6 − 3 | 5 + 24 − 3 = 26 |
| 5 − 4 + 6 × 3 | 5 − 4 + 18 = 19 |
| 5 − 4 × 6 + 3 | 5 − 24 + 3 = −16 |
| 5 × 4 + 6 − 3 | 20 + 6 − 3 = 23 |
| 5 × 4 − 6 + 3 | 20 − 6 + 3 = 17 |

From choices {9, 10, 15, 16, 19}, only **19** is achievable.

ANSWER 12: E

**Problem 1:**  
We have nine distinct integers, six of which are known: 2, 3, 4, 6, 9, 14.  
The median of a nine‑element set is the 5th smallest number.  

Let the unknown integers be \(x,y,z\) (all distinct and distinct from the known six).  
When the numbers are sorted, the median is the 5th term.  

For a known number \(k_i\) (the \(i\)‑th smallest of the known six) to be the median, we need exactly four numbers smaller than it.  
If \(i-1\) known numbers are smaller, we need \(L\) unknown numbers smaller with  
\((i-1)+L=4\Rightarrow L=5-i\).  
Since we have only three unknowns, \(L\ge0\) gives \(i\le5\); also \(L\le3\) gives \(i\ge2\).  
Thus \(i=2,3,4,5\) → medians could be 3, 4, 6, 9.

Can the median be an unknown integer?  
Let the median be an unknown \(u\).  
If \(u\) lies between the known numbers, we need exactly four numbers below it.  
With two other unknowns, the possibilities are:
- \(u\) between 4 and 6 → \(u=5\);
- \(u\) between 6 and 9 → \(u=7\) or \(u=8\).

All three of these can be realized while keeping the numbers distinct.  
Thus the possible median values are \(\{3,4,5,6,7,8,9\}\), a total of **7** values.

**Answer:** 7 → Choice **D**.  

ANSWER 1: D  

---

**Problem 2:**  
We start with the list 2, 6, 7, 7, 28 and insert a number \(N\).  
Let the six numbers be sorted. The median of six numbers is the average of the 3rd and 4th smallest values.

Let the sorted list after insertion be \(a_1\le a_2\le a_3\le a_4\le a_5\le a_6\).  
The median is \(\dfrac{a_3+a_4}{2}\).  
The mean is \(\dfrac{\text{sum}}{6}\).  
Given that the mean equals twice the median:
\[
\frac{\text{sum}}{6}=2\cdot\frac{a_3+a_4}{2}=a_3+a_4 .
\]
Hence \(\text{sum}=6(a_3+a_4)\).

Consider where \(N\) can fall:

*If \(N\le2\):* \(a_3=6,\ a_4=7\) → median \(=6.5\).  
Sum \(=N+50\). Equation gives \(N=28\), contradicting \(N\le2\).

*If \(2<N\le6\):* \(a_3=6,\ a_4=7\) → same median \(=6.5\).  
Again \(N=28\), impossible.

*If \(6<N\le7\):* only integer possibility \(N=7\).  
List becomes 2, 6, 7, 7, 7, 28 → median \(=7\).  
Sum \(=57\) → mean \(=9.5\neq14\).

*If \(7<N<28\):* \(a_3=7,\ a_4=7\) → median \(=7\).  
Sum \(=N+50\). Equation \(N+50=84\) gives \(N=34\), which is not \(<28\).

*If \(N=28\):* median still \(=7\). Sum \(=78\) → mean \(=13\neq14\).

*If \(N>28\):* \(a_3=7,\ a_4=7\) → median \(=7\).  
Equation gives \(N+50=84\) → \(N=34\). This works (34 > 28).

Thus the only possible value is **34**.

**Answer:** 34 → Choice **E**.  

ANSWER 2: E  

---

**Problem 3:**  
A four‑digit integer with all digits even can use digits \{0,2,4,6,8\} for each place, except the thousands digit cannot be 0.  

Number of such integers: \(4\times5\times5\times5=500\).

A number is divisible by 4 iff its last two digits form a number divisible by 4.  
List all two‑digit endings with even digits (25 possibilities).  

Checking each yields 15 endings that are multiples of 4:  
00, 04, 08, 20, 24, 28, 40, 44, 48, 60, 64, 68, 80, 84, 88.

Thus the proportion is \(\frac{15}{25}=\frac{3}{5}\).

**Answer:** \(\frac{3}{5}\) → Choice **D**.  

ANSWER 3: D  

---

**Problem 4:**  
6 % tax on \$20 = \(0.06\times20 = \$1.20\).  
6.5 % tax on \$20 = \(0.065\times20 = \$1.30\).  

Difference = \$1.30 − \$1.20 = **\$0.10**.

**Answer:** \$0.10 → Choice **B**.  

ANSWER 4: B  

---

**Problem 5:**  
Sum from 1901 to 1993 (inclusive):  
Number of terms = \(1993-1901+1 = 93\).  
Sum = \(\frac{(1901+1993)\times93}{2}= \frac{3894\times93}{2}=181\,071\).

Sum from 101 to 193 (inclusive):  
Number of terms = \(193-101+1 = 93\).  
Sum = \(\frac{(101+193)\times93}{2}= \frac{294\times93}{2}=13\,671\).

Difference = \(181\,071-13\,671 = 167\,400\).

**Answer:** 167 400 → Choice **A**.  

ANSWER 5: A  

---

**Problem 6:**  
The sign pattern for the numbers \(n\) is:
- \(n\equiv1\pmod4\) → +,
- \(n\equiv2\pmod4\) → −,
- \(n\equiv3\pmod4\) → −,
- \(n\equiv0\pmod4\) → +.

Thus each block of four consecutive integers \(\{4k+1,4k+2,4k+3,4k+4\}\) contributes  
\((4k+1)-(4k+2)-(4k+3)+(4k+4)=0\).

From 1 through 1996 there are \(1996/4=499\) complete blocks, so the total sum is **0**.

**Answer:** 0 → Choice **C**.  

ANSWER 6: C  

---

**Problem 7:**  
Let \(S\) = people wearing sunglasses, \(|S|=50\).  
Let \(C\) = people wearing caps, \(|C|=35\).  
Let \(x=|S\cap C|\).

Given \(P(S\mid C)=\frac{x}{35}= \frac{2}{5}\).  
Thus \(x = \frac{2}{5}\times35 = 14\).

We need \(P(C\mid S)=\frac{x}{50}= \frac{14}{50}= \frac{7}{25}\).

**Answer:** \(\frac{7}{25}\) → Choice **B**.  

ANSWER 7: B  

---

**Problem 8:**  
A three‑digit number with digit sum 26 must use digits that sum to 26.  
The only triple of single‑digit numbers with sum 26 is \(\{9,9,8\}\).

The number is even only if its units digit is 8.  
Among the permutations of 9, 9, 8, only 998 ends in 8.  

Thus exactly **1** such number exists.

**Answer:** 1 → Choice **A**.  

ANSWER 8: A  

---

**Problem 9:**  
Let the digits be \(a,b,c\) (hundreds, tens, units) with  
\(1\le a\le9,\ 0\le b,c\le9,\ a+b+c=25\).

Set \(a'=a-1\) (so \(0\le a'\le8\)). Then  
\(a'+b+c=24\).

Count unrestricted non‑negative solutions: \(\binom{24+3-1}{2}= \binom{26}{2}=325\).

Subtract solutions where a digit exceeds its bound:
- \(a'\ge9\): treat \(a''=a'-9\) → \(a''+b+c=15\) → \(\binom{17}{2}=136\).
- \(b\ge10\): \(b'=b-10\) → \(a'+b'+c=14\) → \(\binom{16}{2}=120\).
- \(c\ge10\): similarly 120.

Intersections:
- \(a'\ge9,\ b\ge10\): \(a''+b'+c=5\) → \(\binom{7}{2}=21\).
- \(a'\ge9,\ c\ge10\): 21.
- \(b\ge10,\ c\ge10\): \(a'+b'+c'=4\) → \(\binom{6}{2}=15\).

No triple intersection (would require sum –5).

By inclusion–exclusion, invalid = \(136+120+120-21-21-15 = 319\).  
Valid = \(325-319 = 6\).

**Answer:** 6 → Choice **C**.  

ANSWER 9: C  

---

**Problem 10:**  
Exits numbered 5 through 26 give 22 exits, hence 21 intervals.  
Let the distances between consecutive exits be \(d_1,\dots,d_{21}\).  
We know \(\sum d_i = 118\) miles, and each \(d_i\ge5\).

To maximize the largest interval, make the othe

Problem 1:  
We are given a set $ R $ of nine distinct integers, six of which are $ 2, 3, 4, 6, 9, 14 $. The median of a set of nine numbers is the fifth number when sorted. To find the number of possible values for the median, we consider the possible placements of the three unknown numbers relative to the known ones. The known numbers sorted are $ 2, 3, 4, 6, 9, 14 $. The three unknown numbers can be placed in various positions to shift the median. The possible medians occur when the additional numbers are placed such that the fifth number in the sorted list can be $ 4, 5, 6, 7, 8, $ or $ 9 $. This gives six possible values for the median.  
ANSWER 1: D (corrected to C? Wait, no: the possible medians are 4, 5, 6, 7, 8, 9 — that’s 6 values, so C)  
**Wait, let's re-evaluate:** The known numbers are 2,3,4,6,9,14. Adding three numbers. The median is the 5th number. To get 4 as median: three numbers ≤4, but only 2,3,4 are available — but they are distinct, so we can add numbers like 1,5? Wait no, the three unknown numbers can be placed in different positions. For example:  
- If all three unknown numbers are ≤4: but only three such numbers exist (2,3,4) but they are already in the set. So we cannot add more. Thus, the unknown numbers must be placed in other positions.  
Actually, the unknown numbers can be:  
- All less than 2: median could be 4 if three numbers are added <4, but only 2,3,4 are already present. We cannot add numbers less than 2 without repeating? Wait, the set has distinct integers. So we can add numbers like 1,0,-1, but the problem doesn’t restrict to positive. However, the known numbers are positive. But the problem says "distinct integers", so negatives are allowed. But the median values: if we add numbers less than 2, the sorted order becomes [new numbers, 2,3,4,6,9,14]. With three new numbers, the sorted list has 9 numbers. The fifth is the median. If we add three numbers less than 2, the sorted list is [a,b,c,2,3,4,6,9,14] — no, wait, adding three numbers less than 2: the sorted list would be [a,b,c,2,3,4,6,9,14] — but a,b,c are less than 2, so the first three, then 2 is fourth, 3 is fifth? Wait, no: the original known numbers are 2,3,4,6,9,14. Adding three numbers, say x,y,z. The full sorted list will be the combination. To get median=4, we need the fifth number to be 4. So in the sorted list of 9, the fifth must be 4. That requires that four numbers are ≤4 and four are ≥4? No: the fifth is the middle. For the fifth to be 4, we need at least five numbers ≤4? No: in a sorted list of 9, the fifth is the median. For the fifth to be 4, we need at least five numbers ≤4? Actually, no: the fifth number is the one where four are less and four are greater? No: in a sorted list of 9, the fifth is the median, meaning four are ≤ median and four are ≥ median. But since numbers can be equal, but here all distinct. So for 4 to be the median, we need exactly four numbers less than 4 and four numbers greater than 4? But 4 is in the set. So the sorted list must have four numbers <4, then 4, then four numbers >4. But the known numbers include 2,3,4,6,9,14. So 2,3 are <4, and 6,9,14 are >4. So currently, without unknowns, we have two <4 and three >4. Adding three unknowns. To have four <4 and four >4, we need to add two numbers <4 and one number >4? But then the sorted list would be [two new <4, 2,3,4, then the new >4, then 6,9,14] — but that’s only three numbers after 4? Wait, no: the sorted list would be [new <4 (two), 2,3,4, new >4, 6,9,14] — that’s 9 numbers? 2+3+1+3=9? Wait, no: the two new <4, then 2,3,4 (three numbers), then the new >4, then 6,9,14 (three numbers). So total: 2+3+1+3=9. The fifth number is 4. So yes, we can achieve median=4 by adding two numbers <4 and one number >4. Similarly, we can achieve:  
- Median=5: by adding numbers such that the fifth number is 5. But 5 is not in the set. So we need the fifth number to be 5. This requires that four numbers are <5 and four are >5. But 5 is not present. So we need to add 5? But the set must consist of distinct integers, and 5 is not in the given. So we can add 5. Then the sorted list would include 5. For example, add 5, and two other numbers. To have four <5 and four >5: currently, without unknowns, numbers <5: 2,3,4 (three), numbers >5: 6,9,14 (three). So to have four <5, we need to add one number <5. To have four >5, we need to add one number >5. But we are adding three numbers. So we can add 5, and say 1 and 10. Then sorted: 1,2,3,4,5,6,9,10,14. Median is 5. So median=5 is achievable. Similarly, median=6: we can add numbers such that the fifth is 6. Currently, numbers <6: 2,3,4 (three), numbers >6: 9,14 (two). So to have four <6 and four >6, we need to add one number <6 and two numbers >6. For example, add 5,10,15. Then sorted: 2,3,4,5,6,9,10,14,15. Median=6. Similarly, median=7: add numbers so fifth is 7. Currently, <7: 2,3,4,6 (four), >7: 9,14 (two). So to have four <7 and four >7, we need to add zero <7 (since already four) and two >7. But we are adding three numbers. So we can add 7,8,15. Then sorted: 2,3,4,6,7,8,9,14,15. Median=7. Similarly, median=8: add 8, and two numbers, say 10,11. Then sorted: 2,3,4,6,8,9,10,11,14. Median=8? The fifth is 8. But to have four <8 and four >8: currently, without unknowns, <8: 2,3,4,6 (four), >8:9,14 (two). So to have four >8, we need to add two numbers >8. But we are adding three numbers. So we can add 8,9,10? But 9 is already there. Add 8,10,11. Then sorted: 2,3,4,6,8,9,10,11,14. The fifth is 8. So yes. Similarly, median=9: add numbers so fifth is 9. Currently, <9: 2,3,4,6 (four), >9:14 (one). So to have four >9, we need to add three numbers >9. But we are adding three numbers. So add 10,11,12. Then sorted: 2,3,4,6,9,10,11,12,14. Median=9. So possible medians: 4,5,6,7,8,9. That’s six values.  
ANSWER 1: C  

Problem 2:  
We are given the list 2,6,7,7,28. A number $ N $ is added. The mean becomes twice the median. The original list has five numbers. After adding $ N $, there are 

### Problem 1:
We have a set of 9 distinct integers, so the median (middle value when sorted) is the 5th element. The 6 known numbers sorted are [2,3,4,6,9,14], with 3 additional distinct integers to add. To find possible median values:
- The minimum possible median is 3: add 3 numbers smaller than 2, sorted set becomes [a,b,c,2,3,4,6,9,14], 5th term=3.
- The maximum possible median is 9: add 3 numbers larger than 14, sorted set becomes [2,3,4,6,9,14,x,y,z], 5th term=9.
- All integers between 3 and 9 are valid medians: 3,4,5,6,7,8,9, for 7 total possible values.
Tempting wrong choices: any count less than 7, from missing intermediate medians that can be formed by adding integers between the known values.
ANSWER 1: D

---

### Problem 2:
Original list sum: 2+6+7+7+28=50. Adding N gives a list of 6 elements, sum=50+N, mean=(50+N)/6. The median of 6 elements is the average of the 3rd and 4th sorted terms. The problem states mean=2×median, so 50+N=12×median.
- If N>28, sorted list is [2,6,7,7,28,N], median=(7+7)/2=7. Then 50+N=12×7=84 → N=34, which fits.
Testing other options confirms no other N satisfies the condition.
ANSWER 2: E

---

### Problem 3:
A number is divisible by 4 if its last two digits form a number divisible by 4. For 4-digit numbers with all even digits:
- Total count: 4 (non-zero thousands digits:2,4,6,8) ×5×5×5 (all even digits for other places)=500.
- Valid last two-digit pairs (both even, divisible by 4): For any even tens digit, 10×even is divisible by 4, so only units digits 0,4,8 work, giving 3 valid units digits per tens digit: 5×3=15 valid pairs.
- Fraction: (4×5×15)/500=300/500=3/5.
ANSWER 3: D

---

### Problem 4:
The tax rate difference is 6.5%-6%=0.5%. 0.5% of $20 is 0.005×20=$0.10.
ANSWER 4: B

---

### Problem 5:
Both sums have the same number of terms: 1993-1901+1=93 terms, 193-101+1=93 terms. Subtract term-by-term: (1901-101)+(1902-102)+...+(1993-193)=1800×93=167400.
ANSWER 5: A

---

### Problem 6:
Group terms in sets of 4: (1-2-3+4)+(5-6-7+8)+... each group sums to 0. There are 1996/4=499 groups, total sum=0.
ANSWER 6: C

---

### Problem 7:
Number of people with both caps and sunglasses: 35 cap-wearers × 2/5=14. Probability a sunglass-wearer has a cap:14/50=7/25.
ANSWER 7: B

---

### Problem 8:
An even 3-digit number with digit sum 26 must have an even units digit. The only possible even units digit is 8 (any smaller even digit leaves a sum of the first two digits >18, impossible). The only number is 998, so 1 valid number.
ANSWER 8: A

---

### Problem 9:
Three-digit numbers with digit sum 25 are all permutations of (9,9,7) and (9,8,8). There are 3 permutations of (9,9,7) and 3 permutations of (9,8,8), for 6 total numbers.
ANSWER 9: C

---

### Problem 10:
Between exit 5 and 26, there are 21 gaps. To maximize one gap, minimize the other 20 gaps at 5 miles each: 20×5=100 miles. Maximum gap=118-100=18 miles.
ANSWER 10: C

---

### Problem 11:
Total sum of 5 scores=5×90=450. Median=91, so sorted scores [a,b,91,c,d]. Mode=94, so 94 must appear at least twice, so c=d=94, sum c+d=188. Sum of the two lowest scores=450-91-188=171.
ANSWER 11: B

---

### Problem 12:
Using +,-,* exactly once, one valid calculation: 5-4+6×3=1+18=19, which uses all three operations and matches option E.
ANSWER 12: E

### Problem 1:
We need to find the number of possible median values for a set of 9 distinct integers, 6 of which are 2, 3, 4, 6, 9, 14. The median of 9 numbers is the 5th element when sorted in ascending order. We add 3 distinct integers (not in the given set) and analyze possible medians:
1.  **Case 1: All 3 new numbers < 2**: Sorted list is [a,b,c,2,3,4,6,9,14], median = 3.
2.  **Case 2: 2 new numbers < 2, 1 new number ≥4**: Sorted list is [a,b,2,3,4,...], median =4.
3.  **Case 3: 1 new number <2**:
    - If we add 5 (the only integer between 4 and 6): Sorted list is [a,2,3,4,5,6,...], median=5.
    - If we do not add 5: Sorted list is [a,2,3,4,6,...], median=6.
4.  **Case 4: 0 new numbers <2 (all new numbers ≥5)**:
    - Add 5: Sorted list [2,3,4,5,6,...], median=6.
    - Add a number between 6 and9 (7 or 8): Sorted list [2,3,4,6,7,...] or [2,3,4,6,8,...], median=7 or 8.
    - Add all new numbers ≥9: Sorted list [2,3,4,6,9,...], median=9.
The only impossible medians are 2 (needs 5 numbers ≤2, only 4 max) and values >9 (existing numbers already have 5 elements ≤9). The possible medians are 3,4,5,6,7,8,9: 7 values.
ANSWER 1: D
---
### Problem 2:
Original list: [2,6,7,7,28], sum = 2+6+7+7+28=50. Insert N to make 6 elements. The median of 6 elements is the average of the 3rd and 4th sorted elements. The mean is $\frac{50+N}{6}$, which equals twice the median:
$$\frac{50+N}{6} = 2 \times \text{median} \implies 50 + N = 12 \times \text{median} \implies N = 12 \times \text{median} -50.$$
Test possible median values from the sorted list with N:
- If N ≤7: Sorted list has 3rd=7, 4th=7, median=7. Then $N=12*7 -50=84-50=34$, which is >7, valid.
- Other median values (e.g., 6.5) give N=28, which does not fit the required sorted position.
Verify N=34: Sorted list [2,6,7,7,28,34], median=(7+7)/2=7, mean=(50+34)/6=14, which is twice 7.
ANSWER 2: E
---
### Problem 3:
Positive 4-digit numbers with all even digits:
- Thousands place: 2,4,6,8 (4 choices, no 0).
- Hundreds, tens, units: 0,2,4,6,8 (5 choices each).
Total numbers: $4 \times 5 \times 5 \times5=500$.
A number is divisible by 4 iff its last two digits form a number divisible by 4. For even tens digits (0,2,4,6,8), a two-digit number $10a + b$ is divisible by 4 iff $b$ (units digit) is 0,4,8 (since $10a \equiv 0 \mod4$ for even a). So valid endings: 5 (tens choices) ×3 (units choices)=15.
Total valid numbers: $4 \times5 \times5 \times3=300$. Fraction: $\frac{300}{500}=\frac{3}{5}$.
ANSWER 3: D
---
### Problem 4:
Difference between 6.5% and 6% tax on $20:
$$0.065*20 - 0.06*20 = (0.065-0.06)*20=0.005*20=0.10.$$
ANSWER 4: B
---
### Problem 5:
Both sums have 93 terms (1993-1901+1=93, 193-101+1=93). Each term in the first sum is 1800 more than the corresponding term in the second sum:
Total difference = $93 \times 1800 = 167400$.
ANSWER 5: A
---
### Problem 6:
The sign pattern repeats every 4 terms: $+ - - +$, and each block of 4 sums to 0:
$$(1-2-3+4)+(5-6-7+8)+\dots+(1993-1994-1995+1996)=0.$$
1996 is divisible by 4, so there are 499 such blocks, total sum 0.
ANSWER 6: C
---
### Problem 7:
Let $S$ = sunglass wearers ($|S|=50$), $C$ = cap wearers ($|C|=35$). $P(S|C)=\frac{|S∩C|}{|C|}=\frac{2}{5}$, so $|S∩C|=\frac{2}{5}*35=14$.
The probability a sunglass wearer wears a cap is $P(C|S)=\frac{|S∩C|}{|S|}=\frac{14}{50}=\frac{7}{25}$.
ANSWER 7: B
---
### Problem 8:
3-digit numbers with digit sum 26: maximum digit sum is 27 (999), so 26 is 1 less than maximum, meaning digits are permutations of (9,9,8). The only even permutation is 998 (units digit 8). So 1 number.
ANSWER 8: A
---
### Problem 9:
3-digit numbers with digit sum 25: 25 is 2 less than 27, so digits are permutations of (9,9,7) and (9,8,8):
- Permutations of (9,9,7): 997,979,799 (3 numbers).
- Permutations of (9,8,8): 988,898,889 (3 numbers).
Total 6 numbers.
ANSWER 9: C
---
### Problem 10:
Between exit 5 and 26, there are $26-5=21$ gaps, total distance 118 miles. Each gap ≥5 miles. To maximize one gap, minimize the other 20: $20*5=100$, so maximum gap = $118-100=18$.
ANSWER 10: C
---
### Problem 11:
Five sorted scores: $a≤b≤c≤d≤e$. Median $c=91$, mean 90, so total sum $5*90=450$, so $a+b+d+e=450-91=359$. Mode is 94, so 94 appears at least twice. Since $c=91$, 94 can only be in $d$ and $e$, so $d=e=94$. Then $a+b=359-94-94=171$. This sum is fixed (valid examples exist, e.g., 81+90=171).
ANSWER 11: B
---
### Problem 12:
Permute +, -, × in 5 __4 __6 __3, using standard order of operations (multiply first):
- $5 -4 +6×3 = 5-4+18=19$, which is an option.
All other permutations give values not in the choices.
ANSWER 12: E

I'll solve each problem step by step.

---

**Problem 1:**

We have 6 elements: {2, 3, 4, 6, 9, 14}, and need to add 3 more distinct integers to make 9 total.

The median of 9 numbers is the 5th element when arranged in order.

Currently we have 6 numbers. By strategically placing 3 new numbers, we can shift which value becomes the median.

Let me consider what values can be the median (5th position):
- If we add 3 numbers all ≤ 2, the sorted list has our 6 original numbers in positions 4-9, so median is 4
- If we add numbers strategically between and around the original set, we can make the median be: 2, 3, 4, 6, 9, 14, or a new number
- If we add 3 numbers all ≥ 14, the sorted list has our 6 original numbers in positions 1-6, so median is 9

Testing systematically:
- Median = 2: add three numbers < 2 (e.g., -2, -1, 0) ✓
- Median = 3: add two < 2, one between 3 and 4 ✓
- Median = 4: add three < 2, or arranged appropriately ✓
- Median = 6: add one or two strategically ✓
- Median = 9: add three > 14, or arranged appropriately ✓
- Median = 14: add three > 14 ✓
- Median = a new value between existing numbers ✓

The possible medians are: 2, 3, 4, 6, 9, 14, and any new integer we insert. We can get 7 different values.

**ANSWER 1: D**

---

**Problem 2:**

Original list: 2, 6, 7, 7, 28. Sum = 50.

When N is inserted, we have 6 numbers. Let's find the median of 6 numbers (average of 3rd and 4th when sorted).

Sum of all 6 numbers = 50 + N
Mean = (50 + N)/6

We need: Mean = 2 × Median

Let me test by considering where N falls:
- If N ≤ 2: sorted list is N, 2, 6, 7, 7, 28. Median = (6+7)/2 = 6.5. Mean = 2×6.5 = 13, so 50+N = 78, N = 28. Contradiction.
- If 2 < N ≤ 6: sorted list is 2, N, 6, 7, 7, 28. Median = (6+7)/2 = 6.5. Mean = 13, so N = 28. Contradiction.
- If 6 < N ≤ 7: sorted list is 2, 6, N, 7, 7, 28. Median = (N+7)/2. Mean = 2×(N+7)/2 = N+7. So (50+N)/6 = N+7, giving 50+N = 6N+42, so 5N = 8, N = 1.6. Not in range.
- If 7 < N ≤ 28: sorted list is 2, 6, 7, N, 7, 28 or 2, 6, 7, 7, N, 28. Median = (7+N)/2. Mean = 2×(7+N)/2 = 7+N. So (50+N)/6 = 7+N, giving 50+N = 42+6N, so 5N = 8, N = 1.6. Not valid.
- If N > 28: sorted list is 2, 6, 7, 7, 28, N. Median = (7+7)/2 = 7. Mean = 14. So 50+N = 84, N = 34. ✓

**ANSWER 2: E**

---

**Problem 3:**

Four-digit numbers with all even digits: first digit ∈ {2,4,6,8}, others ∈ {0,2,4,6,8}.

Total count: 4 × 5 × 5 × 5 = 500

For divisibility by 4, the last two digits must form a number divisible by 4.

Last two digits from {00, 02, 04, 06, 08, 20, 22, 24, 26, 28, 40, 42, 44, 46, 48, 60, 62, 64, 66, 68, 80, 82, 84, 86, 88}

Divisible by 4: 00, 04, 08, 20, 24, 28, 40, 44, 48, 60, 64, 68, 80, 84, 88 (15 out of 25)

Count divisible by 4: 4 × 5 × 15 = 300

But wait, the second digit is 5 choices, third is which determines divisibility by 4.

Actually: 4 × 5 × (number of even two-digit combos divisible by 4) = 4 × 5 × 15 = 300

Hmm, let me recalculate. First digit: 4 choices. Second digit: 5 choices. Last two digits must be divisible by 4: check all pairs from {0,2,4,6,8}×{0,2,4,6,8} giving 15 valid pairs.

Count = 4 × 5 × 15 = 300. Fraction = 300/500 = 3/5.

**ANSWER 3: D**

---

**Problem 4:**

6.5% tax on $20: 0.065 × 20 = $1.30
6% tax on $20: 0.06 × 20 = $1.20
Difference: $1.30 - $1.20 = $0.10

**ANSWER 4: B**

---

**Problem 5:**

Sum from 1901 to 1993: This is 93 consecutive integers starting at 1901.
Sum = (1901 + 1993)/2 × 93 = 1947 × 93 = 181,071

Sum from 101 to 193: This is 93 consecutive integers starting at 101.
Sum = (101 + 193)/2 × 93 = 147 × 93 = 13,671

Difference = 181,071 - 13,671 = 167,400

**ANSWER 5: A**

---

**Problem 6:**

Pattern: 1 - 2 - 3 + 4 + 5 - 6 - 7 + 8 + 9 - 10 - 11 + ...

Group by 4s: (1-2-3+4) + (5-6-7+8) + (9-10-11+12) + ... + (1993-1994-1995+1996)

Each group of 4: (4k+1) - (4k+2) - (4k+3) + (4k+4) = 4k+1-4k-2-4k-3+4k+4 = 0

From 1 to 1996, we have 1996/4 = 499 complete groups, each summing to 0.

Total = 0

**ANSWER 6: C**

---

**Problem 7:**

Let x = people wearing both sunglasses and caps.

P(sunglasses | cap) = x/35 = 2/5, so x = 14.

P(cap | sunglasses) = 14/50 = 7/25.

**ANSWER 7: B**

---

**Problem 8:**

Three-digit even numbers with digit sum 26.

For an even number, last digit ∈ {0, 2, 4, 6, 8}.

If last digit = 0: first two digits sum to 26. Max is 9+9=18. Impossible.
If last digit = 2: first two digits sum to 24. Only 9+9=18 < 24. Impossible.
If last digit = 4: first two digits sum to 22. Impossible (max 18).
If last digit = 6: first two digits sum to 20. Impossible.
If last digit = 8: first two digits sum to 18. This requires 9+9=18. Number: 998. ✓

Only 1 number.

**ANSWER 8: A**

---

**Problem 9:**

Three-digit numbers abc where a+b+c=25, with a ∈ {1,...,9}, b,c ∈ {0,...,9}.

a+b+c = 25 means b+c = 25-a.

For valid solutions: 0 ≤ b+c ≤ 18, so 7 ≤ a ≤ 9.

- a=7: b+c=18. Solutions: (9,9) → 1 way
- a=8: b+c=17. Solutions: (8,9), (9,8) → 2 ways
- a=9: b+c=16. Solutions: (7,9), (8,8), (9,7) → 3 ways

Total = 1 + 2 + 3 = 6

**ANSWER 9: C**

---

**Problem 10:**

Exits from 5th to 26th = 26-5 = 21 segments between consecutive exits.

Each segment ≥ 5 miles. Total distance = 118 miles.

If 20 segments are 5 miles each: 20 × 5 = 100 miles.
Remaining for the largest segment: 118 - 100 = 18 miles.

**ANSWER 10: C**

---

**Problem 11:**

Five scores: mean = 90 (sum = 450), median = 91, mode = 94.

Arranged in order: a, b, c, d, e where c = 91 (median).

Mode = 94 means 94 appears at least twice. Since median is 91, we must have 94 appearing in {d, e}.

Case 1: d = e = 94. Then a + b + 91 + 94 + 94 = 450, so a + b = 171.

Check consistency: we need a ≤ b ≤ 91 and mode is 94 (appears twice). If a or b equals 94, mode wouldn't be unique. So a, b < 91 works. ✓

**ANSWER 11: B**

---

**Problem 12:**

We need to place +, −, × in three blanks: 5 __ 4 __ 6 __ 3

Trying combinations:
- 5 + 4 × 6 − 3 = 5 + 24 − 3 = 26
- 5 × 4 + 6 − 3 = 20 + 6 − 3 = 23
- 5 × 4 − 6 + 3 = 20 − 6 + 3 = 17
- 5 + 4 − 6 × 3 = 5 + 4 − 18 = −9
- 5 − 4 

Problem 1:

We are given a set of 9 distinct integers. Six of them are
\[
2,3,4,6,9,14.
\]
We want the number of possible values of the median of the 9 numbers.

For 9 numbers, the median is the 5th number when the list is ordered.

Now look at the six given numbers in sorted order:
\[
2,3,4,6,9,14.
\]
Since there are 3 other distinct integers, the median can be any number that ends up in the 5th position after inserting those 3 numbers.

Let’s see what values are possible:

- If we place 3 new numbers all very small, then the 5th number could be 4.
- It could also be 6, or 9, or even 14, depending on how many new numbers are inserted before each.
- It cannot be 2 or 3, because there would need to be at least 4 numbers less than or equal to it, but only 3 extra numbers are available and the given numbers already limit this.
- It cannot be any number not among the given six if the median must be one of the existing numbers? Let’s check carefully: since the 3 extra numbers are distinct integers and can be placed anywhere, the 5th number could indeed be one of the inserted numbers, but then it would have to lie between the 4th and 6th numbers. Since the six given numbers already occupy several possible positions, the median can also be a new integer. But because the set is of distinct integers and the question asks how many possible values, we should count all possible medians that can occur.

A cleaner way: the median of 9 distinct integers is the middle element. With 6 fixed numbers, the 3 added numbers can be arranged so that the median is any integer between the 4th smallest and 6th smallest of the fixed numbers, inclusive, but here the possible distinct medians are:
\[
4,5,6,7,8,9
\]
Wait — can all of these really happen? Let’s test.

- Median 4: choose 3 new numbers less than 4.
- Median 5: choose 3 new numbers so that 5 is between 4 and 6 and becomes 5th.
- Median 6: choose 3 new numbers with 6 as middle.
- Median 7, 8, 9 similarly.

So the possible medians are the integers from 4 through 9, which is 6 values.

However, we must match the provided choices. The intended count is the number of possible median values among the listed integers and inserted integers. Since the only restriction is distinct integers, every integer from 4 through 9 can occur.

That gives 6 possible values.

ANSWER 1: C

---

Problem 2:

We insert a number \(N\) into the list
\[
2,6,7,7,28.
\]
Then the mean becomes twice the median.

Let’s analyze the median after inserting \(N\) into 6 numbers total.

The original sorted list is:
\[
2,6,7,7,28.
\]
There are 5 numbers. After inserting one number, there are 6 numbers, so the median is the average of the 3rd and 4th numbers in the sorted list.

We try the choices.

If \(N=7\), the list becomes
\[
2,6,7,7,7,28.
\]
Median:
\[
\frac{7+7}{2}=7.
\]
Mean:
\[
\frac{2+6+7+7+7+28}{6}=\frac{57}{6}=9.5.
\]
Twice the median would be 14, not 9.5, so not 7.

Try \(N=14\):
\[
2,6,7,7,14,28.
\]
Median:
\[
\frac{7+7}{2}=7.
\]
Mean:
\[
\frac{2+6+7+7+14+28}{6}=\frac{64}{6}=\frac{32}{3},
\]
not 14.

Try \(N=20\):
\[
2,6,7,7,20,28.
\]
Median is still 7.
Mean:
\[
\frac{2+6+7+7+20+28}{6}=\frac{70}{6}=\frac{35}{3},
\]
not 14.

Try \(N=28\):
\[
2,6,7,7,28,28.
\]
Median is 7.
Mean:
\[
\frac{2+6+7+7+28+28}{6}=\frac{78}{6}=13,
\]
not 14.

Try \(N=34\):
\[
2,6,7,7,28,34.
\]
Median is 7.
Mean:
\[
\frac{2+6+7+7+28+34}{6}=\frac{84}{6}=14.
\]
Twice the median is
\[
2\cdot 7=14.
\]
This works.

So \(N=34\).

ANSWER 2: E

---

Problem 3:

We consider all positive four-digit integers whose digits are all even.

Even digits are
\[
0,2,4,6,8.
\]
A four-digit number cannot start with 0, so the thousands digit has 4 choices:
\[
2,4,6,8.
\]
Each of the other three digits has 5 choices, so the total number is
\[
4\cdot 5^3.
\]

Now determine how many are divisible by 4.

A number is divisible by 4 if its last two digits form a number divisible by 4.

So we only need to count ordered pairs of even digits \((t,u)\) from \(\{0,2,4,6,8\}\) such that \(10t+u\) is divisible by 4.

Check the possible last two digits:

- \(00\) yes
- \(02\) no
- \(04\) yes
- \(06\) no
- \(08\) yes

- \(20\) yes
- \(22\) no
- \(24\) yes
- \(26\) no
- \(28\) yes

- \(40\) yes
- \(42\) no
- \(44\) yes
- \(46\) no
- \(48\) yes

- \(60\) yes
- \(62\) no
- \(64\) yes
- \(66\) no
- \(68\) yes

- \(80\) yes
- \(82\) no
- \(84\) yes
- \(86\) no
- \(88\) yes

For each tens digit, 3 of the 5 possibilities work, so there are
\[
5\cdot 3=15
\]
valid last-two-digit pairs.

Since the first two digits can be anything allowed:
- thousands digit: 4 choices
- hundreds digit: 5 choices

Total divisible by 4:
\[
4\cdot 5\cdot 15
\]
Total numbers:
\[
4\cdot 5^3 = 4\cdot 125
\]
So the fraction is
\[
\frac{4\cdot 5\cdot 15}{4\cdot 5^3}=\frac{15}{25}=\frac35.
\]

ANSWER 3: D

---

Problem 4:

An item costs \$20 before tax.

Difference between 6.5% tax and 6% tax is
\[
0.5\% = 0.005.
\]
So the difference in tax is
\[
0.005 \times 20 = 0.1
\]
dollar, which is \(\$0.10\).

ANSWER 4: B

---

Problem 5:

We need to compute
\[
(1901+1902+\cdots+1993) - (101+102+\cdots+193).
\]

Notice each term in the second sum is exactly 1800 less than the corresponding term in the first sum:
\[
1901-101=1800,\quad 1902-102=1800,\ \ldots,\ 1993-193=1800.
\]
How many terms are there?

From 1901 to 1993 inclusive:
\[
1993-1901+1=93
\]
terms.

So the difference is
\[
93 \cdot 1800 = 167400.
\]

ANSWER 5: A

---

Problem 6:

We evaluate
\[
1-2-3+4+5-6-7+8+9-10-11+\cdots+1992+1993-1994-1995+1996.
\]

Look at the pattern in blocks of 4:
\[
(1-2-3+4),\ (5-6-7+8),\ (9-10-11+12),\ \ldots
\]
Each block of 4 consecutive integers has sum
\[
n-(n+1)-(n+2)+(n+3)=0
\]
when \(n\equiv 1 \pmod 4\).

So all full blocks cancel.

Now check the last block:
The pattern continues through \(1992+1993-1994-1995+1996\), which is not a full 4-term block. But we can group from 1 to 1996 in blocks of 4:
\[
(1-2-3+4)+(5-6-7+8)+\cdots+(1993-1994-1995+1996).
\]
Each block sums to 0.

Therefore the entire 

Problem 1:  
We have 9 distinct integers. Six of them are fixed: 2, 3, 4, 6, 9, 14. The other 3 are distinct integers of our choice.  
With 9 numbers, the median is the 5th number when sorted.

So the median depends only on how many of the fixed numbers fall below/at various positions, and where we insert the 3 new numbers.

Let’s analyze possible 5th elements.

### Step 1: Sort the fixed numbers
Fixed numbers in increasing order:
2, 3, 4, 6, 9, 14

The median (5th) could be one of these fixed numbers, or one of the new ones.

### Casework by choosing what the 5th number is

#### Can the median be 2?
For 2 to be the median, we need at least 4 numbers ≤ 2, but only 2 itself is ≤ 2 among the fixed set (and new numbers would have to be distinct integers ≤ 2).  
That means we’d need 4 numbers less than or equal to 2, including 2, so we’d need 3 additional distinct integers ≤ 2 (other than 2). That’s possible (e.g., 0, −1, 1).  
But then 2 would be the 5th, meaning exactly 4 numbers are <2. Can we guarantee? Actually, since we can choose any 3 extra integers, we can make exactly 4 numbers less than 2 by picking 3 integers <2 plus none from fixed besides 2. That would make 2 the 5th. However, note 2 is already one of the 9 distinct integers, so it can be the median.  
So potentially 2 is possible.

But the problem asks for the number of possible median values, and the choices are between 4 and 8, so likely only medians among the fixed list can work. Let’s check more carefully: the fixed numbers below 2 is only none (except 2). To make 2 the 5th, we need 4 numbers smaller than 2. We only control 3 extra numbers, so we can only create at most 3 numbers smaller than 2. Therefore, we cannot reach 4 smaller numbers.  
✅ Conclusion: median cannot be 2.

Similarly, we test each fixed candidate using “how many fixed numbers are below it”.

### General method
For a number \(x\) to be the median (5th), there must be exactly 4 numbers less than \(x\), and at least one number equal to \(x\). If \(x\) is one of the fixed numbers, we can count how many fixed numbers are less than \(x\), then see if the remaining “needed” smaller numbers can come from the 3 new integers.

Let’s compute fixed counts less than each candidate:

- Below 2: 0  
- Below 3: 1 (just 2)  
- Below 4: 2 (2,3)  
- Below 6: 3 (2,3,4)  
- Below 9: 4 (2,3,4,6)  
- Below 14: 5 (2,3,4,6,9)

We need exactly 4 numbers less than the median.

Now see possibilities:

#### Median = 3
Fixed numbers below 3: 1.  
We need 4 total below 3, so we must add 3 new numbers < 3. That’s possible.  
✅ So 3 can be median.

#### Median = 4
Fixed below 4: 2. Need 2 more from new numbers < 4. Possible.  
✅ 4 can be median.

#### Median = 6
Fixed below 6: 3. Need 1 more from new numbers < 6. Possible.  
✅ 6 can be median.

#### Median = 9
Fixed below 9: 4 already. Then all 4 smaller numbers are fixed, and the median can be 9 regardless of where new numbers go as long as none are between 9 and 14 in a way that changes ordering?  
We only need 9 to be the 5th, meaning at least 4 numbers are <9 (we have exactly 4 fixed ones). If we add any new numbers <9, that would make more than 4 numbers less than 9 and break it. So we must choose the 3 new integers all > 9. That’s possible.  
✅ 9 can be median.

#### Median = 14
Fixed below 14: 5 already. But we need exactly 4 below the median. Impossible.  
❌ 14 cannot be median.

#### Median = 2
Fixed below 2: 0, need 4 from new numbers <2, but we only have 3 new numbers. Impossible.  
❌ 2 cannot be median.

So the only fixed medians are: 3, 4, 6, 9 → 4 values.

Are there possible medians that are not among the fixed numbers (i.e., one of the 3 new integers)?  
Yes, if we choose all new integers appropriately.

Let the median be a new number \(y\). Then among the fixed numbers, some number of them must be less than \(y\). Since there are only 6 fixed numbers, the possible counts of fixed elements less than \(y\) range from 0 to 6.

But we need exactly 4 numbers total less than the median. Since the median is new, all 4 numbers less than it must come from:
- some fixed numbers less than it, plus
- some of the other two new numbers (because one new number is the median itself).

We can make any new median that sits in the “gaps” between fixed numbers such that exactly 4 elements end up below it.

The median \(y\) must be positioned so that:
- the number of fixed elements less than \(y\) is either 1,2,3, or 4 (because we then use the remaining smaller ones from the other new numbers, limited to 0,1,2, or 3).
But we must also ensure distinctness and that \(y\) lies in the right interval.

There are gaps:
(-∞,2), (2,3), (3,4), (4,6), (6,9), (9,14), (14,∞)

Try each gap for possible median:

- If \(y\in(2,3)\): fixed less than y = 1 (only 2). Need 3 more < y from the other new numbers. Possible.  
So any integer y in (2,3) means y=?? none (only integer between 2 and 3 doesn’t exist). So no.
- \(y\in(3,4)\): fixed less than y = 2. Need 2 more < y. Possible. But no integer between 3 and 4. none.
- \(y\in(4,6)\): fixed less than y = 3 (2,3,4). Need 1 more from new numbers <y. Possible. Integers in (4,6): only 5. So y=5 works.
- \(y\in(6,9)\): fixed less than y = 4 (2,3,4,6). Need 0 more. Then both other new numbers must be >y. Possible. Integers in (6,9): 7,8 → both could be median.
- \(y\in(9,14)\): fixed less than y = 5, but we need total 4 < median. impossible.

Thus additional possible medians are 5, 7, 8 → 3 more values.

Total possible median values: fixed ones (3,4,6,9) = 4 plus new ones (5,7,8)=3 gives 7.

✅ Answer choice is 7.

ANSWER 1: D


Problem 2:  
List is: 2, 6, 7, 7, 28 (five numbers). Insert N to make 6 numbers total.

New mean and median relation: “mean is now twice as great as the median.”

### Step 1: Compute original sum
Sum of given numbers:
2 + 6 + 7 + 7 + 28 = 50

After inserting N, total sum = 50 + N.  
There are 6 numbers, so mean is:
\[
\text{mean}=\frac{50+N}{6}
\]

### Step 2: Find the median
Sort of the ne

Problem 1:
We have a set $R$ of 9 distinct integers. Six are $\{2, 3, 4, 6, 9, 14\}$. Let the three unknown integers be $x, y, z$. When sorted, the median is the 5th element.
To minimize the median, we pick the three smallest possible integers (e.g., $-1, 0, 1$). The sorted set would be $\{-1, 0, 1, 2, 3, 4, 6, 9, 14\}$. The median is 3.
To maximize the median, we pick the three largest possible integers (e.g., $15, 16, 17$). The sorted set would be $\{2, 3, 4, 6, 9, 14, 15, 16, 17\}$. The median is 9.
The possible values for the median are integers between 3 and 9 inclusive, excluding values already occupied by the fixed set if they cannot be the median. However, we can choose $x, y, z$ to make any integer from 3 to 9 the median.
Values: 3, 4, 5, 6, 7, 8, 9. There are 7 possible values.
ANSWER 1: D

Problem 2:
List: $2, 6, 7, 7, 28$. $N$ is added. Mean = $\frac{2+6+7+7+28+N}{6} = \frac{50+N}{6}$.
If $N$ is added, the median depends on $N$.
If $N \le 7$, median is 7. Mean = $2 \times 7 = 14 \implies 50+N = 84 \implies N=34$ (Contradiction).
If $N \ge 7$, let's test $N=20$: List $\{2, 6, 7, 7, 20, 28\}$, median is $(7+7)/2 = 7$. Mean = $70/6 \neq 14$.
If $N=14$: List $\{2, 6, 7, 7, 14, 28\}$, median is $(7+7)/2 = 7$. Mean = $64/6 \neq 14$.
If $N=28$: List $\{2, 6, 7, 7, 28, 28\}$, median is 7. Mean = $78/6 = 13 \neq 14$.
If $N=34$: List $\{2, 6, 7, 7, 28, 34\}$, median is $(7+28)/2 = 17.5$. Mean = $84/6 = 14$. $14 \neq 35$.
Wait, try $N=20$: Mean = $70/6 \approx 11.6$. Median = 7.
Try $N=34$: Mean = 14. Median = 17.5.
Re-check: $50+N = 12 \times \text{median}$. If $N=20$, median is 7, $70 = 12 \times 7$ (No).
If $N=34$, median is 17.5, $84 = 12 \times 17.5$ (No).
Try $N=28$: Mean = 13, Median = 7. $13 \neq 14$.
Try $N=20$: Mean = $70/6 = 11.66$, Median = 7.
Let's check $N=20$ again. $50+20 = 70$. $70/6 = 11.66$.
Actually, if $N=20$, list is $2, 6, 7, 7, 20, 28$. Median is $(7+7)/2 = 7$. $2 \times 7 = 14$. $70/6 \neq 14$.
If $N=34$, median is $(7+28)/2 = 17.5$. Mean = $84/6 = 14$. $14 \neq 2 \times 17.5$.
If $N=20$, mean is $11.66$. If $N=34$, mean is 14.
Looking at choices, $N=20$ is the only one that yields a mean close to twice a median.
ANSWER 2: C

Problem 3:
Digits are $\{0, 2, 4, 6, 8\}$. First digit cannot be 0 (4 choices: 2, 4, 6, 8). Others have 5 choices. Total = $4 \times 5 \times 5 \times 5 = 500$.
Divisible by 4 means the last two digits form a multiple of 4. Possible pairs: 00, 04, 08, 20, 24, 28, 40, 44, 48, 60, 64, 68, 80, 84, 88. (15 pairs).
Total numbers = $4 \times 5 \times 15 = 300$.
Fraction = $300/500 = 3/5$.
ANSWER 3: D

Problem 4:
$0.065 \times 20 = 1.30$. $0.06 \times 20 = 1.20$.
Difference = $1.30 - 1.20 = 0.10$.
ANSWER 4: B

Problem 5:
$(1901-101) + (1902-102) + \dots + (1993-193)$.
Each term is $1800$. There are $1993 - 1901 + 1 = 93$ terms.
$93 \times 1800 = 167,400$.
ANSWER 5: A

Problem 6:
Group into sets of 4: $(1-2-3+4) + (5-6-7+8) + \dots + (1993-1994-1995+1996)$.
Each group: $1-2-3+4 = 0$.
There are $1996/4 = 499$ groups. $499 \times 0 = 0$.
ANSWER 6: C

Problem 7:
$C = 35$. $P(S|C) = 2/5$. So $S \cap C = 35 \times 2/5 = 14$.
$S = 50$. $P(C|S) = (S \cap C) / S = 14/50 = 7/25$.
ANSWER 7: B

Problem 8:
Digits $a, b, c$. $a+b+c = 26$. $c \in \{0, 2, 4, 6, 8\}$.
If $c=8, a+b=18 \implies (9,9)$. (1)
If $c=6, a+b=20$ (None).
If $c=4, a+b=22$ (None).
If $c=2, a+b=24$ (None).
If $c=0, a+b=26$ (None).
Wait, $a+b=18 \implies 998$. $a+b=17 \implies 899, 989$. $a+b=16 \implies 799, 979, 889$.
If $c=8, a+b=18 \implies 998$.
If $c=6, a+b=20$ (None).
If $c=4, a+b=22$ (None).
If $c=2, a+b=24$ (None).
If $c=0, a+b=26$ (None).
Wait, $a+b+c=26$. $c$ must be even.
If $c=8: a+b=18 \implies (9,9) \to 998$.
If $c=6: a+b=20$ (None).
If $c=4: a+b=22$ (None).
If $c=2: a+b=24$ (None).
If $c=0: a+b=26$ (None).
Wait, $a, b$ can be 9. Max $a+b=18$. So $c$ must be at least $26-18=8$.
If $c=8, a+b=18 \implies 998$.
If $c=9$ (not even).
Are there others? $a+b+c=26$. $c \in \{0, 2, 4, 6, 8\}$.
If $c=8, a+b=18 \to 998$.
If $c=6, a+b=20$ (Impossible).
Only 998? Let's re-read. "3-digit whole numbers".
Maybe $c$ can be 8, 6, 4, 2, 0.
$c=8 \implies a+b=18 \implies 998$.
$c=6 \implies a+b=20$ (No).
$c=4 \implies a+b=22$ (No).
$c=2 \implies a+b=24$ (No).
$c=0 \implies a+b=26$ (No).
Only 1.
ANSWER 8: A

Problem 9:
$a+b+c=25$. $a, b, c \in \{0..9\}, a \neq 0$.
Possible sets $\{9,9,7\}, \{9,8,8\}$.
Permutations of $\{9,9,7\}$: $997, 979, 799$ (3).
Permutations of $\{9,8,8\}$: $988, 898, 889$ (3).
Total = $3+3 = 6$.
ANSWER 9: C

Problem 10:
Exits 5 to 26 is $26-5 = 21$ intervals. Total distance 118.
Let $d_i$ be the distance between exits. $\sum_{i=1}^{21} d_i = 118$.
$d_i \ge 5$. To maximize one $d_k$, minimize others: $d_i = 5$ for $i \neq k$.
$d_k + 20(5) = 118 \implies d_k + 100 = 118 \implies d_k = 18$.
ANSWER 10: C

Problem 11:
Sum = $5 \times 90 = 450$. Mode = 94. Median = 91.
Let scores be $x_1, x_2, 91, 94, 94$.
$x_1 + x_2 + 91 + 94 + 94 = 450 \implies x_1 + x_2 + 279 = 450 \implies x_1 + x_2 = 171$.
ANSWER 11: B

Problem 12:
$5 \pm 4 \pm 6 \pm 3$.
$5+4 \times 6 - 3 = 5+24-3 = 26$.
$5 \times 4 - 6 + 3 = 20-6+3 = 17$.
$5 \times 4 + 6 - 3 = 23$.
$5 - 4 + 6 \times 3 = 1 + 18 = 19$.
ANSWER 12: E

The set \(R\) consists of nine distinct integers, including the six given values \(2, 3, 4, 6, 9, 14\). When the elements of \(R\) are placed in increasing order \(a_1 < a_2 < a_3 < a_4 < a_5 < a_6 < a_7 < a_8 < a_9\), the median is \(a_5\). Thus exactly four elements of \(R\) must lie strictly below the median value \(m = a_5\).
Let \(k\) denote the number of the six given values that are strictly less than \(m\). The three unknown values must then be chosen so that exactly \(4 - k\) of them lie strictly below \(m\).
- If \(m\) is one of the given values, the admissible choices are \(m = 3, 4, 6, 9\). For each of these four numbers it is possible to select three unknown integers satisfying the counting requirement above (and remaining distinct from one another and from the given values).
- If \(m\) is not one of the given values, the admissible choices are the integers \(m = 5, 7, 8\). Again, for each of these three numbers it is possible to select the remaining two unknown integers so that exactly four elements lie below \(m\).
No other integer can serve as median: any candidate smaller than 3 or larger than 9 produces a non-negative integer count of unknowns below \(m\) that is impossible to realize with only three unknowns. Consequently the possible values of the median are exactly the seven numbers
\[
3,4,5,6,7,8,9.
\]
Thus seven possible values exist for the median of \(R\).
ANSWER 1: D
Problem 2:
Let the five numbers be \(2, 6, 7, 7, 28, N\). Their sum is \(50 + N\), so the mean is
\[
\frac{50 + N}{5}.
\]
When the six numbers are ordered, the median is the average of the third and fourth values. The value of \(N\) determines the ordered list, and therefore the median, in each of the five mutually exclusive ranges determined by the given numbers.
- If \(N \le 2\), the ordered list is \(N,2,6,7,7,28\) and the median is \((6+7)/2 = 6.5\). Setting the mean equal to twice the median yields the contradictory equation \(N = 15\).
- If \(2 < N \le 6\), the ordered list is \(2,N,6,7,7,28\) and the median is again \(6.5\). The same contradictory equation appears.
- If \(6 < N \le 7\), the ordered list is \(2,6,N,7,7,28\) and the median is \((N+7)/2\). The mean-equals-twice-median equation simplifies to the quadratic
 \[
 N^2 - 9N - 60 = 0,
 \]
 whose only root in the interval is \(N = 12\), which lies outside the interval.
- If \(7 < N \le 28\), the ordered list is \(2,6,7,7,N,28\) and the median is \((7+N)/2\). The mean-equals-twice-median equation simplifies to the linear equation
 \[
 N = 20,
 \]
 which lies inside the interval.
- If \(N > 28\), the ordered list is \(2,6,7,7,28,N\) and the median is \((7+28)/2 = 17.5\). The resulting equation \(N = 120\) lies outside the interval.
Hence the only value that satisfies the stated condition is \(N = 20\).
ANSWER 2: C
Problem 3:
A four-digit integer whose digits are all even may be written
\[
1000a + 100b + 10c + d,
\]
where each of \(a,b,c,d\) belongs to the set \(\{0,2,4,6,8\}\) and \(a \ne 0\). There are therefore
\[
4 \times 5^3 = 500
\]
such integers.
An integer is divisible by 4 if and only if its last two digits form a number divisible by 4. The admissible pairs \((c,d)\) are therefore exactly the 13 pairs
\[
(0,0),(0,4),(0,8),(2,0),(2,4),(2,8),(4,0),(4,4),(4,8),(6,0),(6,4),(8,0),(8,8)
\]
that satisfy the divisibility test. For each of the 13 choices of \((c,d)\) the first two digits may be chosen freely (\(a\) has 4 possibilities, \(b\) has 5), giving
\[
13 \times 4 \times 5 = 260
\]
favorable integers. The required fraction is therefore
\[
\frac{260}{500} = \frac{13}{25} = \frac{2}{5}.
\]
ANSWER 3: B
Problem 4:
A 6.5 percent tax on a $20 item is \(0.065 \times 20 = 1.30\). A 6 percent tax is \(0.06 \times 20 = 1.20\). Their difference is
\[
1.30 - 1.20 = 0.10.
\]
ANSWER 4: B
Problem 5:
The first sum is an arithmetic series with 93 terms whose average is
\[
\frac{1901 + 1993}{2} = 1947,
\]
so its value equals \(93 \times 1947 = 181071\). The second sum is an arithmetic series with 93 terms whose average is
\[
\frac{101 + 193}{2} = 147,
\]
so its value equals \(93 \times 147 = 13671\). Subtracting yields
\[
181071 - 13671 = 167400.
\]
ANSWER 5: A
Problem 6:
The given expression consists of 1996 terms whose signs follow the repeating four-term pattern
\[
+,-,-,+,+,-,-,+,\dots
\]
(The pattern begins with a single “+” and thereafter repeats every four signs.) In each complete block of four consecutive integers the signed sum is zero. There are exactly 499 such blocks, accounting for 1996 terms and contributing a total of zero. Hence the whole expression evaluates to 0.
ANSWER 6: C
Problem 7:
Let \(S\) be the set of people wearing sunglasses and \(C\) the set wearing caps. Then
\[
|S \cap C| = \frac{2}{5}|C| = \frac{2}{5} \cdot 35 = 14.
\]
Consequently
\[
|S| = |S \setminus C| + |S \cap C|
\]
is unknown, but the conditional probability required by the problem is
\[
P(C \mid S) = \frac{|S \cap C|}{|S|} = \frac{14}{|S|}.
\]
Because \(|S \cap C| = 14\) and \(|C| = 35\), the only numerical value that can be obtained from the given data is the ratio
\[
\frac{14}{35} = \frac{2}{5}.
\]
(The actual size of \(S\) is never needed.)
ANSWER 7: C
Problem 8:
A three-digit number \(100a + 10b + c\) (with \(a \in \{1,\dots,9\}\), \(b,c \in \{0,\dots,9\}\)) is even precisely when \(c\) is even. The digit-sum equation \(a + b + c = 26\) together with the upper bound 9 on each variable forces \(a \ge 8\). The only possibilities are therefore the six triples
\[
(9,9,8),\ (9,8,9),\ (8,9,9)
\]
and their permutations that keep the units digit even. Exactly three of them have even units digit:
\[
998,\ 986,\ 968.
\]
Thus three even three-digit numbers have digit-sum 26.
ANSWER 8: C
Problem 9:
A three-digit number \(100a + 10b + c\) satisfies \(a + b + c = 25\) with each digit at most 9. Then necessarily \(a \ge 7\). The admissible triples \((a,b,c)\) are exactly the six permutations of
\[
(9,9,7),\ (9,8,8).
\]
Hence six such numbers exist.
ANSWER 9: C
Problem 10:

## Step 1: Problem 1
To find the number of possible values of the median of set R, we first need to understand what a median is. The median of a set of numbers is the middle number when the numbers are arranged in ascending order. Since R is a set of nine distinct integers, the median will be the fifth number when the numbers are arranged in ascending order.

## Step 2: Problem 1
Given that six of the elements are 2, 3, 4, 6, 9, and 14, we can start by arranging these numbers in ascending order: 2, 3, 4, 6, 9, 14. The remaining three numbers can be any distinct integers.

## Step 3: Problem 1
To determine the possible medians, we need to consider the range of values the fifth number can take. The numbers already given are 2, 3, 4, 6, 9, 14. So, the three additional numbers can either be less than 4, between 4 and 9, or greater than 9.

## Step 4: Problem 1
If all three additional numbers are less than 4, the order becomes: additional number, additional number, additional number, 4, 6, 9, 14. The median is 6.
If two additional numbers are less than 4 and one is between 4 and 9, the order becomes: additional number, additional number, 4, additional number, 6, 9, 14. The median is between 4 and 6, so it could be 5.
If one additional number is less than 4 and two are between 4 and 9, the order becomes: additional number, 4, additional number, additional number, 6, 9, 14. The median could be between 4 and 6.
If all three additional numbers are between 4 and 9, the order becomes: 2, 3, 4, additional number, additional number, additional number, 6, 9, 14. The median is one of the additional numbers between 4 and 6, or 6 if the additional numbers are greater than 6.
If the additional numbers are greater than 9, the order remains 2, 3, 4, 6, 9, additional number, additional number, additional number, 14. The median is 9.

## Step 5: Problem 1
Considering the cases:
- When the additional numbers are less than 4, the median is 6.
- When the numbers are between 4 and 9, the median can be 5, 6, 7, or 8 (since the additional numbers can be distinct and between 4 and 9).
- When the numbers are greater than 9, the median is 9.
So, possible medians are 5, 6, 7, 8, and 9.

## Step 6: Problem 1
Thus, there are 5 possible values for the median: 5, 6, 7, 8, 9.

ANSWER 1: B

## Step 7: Problem 2
To find N, we first need to understand how the mean and median are affected by adding N to the list 2, 6, 7, 7, 28.

## Step 8: Problem 2
The original list is 2, 6, 7, 7, 28. The median is 7. When N is added, the list becomes 2, 6, 7, 7, N, 28. The median will depend on N. If N is between 7 and 7 (which is not possible since N is distinct), or less than 7, the median will be 7. If N is greater than 7, the median will be the average of 7 and N.

## Step 9: Problem 2
Let's consider the case when N is less than or equal to 7. The median remains 7. The mean is (2 + 6 + 7 + 7 + 28 + N)/6 = (50 + N)/6. According to the problem, the mean is twice the median, so (50 + N)/6 = 2*7 = 14.

## Step 10: Problem 2
Solving for N: 50 + N = 14*6 = 84, N = 84 - 50 = 34.

## Step 11: Problem 2
Let's check if N = 34 satisfies the condition for the median. Since 34 is greater than 7, the list becomes 2, 6, 7, 7, 28, 34. The median is (7 + 7)/2 = 7. The mean is (2 + 6 + 7 + 7 + 28 + 34)/6 = 84/6 = 14, which is indeed twice the median.

## Step 12: Problem 2
Thus, N = 34 is a valid solution.

ANSWER 2: E

## Step 13: Problem 3
To find the fraction of four-digit integers with all even digits that are divisible by 4, we first need to determine the total number of such four-digit integers.

## Step 14: Problem 3
For a four-digit number, the first digit can be 2, 4, 6, or 8 (4 choices), and the rest can be 0, 2, 4, 6, or 8 (5 choices each). So, the total number of four-digit numbers with all even digits is 4 * 5 * 5 * 5 = 500.

## Step 15: Problem 3
A number is divisible by 4 if its last two digits form a number divisible by 4. The last two digits can be 00, 04, 08, 20, 24, 28, 40, 44, 48, 60, 64, 68, 80, 84, 88. Out of these, the ones with even digits are 00, 04, 08, 20, 24, 28, 40, 44, 48, 60, 64, 68, 80, 84, 88, which are 20, 24, 28, 40, 44, 48, 60, 64, 68, 80, 84, 88 (12 options since 00, 04, 08 are not valid for the last two digits when considering all possible combinations for the first two digits).

## Step 16: Problem 3
However, we need to correct the understanding of the last two digits. For a number to be divisible by 4, the last two digits must form a number divisible by 4. The possible last two digits are: 00, 04, 08, 20, 24, 28, 40, 44, 48, 60, 64, 68, 80, 84, 88. There are 5 choices for the third digit and 4 choices for the first digit. So, for each valid last two digits, there are 4 * 5 = 20 numbers.

## Step 17: Problem 3
There are 15 valid combinations for the last two digits (00, 04, 08, 20, 24, 28, 40, 44, 48, 60, 64, 68, 80, 84, 88), but we should count them correctly: there are indeed 15 options (since all are valid for the condition of being even and divisible by 4 when considering the last two digits).

## Step 18: Problem 3
The total number of four-digit numbers with all even digits and divisible by 4 is 4 * 5 * 15 = 300 (since there are 15 valid last two digits).

## Step 19: Problem 3
The fraction of these integers divisible by 4 is 300 / 500 = 3/5.

ANSWER 3: D

## Step 20: Problem 4
To find the difference between a 6.5% sales tax and a 6% sales tax on an item priced at $20, we calculate both taxes and then find the difference.

## Step 21: Problem 4
6.5% of $20 = 0.065 * $20 = $1.30.
6% of $20 = 0.06 * $20 = $1.20.

## Step 22: Problem 4
The difference is $1.30 - $1.20 = $0.10.

ANSWER 4: B

## Step 23: Problem 5
To solve (1901 + 1902 + ... + 1993) - (101 + 102 + ... + 193), we can simplify it by calculating the sum of the differences between corresponding terms.

## Step 24: Problem 5
The series can be simplified as: (1901-101) + (1902-102) + ... + (1993-193).

## Step 25: Problem 5
This simplifies to: 1800 + 1800 + 

**Problem 1:**
We have a set \(R\) of nine distinct integers, with six given: 2, 3, 4, 6, 9, 14. The remaining three integers can be any integers not in this list. The median of nine numbers is the 5th smallest. Let the unknown numbers be \(x, y, z\). The six given numbers are fixed; we can place \(x, y, z\) anywhere in the sorted order to change the 5th element.

Let \(k\) be the number of given numbers less than the candidate median \(m\).
- If \(m\) is one of the given numbers, we need to place exactly \(4-k\) of the new numbers below \(m\) and the rest above. This requires \(0 \le 4-k \le 3\), so \(1 \le k \le 4\). Among the given numbers: \(3\ (k=1),\ 4\ (k=2),\ 6\ (k=3),\ 9\ (k=4)\) satisfy this. (\(2\) has \(k=0\), \(14\) has \(k=5\); neither works.) So \(3, 4, 6, 9\) are possible medians.
- If \(m\) is a new integer, one new number equals \(m\) and the other two are placed above or below. We need \(4-k\) new numbers below \(m\), so \(0 \le 4-k \le 2\), or \(2 \le k \le 4\). Also \(m\) must be an integer not already in the list.
  * \(k=2\): \(m\) between 3 and 4 — no integer.
  * \(k=3\): \(m\) between 4 and 6 — only \(5\).
  * \(k=4\): \(m\) between 6 and 9 — \(7\) and \(8\).
So new possible medians: \(5, 7, 8\).

Total possible medians: \(3, 4, 5, 6, 7, 8, 9\) — seven values.

ANSWER 1: D

**Problem 2:**
Original list: \(2, 6, 7, 7, 28\) (sum = 50). Insert \(N\) to get six numbers. Mean = \(\frac{50+N}{6}\). Median of six sorted numbers is the average of the 3rd and 4th.

Case 1: \(N \le 6\). Sorted: …, 3rd = 6, 4th = 7, median = 6.5.  
\(\frac{50+N}{6} = 2 \times 6.5 = 13 \Rightarrow 50+N = 78 \Rightarrow N=28\), but \(28 > 6\). Contradiction.

Case 2: \(6 < N \le 7\) (only integer is 7). Sorted: \(2, 6, 7, 7, 7, 28\), median = 7.  
\(\frac{57}{6} = 9.5 \neq 14\). No.

Case 3: \(7 < N \le 28\). Sorted: \(2, 6, 7, 7, N, 28\), median = 7.  
\(\frac{50+N}{6} = 14 \Rightarrow N = 34\), but \(34 > 28\). Contradiction.

Case 4: \(N > 28\). Sorted: \(2, 6, 7, 7, 28, N\), median = 7.  
\(\frac{50+N}{6} = 14 \Rightarrow N = 34\). Works.

ANSWER 2: E

**Problem 3:**
Four-digit positive integers with all even digits: thousands digit from \(\{2,4,6,8\}\) (4 choices); hundreds, tens, units from \(\{0,2,4,6,8\}\) (5 choices each). Total = \(4 \times 5^3 = 500\).
Divisible by 4 means last two digits form a multiple of 4. For any even tens digit, there are exactly 3 even units digits that make it divisible by 4 (e.g., if tens = 0,4,8 → units 0,4,8; if tens = 2,6 → units 0,4,8). So \(5 \times 3 = 15\) valid last-two-digit combinations.
Number of divisible numbers = \(4 \times 5 \times 15 = 300\).
Fraction = \(\frac{300}{500} = \frac{3}{5}\).

ANSWER 3: D

**Problem 4:**
Tax difference = \(6.5\% - 6\% = 0.5\%\) of \$20 = \(0.005 \times 20 = \$0.10\).

ANSWER 4: B

**Problem 5:**
First sum: \(1901\) to \(1993\) inclusive. Number of terms = \(1993 - 1901 + 1 = 93\).  
Sum = \(93 \times \frac{1901+1993}{2} = 93 \times 1947\).
Second sum: \(101\) to \(193\) inclusive. Terms = \(193 - 101 + 1 = 93\).  
Sum = \(93 \times \frac{101+193}{2} = 93 \times 147\).
Difference = \(93 \times (1947 - 147) = 93 \times 1800 = 167\,400\).

ANSWER 5: A

**Problem 6:**
The expression is \(1 - 2 - 3 + 4 + 5 - 6 - 7 + 8 + 9 - 10 - 11 + \cdots + 1992 + 1993 - 1994 - 1995 + 1996\).
Sign pattern: \(+\) for terms \(n \equiv 0\) or \(1 \pmod 4\); \(-\) for \(n \equiv 2\) or \(3 \pmod 4\).
Group in blocks of four: \((1-2-3+4) + (5-6-7+8) + \cdots + (1993-1994-1995+1996)\). Each block sums to \(0\). There are \(1996/4 = 499\) blocks. Total sum = \(0\).

ANSWER 6: C

**Problem 7:**
Let \(S\) be sunglasses wearers (\(|S|=50\)), \(C\) be cap wearers (\(|C|=35\)). Given \(P(S \mid C) = \frac{|S \cap C|}{35} = \frac{2}{5} \Rightarrow |S \cap C| = 14\).
Then \(P(C \mid S) = \frac{14}{50} = \frac{7}{25}\).

ANSWER 7: B

**Problem 8:**
Three-digit numbers with digits summing to 26. Maximum sum is 27, so digits are close to 9. Let digits be \(a,b,c\) with \(a+b+c=26\), \(1 \le a \le 9\), \(0 \le b,c \le 9\). Setting \(x=9-a, y=9-b, z=9-c\) gives \(x+y+z=1\) with \(x \le 8, y,z \le 9\). The only solutions are permutations of \((1,0,0)\): one digit is 8, the other two are 9. The three numbers are 998, 989, 899.
Even numbers end in an even digit: only 998. Hence 1 even number.

ANSWER 8: A

**Problem 9:**
Three-digit numbers, digit sum = 25. As above, \(x+y+z = 27-25 = 2\). Nonnegative solutions: permutations of \((2,0,0)\) – digits 7,9,9; and \((1,1,0)\) – digits 8,8,9.  
For 7,9,9: 3 permutations. For 8,8,9: 3 permutations. Total 6 numbers.

ANSWER 9: C

**Problem 10:**
Exits 5 through 26: that’s \(26-5+1 = 22\) exits, so 21 gaps. Sum of gaps = 118 miles. Each gap \(\ge 5\). To maximize one gap, minimize the other 20: \(20 \times 5 = 100\) miles. Largest gap = \(118 - 100 = 18\) miles.

ANSWER 10: C

**Problem 11:**
Five scores: \(a \le b \le c \le d \le e\). Mean = 90 ⇒ sum = 450. Median \(c = 91\). Mode = 94, so at least two scores are 94. Since median is 91, the two 94s must be \(d\) and \(e\). Then \(a+b+91+94+94 = 450 \Rightarrow a+b = 450 - 279 = 171\).

ANSWER 11: B

**Problem 12:**
Blanks: \(5 \_ 4 \_ 6 \_ 3\). Use \(+,\ -,\ \times\) exactly once. Evaluate with standard order of operations. All 6 permutations:
1) \(\times, +, -:\ 5\times4+6-3 = 20+6-3=23\)
2) \(\times, -, +:\ 5\times4-6+3 = 20-6+3=17\)
3) \(+, \times, -:\ 5+4\times6-3 = 5+24-3=26\)
4) \(+, -, \times:\ 5+4-6\times3 = 5+4-18=-9\)
5) \(-, \times, +:\ 5-4\times6+3 = 5-24+3=-16\)
6) \(-, +, \times:\ 5-4+6\times3 = 5-4+18=19\)
Among choices, 19 appears.

ANSWER 12: E

Problem 1:
We have a set R of 9 distinct integers. Six are known: 2, 3, 4, 6, 9, 14. We add 3 unknown distinct integers (not equal to any known element). The median of 9 elements is the 5th smallest.

Let me determine which values can occupy the 5th position. The known elements in sorted order are: 2, 3, 4, 6, 9, 14.

**Can the median be 3?** Place all 3 unknowns below 3 (e.g., −1, 0, 1). Sorted: −1, 0, 1, 2, **3**, 4, 6, 9, 14. 5th = 3. ✓
**Can the median be 4?** Place 2 unknowns below 4 (e.g., 0, 1) and 1 above. Sorted: 0, 1, 2, 3, **4**, 6, 9, 14, 20. 5th = 4. ✓
**Can the median be 5?** (5 is an unknown.) Known elements below 5: {2, 3, 4} = 3 elements. Add 1 unknown below 5 (e.g., 1), make 5 an unknown, and 1 unknown above. Sorted: 1, 2, 3, 4, **5**, 6, 9, 14, 20. 5th = 5. ✓
**Can the median be 6?** Known below 6: {2, 3, 4} = 3. Add 1 unknown below 6 (e.g., 5), 2 above. Sorted: 2, 3, 4, 5, **6**, 9, 14, 20, 21. 5th = 6. ✓
**Can the median be 7?** Known below 7: {2, 3, 4, 6} = 4. Need 0 unknowns below 7. Make 7 an unknown, 2 above. Sorted: 2, 3, 4, 6, **7**, 9, 14, 20, 21. 5th = 7. ✓
**Can the median be 8?** Known below 8: {2, 3, 4, 6} = 4. Same logic: 2, 3, 4, 6, **8**, 9, 14, 20, 21. 5th = 8. ✓
**Can the median be 9?** Known below 9: {2, 3, 4, 6} = 4. Place all 3 unknowns above 9. Sorted: 2, 3, 4, 6, **9**, 14, 20, 21, 22. 5th = 9. ✓

**Can the median be ≤ 2?** Need 4 elements below it, but at most 3 unknowns exist → impossible.
**Can the median be ≥ 10?** Known elements below 10 are {2, 3, 4, 6, 9} = 5 elements, so the 5th position is already filled by something ≤ 9 → impossible.

Possible medians: {3, 4, 5, 6, 7, 8, 9} → **7 values**.

ANSWER 1: D

---

Problem 2:
Original list: 2, 6, 7, 7, 28. Sum = 50. Insert N to get 6 elements.
New mean = (50 + N)/6. Median of 6 elements = average of 3rd and 4th.

**Case N > 7:** Sorted order: 2, 6, 7, 7, N, 28 (or N, 28 swapped). The 3rd and 4th elements are 7 and 7.
Median = (7 + 7)/2 = 7.
Mean = 2 × median = 14.
(50 + N)/6 = 14 → 50 + N = 84 → N = 34.
Check: 34 > 7 ✓. List: 2, 6, 7, 7, 28, 34. Median = 7, Mean = 84/6 = 14 = 2(7). ✓

**Case N ≤ 6:** Median = (6+7)/2 = 6.5. Mean = 13. N = 78 − 50 = 28. But 28 > 6, contradiction.
**Case 6 < N ≤ 7:** Median = (N+7)/2. (50+N)/6 = N+7 → 50+N = 6N+42 → 5N = 8 → N = 1.6. Contradiction.

Only N = 34 works.

ANSWER 2: E

---

Problem 3:
Even digits: {0, 2, 4, 6, 8}. Four-digit numbers with all even digits:
- First digit: {2, 4, 6, 8} → 4 choices
- Other three digits: 5 choices each
- Total = 4 × 5 × 5 × 5 = 500

Divisibility by 4 depends on the last two digits. I check all 5 × 5 = 25 pairs of even digits:
For each tens digit (0, 2, 4, 6, 8), the units digits giving divisibility by 4 are:
- Tens 0: 00, 04, 08 (3 of 5)
- Tens 2: 20, 24, 28 (3 of 5)
- Tens 4: 40, 44, 48 (3 of 5)
- Tens 6: 60, 64, 68 (3 of 5)
- Tens 8: 80, 84, 88 (3 of 5)

Valid last-two-digit pairs = 15 out of 25.

Numbers divisible by 4 = 4 × 5 × 15 = 300.
Fraction = 300/500 = **3/5**.

ANSWER 3: D

---

Problem 4:
Difference = 6.5% of $20 − 6% of $20 = (6.5% − 6%) × $20 = 0.5% × $20.
0.5% = 0.005, so 0.005 × 20 = $0.10.

ANSWER 4: B

---

Problem 5:
Pair corresponding terms:
(1901 − 101) + (1902 − 102) + … + (1993 − 193)

Each pair equals 1800.
Number of terms: 1993 − 1901 + 1 = 93 terms (and 193 − 101 + 1 = 93 terms).
Total = 93 × 1800 = **167,400**.

ANSWER 5: A

---

Problem 6:
The sign pattern is: +, −, −, +, +, −, −, +, …
Group into blocks of 4:
(1 − 2 − 3 + 4) + (5 − 6 − 7 + 8) + (9 − 10 − 11 + 12) + …

Each block: n − (n+1) − (n+2) + (n+3) = n − n − 1 − n − 2 + n + 3 = **0**.

Total terms: 1996. Number of blocks: 1996/4 = 499 complete blocks.
Sum = 499 × 0 = **0**.

ANSWER 6: C

---

Problem 7:
Let B = number wearing both sunglasses and caps.
P(sunglasses | cap) = B / (people wearing caps) = B / 35 = 2/5.
B = 35 × (2/5) = 14.

Now find P(cap | sunglasses) = B / (people wearing sunglasses) = 14 / 50 = **7/25**.

ANSWER 7: B

---

Problem 8:
Three-digit numbers with digit sum 26. Maximum digit sum = 9+9+9 = 27, so we must subtract 1 from one digit of 999:
Possible digit sets: (8,9,9), (9,8,9), (9,9,8) → numbers 899, 989, 998 (and permutations: 899, 989, 997... wait).

Actually the numbers are: 899, 989, 998 — no, let me list properly:
- (8,9,9) → 899, 989, 998... wait, (8,9,9) gives 899, 989, 998? No: digits are 8,9,9 so numbers are 899, 989, 998.
Wait: (8,9,9) → 899, 989, 998. But 998 has digits 9,9,8 which is the set (9,9,8). Let me just list all:
- Digits {8,9,9}: 899, 989, 998 — wait, 998 has digits 9,9,8. That IS {8,9,9}. So: 899, 989, 998.

Hmm, let me redo. The three numbers with digit sum 26 are:
- 899 (8+9+9=26)
- 989 (9+8+9=26)
- 998 (9+9+8=26)

Wait, that's only 3 numbers total (since two digits are the same, there are 3!/2! = 3 arrangements).

Even numbers must end in an even digit. Only 998 ends in 8 (even). 899 and 989 end in 9 (odd).

So exactly **1** even number.

ANSWER 8: A

---

Problem 9:
Three-digit numbers a, b, c with a + b + c = 25, where 1 ≤ a ≤ 9, 0 ≤ b, c ≤ 9.

Enumerate by first digit a:
- a = 9: b + c = 16 → (7,9), (8,8), (9,7) → 979, 988, 997 (3 numbers)
- a = 8: b + c = 17 → (8,9), (9,8) → 889, 898 (2 numbers)
- a = 7: b + c = 18 → (9,9) → 799 (1 number)
- a ≤ 6: b + c ≥ 19, but max b + c = 18 → impossible

Total = 3 + 2 + 1 = **6**.

ANSWER 9: C

---

Problem 10:
Exits 5 through 26 give 22 exits and 21 consecutive gaps. Total distance = 118 miles. Each gap ≥ 5 miles.

To maximize one gap, minimize the other 20 gaps to 5 miles each:
Max gap + 20(5) = 118
Max gap + 100 = 118
Max gap = **18** miles.

ANSWER 10: C

---

Problem 11:
Five scores sorted: a ≤ b ≤ c ≤ d ≤ e.
- Mean = 90 → sum = 450
- Median = 91 → c = 91
- Mode = 94 → 94 appears most frequently (at least twice)

Since a ≤ b ≤ c = 91, neither a nor b can be 94. So 94 must appear among d and e. For 94 to be the mode, we need d = e = 94 (appearing twice).

Sum: a + b + 91 + 94 + 94 = 450
a + b = 450 − 279 = **171**

(We can verify valid scores 

 Problem 1:
We are given a set \(R\) of nine distinct integers, six of which are \(2,3,4,6,9,14\). We must determine how many different values are possible for the median of \(R\).

Because there are nine distinct integers, the median is the fifth smallest element when the set is listed in increasing order. Let the three unknown distinct integers be chosen freely (as long as they are not equal to any of the six given numbers). We examine two cases: when the median is one of the six known numbers and when it is one of the three unknown numbers.

**Case 1: The median is a known number.**
Let the median be \(m\). Of the six known numbers, let \(L\) be how many are less than \(m\). Then there are \(5-L\) known numbers greater than \(m\). To make \(m\) the fifth element, we need exactly four numbers smaller than \(m\) and four larger. Thus we need \(4-L\) unknown numbers below \(m\) and \(4-(5-L)=L-1\) unknown numbers above \(m\). This uses exactly \((4-L)+(L-1)=3\) unknowns, which is perfect. We need \(4-L\ge 0\) and \(L-1\ge 0\), i.e. \(1\le L\le 4\).

- \(m=2\): \(L=0\) → impossible.
- \(m=3\): \(L=1\) → need 3 unknowns \(<3\), 0 above. Possible (e.g. \(-1,0,1\)).
- \(m=4\): \(L=2\) → need 2 unknowns \(<4\), 1 above. Possible (e.g. \(0,1,5\)).
- \(m=6\): \(L=3\) → need 1 unknown \(<6\), 2 above. Possible (e.g. \(5,7,8\)).
- \(m=9\): \(L=4\) → need 0 unknowns \(<9\), 3 above. Possible (e.g. \(10,11,12\)).
- \(m=14\): \(L=5\) → need \(-1\) unknowns below; impossible.

So the known possible medians are \(3,4,6,9\).

**Case 2: The median is an unknown number \(x\).**
Then \(x\) itself uses one of the three unknown slots. We need four numbers below \(x\) and four above. With \(L\) known numbers below \(x\), we need \(4-L\) other unknowns below \(x\) and \(4-(6-L)=L-2\) other unknowns above \(x\). The total other unknowns used is \((4-L)+(L-2)=2\), which matches the two remaining unknowns. We need \(4-L\ge 0\) and \(L-2\ge 0\), so \(2\le L\le 4\).

- \(L=2\): known below are \(2,3\); known above are \(4,6,9,14\). Then \(3<x<4\), impossible for an integer.
- \(L=3\): known below are \(2,3,4\); known above are \(6,9,14\). Then \(4<x<6\), so \(x=5\). This works (e.g. add \(1,5,7\)).
- \(L=4\): known below are \(2,3,4,6\); known above are \(9,14\). Then \(6<x<9\), so \(x=7\) or \(x=8\). Both work (e.g. add \(7,10,11\) or \(8,10,11\)).

Thus the unknown possible medians are \(5,7,8\).

Collecting all possibilities: \(\{3,4,5,6,7,8,9\}\). That is **7** different values. The tempting answer of 4 counts only the known medians, and 2 or 14 are impossible because too many numbers would have to be forced to one side.

ANSWER 1: D

---

Problem 2:
We insert a number \(N\) into the list \(2,6,7,7,28\) so that the new mean of the six numbers is twice the new median. Find \(N\).

The original sum is \(2+6+7+7+28=50\). After insertion, the sum is \(50+N\) and the mean is \(\frac{50+N}{6}\).

The median depends on the size of \(N\) relative to the ordered list. Because the list contains six numbers, the median is the average of the third and fourth values.

- If \(N\le 6\) (as an integer), the ordered list begins with \(N\) or \(2\), then the third and fourth terms are \(6\) and \(7\), giving median \(\frac{6+7}{2}=6.5\).
  Setting mean \(=2\times 6.5=13\): \(\frac{50+N}{6}=13\Rightarrow N=28\). But \(N=28\) does not satisfy \(N\le 6\), so this case yields no valid solution.

- If \(N\ge 7\) (integer), the third and fourth terms are both \(7\) (since the two \(7\)’s occupy those middle slots, or \(N=7\) makes three \(7\)’s), so the median is \(7\).
  Setting mean \(=2\times 7=14\): \(\frac{50+N}{6}=14\Rightarrow 50+N=84\Rightarrow N=34\).

Check: the list becomes \(2,6,7,7,28,34\). The median is \(\frac{7+7}{2}=7\). The mean is \(\frac{84}{6}=14\), which is indeed twice the median. The other choices fail: \(N=28\) gives mean \(13\) and median \(7\); \(N=14\) gives mean \(\frac{64}{6}\neq 14\).

ANSWER 2: E

---

Problem 3:
We consider all positive four-digit integers whose digits are all even. We need the fraction of them that are divisible by \(4\).

Even digits are \(\{0,2,4,6,8\}\). For a four-digit number \(ABCD\):
- \(A\) (thousands) has 4 choices: \(\{2,4,6,8\}\).
- \(B,C,D\) each have 5 choices.

Total count: \(4\cdot 5\cdot 5\cdot 5 = 500\).

A number is divisible by \(4\) exactly when its last two digits form a multiple of \(4\). Here \(C\) and \(D\) are both even. Consider \(10C+D \pmod 4\). Since \(C\) is even, write \(C=2k\); then \(10C=20k\equiv 0\pmod 4\). Thus \(10C+D\equiv D\pmod 4\). So we simply need \(D\equiv 0\pmod 4\).

Among the even digits, those divisible by \(4\) are \(0,4,8\): **3 choices** for \(D\). The digits \(A,B,C\) are unrestricted (4, 5, and 5 choices).

Count of favorable numbers: \(4\cdot 5\cdot 5\cdot 3 = 300\).

The fraction is \(\frac{300}{500}=\frac{3}{5}\). One might mistakenly think the fraction is \(\frac12\) because half of all integers are divisible by \(4\), but the restriction to even digits changes the units-digit distribution.

ANSWER 3: D

---

Problem 4:
Find the difference between a \(6.5\%\) sales tax and a \(6\%\) sales tax on a \(\$20\) item.

Method 1: Compute each tax.
- \(6.5\%\) of \(\$20 = 0.065\times 20 = \$1.30\).
- \(6\%\) of \(\$20 = 0.06\times 20 = \$1.20\).

Difference: \(\$1.30-\$1.20 = \$0.10\).

Method 2: The difference in rates is \(0.5\%\), and \(0.5\%\) of \(\$20 = 0.005\times 20 = \$0.10\).

The distractors correspond to misplacing the decimal (e.g. \(\$1\) would be \(5\%\) of \(\$20\)).

ANSWER 4: B

---

Problem 5:
Evaluate \((1901+1902+\dots+1993)-(101+102+\dots+193)\).

Each sum has the same number of terms:
\(1993-1901+1 = 93\) terms, and \(193-101+1 = 93\) terms.

Pair the terms:
\((1901-101)+(1902-102)+\dots+(1993-193)\).

Each difference equals \(1800\). With 93 such pairs, the total is
\(93\times 1800 = 93\times 18\times 100 = 1674\times 100 = 167{,}400\).

A common error is miscounting the number of terms as 92, which would give \

Problem 1:
R is a set of nine distinct integers with six known elements: 2, 3, 4, 6, 9, 14. Since R has 9 elements, the median is the 5th element when sorted.

I need to determine which integers can be the 5th element. The median must have exactly 4 elements less than it and 4 elements greater than it.

**Checking each candidate:**

- **Median = 3**: Need 4 elements < 3. Known: {2}. Need 3 unknowns all < 3 (e.g., 1, 0, -1). Elements > 3: {4, 6, 9, 14} = 4. ✓
- **Median = 4**: Need 4 elements < 4. Known: {2, 3}. Need 2 unknowns < 4. Elements > 4: {6, 9, 14} + 1 unknown > 4. ✓
- **Median = 5**: Need 4 elements < 5. Known: {2, 3, 4}. Need 1 unknown < 5. Elements > 5: {6, 9, 14} + 1 unknown > 5. (5 is an unknown) ✓
- **Median = 6**: Need 4 elements < 6. Known: {2, 3, 4}. Need 1 unknown < 6. Elements > 6: {9, 14} + 2 unknowns > 6. ✓
- **Median = 7**: Need 4 elements < 7. Known: {2, 3, 4, 6} = 4. Elements > 7: {9, 14} + 2 unknowns > 7. (7 is an unknown) ✓
- **Median = 8**: Need 4 elements < 8. Known: {2, 3, 4, 6} = 4. Elements > 8: {9, 14} + 2 unknowns > 8. (8 is an unknown) ✓
- **Median = 9**: Need 4 elements < 9. Known: {2, 3, 4, 6} = 4. Elements > 9: {14} + 3 unknowns > 9. ✓

**Values that cannot be median:**
- 2: Need 4 elements < 2, but only 3 unknowns available. ✗
- 10-13: Already have 5 known elements less than these. ✗
- 14: Already have 5 known elements less than 14. ✗

Possible median values: {3, 4, 5, 6, 7, 8, 9} → **7 values**

ANSWER 1: D

---

Problem 2:
List: 2, 6, 7, 7, 28. Insert N so that mean = 2 × median.

Sum of original list = 50, so new mean = (50 + N)/6.

**Finding the median based on N:**
- If N > 7: Sorted list is 2, 6, 7, 7, 28, N (or 2, 6, 7, 7, N, 28 if 7 < N < 28). Median = (7+7)/2 = 7.
- If N ≤ 6: Median = (6+7)/2 = 6.5
- If 6 < N ≤ 7: Median = (N+7)/2

**Solving for N > 7:**
(50 + N)/6 = 2(7) = 14
50 + N = 84
N = 34

Verification: List = 2, 6, 7, 7, 28, 34. Mean = 84/6 = 14. Median = 7. ✓

ANSWER 2: E

---

Problem 3:
Four-digit integers with all even digits. Even digits: {0, 2, 4, 6, 8}.

**Total count:** First digit: 4 choices (2,4,6,8), others: 5 choices each → 4 × 5³ = 500

**Divisible by 4:** Last two digits must form a number divisible by 4.

Listing all 25 pairs of even last digits and checking divisibility by 4:
- 00✓, 02✗, 04✓, 06✗, 08✓
- 20✓, 22✗, 24✓, 26✗, 28✓
- 40✓, 42✗, 44✓, 46✗, 48✓
- 60✓, 62✗, 64✓, 66✗, 68✓
- 80✓, 82✗, 84✓, 86✗, 88✓

That's 15 valid pairs.

**Count divisible by 4:** 4 × 5 × 15 = 300

**Fraction:** 300/500 = 3/5

ANSWER 3: D

---

Problem 4:
6.5% tax on $20 = 0.065 × 20 = $1.30
6% tax on $20 = 0.06 × 20 = $1.20

Difference = $1.30 - $1.20 = $0.10

ANSWER 4: B

---

Problem 5:
(1901 + 1902 + ... + 1993) − (101 + 102 + ... + 193)

Both sums have 93 terms each.

First sum = 93 × (1901 + 1993)/2 = 93 × 1947
Second sum = 93 × (101 + 193)/2 = 93 × 147

Difference = 93 × (1947 − 147) = 93 × 1800 = 167,400

ANSWER 5: A

---

Problem 6:
1 − 2 − 3 + 4 + 5 − 6 − 7 + 8 + ... + 1993 − 1994 − 1995 + 1996

The pattern repeats every 4 terms: +, −, −, +

Group k: (4k−3) − (4k−2) − (4k−1) + 4k = 4k−3−4k+2−4k+1+4k = 0

Each group of 4 sums to 0.

1996 ÷ 4 = 499 complete groups.

Total = 499 × 0 = 0

ANSWER 6: C

---

Problem 7:
50 people wear sunglasses, 35 wear caps.

P(sunglasses | cap) = 2/5 means: (both)/(caps) = 2/5

Both/35 = 2/5 → Both = 14

P(cap | sunglasses) = (both)/(sunglasses) = 14/50 = 7/25

ANSWER 7: B

---

Problem 8:
3-digit numbers with digit-sum 26 that are even.

For number abc: a + b + c = 26, c is even, a ≥ 1, all digits ≤ 9.

- c = 8: a + b = 18 → only (9,9) → 998 ✓
- c = 6: a + b = 20 → impossible (max 9+9=18)
- c = 4, 2, 0: a + b ≥ 22 → impossible

Only 1 number: 998

ANSWER 8: A

---

Problem 9:
Three-digit numbers with digit-sum 25.

a + b + c = 25, a ≥ 1, all digits ≤ 9.

- a = 9: b + c = 16 → (7,9), (8,8), (9,7) → 3 numbers
- a = 8: b + c = 17 → (8,9), (9,8) → 2 numbers
- a = 7: b + c = 18 → (9,9) → 1 number
- a ≤ 6: b + c ≥ 19 → impossible

Total = 3 + 2 + 1 = 6

ANSWER 9: C

---

Problem 10:
Distance from exit 5 to exit 26 is 118 miles. There are 21 gaps between consecutive exits (5→6, 6→7, ..., 25→26).

Each gap ≥ 5 miles. To maximize one gap, minimize all others.

If 20 gaps are minimized at 5 miles each: 20 × 5 = 100 miles.

Maximum single gap = 118 − 100 = 18 miles.

ANSWER 10: C

---

Problem 11:
Five scores: mean = 90, median = 91, mode = 94.

Sum = 5 × 90 = 450

Sorted: a ≤ b ≤ c ≤ d ≤ e
- c = 91 (median)
- Mode = 94, so 94 appears at least twice. Since c = 91, the 94s must be d and e.

a + b + 91 + 94 + 94 = 450
a + b = 171

ANSWER 11: B

---

Problem 12:
Using +, −, × each exactly once in 5 __ 4 __ 6 __ 3:

| Operations | Result |
|------------|--------|
| 5 + 4 − 6 × 3 | 5 + 4 − 18 = −9 |
| 5 + 4 × 6 − 3 | 5 + 24 − 3 = 26 |
| 5 − 4 + 6 × 3 | 5 − 4 + 18 = 19 |
| 5 − 4 × 6 + 3 | 5 − 24 + 3 = −16 |
| 5 × 4 + 6 − 3 | 20 + 6 − 3 = 23 |
| 5 × 4 − 6 + 3 | 20 − 6 + 3 = 17 |

From choices {9, 10, 15, 16, 19}, only **19** is achievable.

ANSWER 12: E

**Problem 1:**  
We have nine distinct integers, six of which are known: 2, 3, 4, 6, 9, 14.  
The median of a nine‑element set is the 5th smallest number.  

Let the unknown integers be \(x,y,z\) (all distinct and distinct from the known six).  
When the numbers are sorted, the median is the 5th term.  

For a known number \(k_i\) (the \(i\)‑th smallest of the known six) to be the median, we need exactly four numbers smaller than it.  
If \(i-1\) known numbers are smaller, we need \(L\) unknown numbers smaller with  
\((i-1)+L=4\Rightarrow L=5-i\).  
Since we have only three unknowns, \(L\ge0\) gives \(i\le5\); also \(L\le3\) gives \(i\ge2\).  
Thus \(i=2,3,4,5\) → medians could be 3, 4, 6, 9.

Can the median be an unknown integer?  
Let the median be an unknown \(u\).  
If \(u\) lies between the known numbers, we need exactly four numbers below it.  
With two other unknowns, the possibilities are:
- \(u\) between 4 and 6 → \(u=5\);
- \(u\) between 6 and 9 → \(u=7\) or \(u=8\).

All three of these can be realized while keeping the numbers distinct.  
Thus the possible median values are \(\{3,4,5,6,7,8,9\}\), a total of **7** values.

**Answer:** 7 → Choice **D**.  

ANSWER 1: D  

---

**Problem 2:**  
We start with the list 2, 6, 7, 7, 28 and insert a number \(N\).  
Let the six numbers be sorted. The median of six numbers is the average of the 3rd and 4th smallest values.

Let the sorted list after insertion be \(a_1\le a_2\le a_3\le a_4\le a_5\le a_6\).  
The median is \(\dfrac{a_3+a_4}{2}\).  
The mean is \(\dfrac{\text{sum}}{6}\).  
Given that the mean equals twice the median:
\[
\frac{\text{sum}}{6}=2\cdot\frac{a_3+a_4}{2}=a_3+a_4 .
\]
Hence \(\text{sum}=6(a_3+a_4)\).

Consider where \(N\) can fall:

*If \(N\le2\):* \(a_3=6,\ a_4=7\) → median \(=6.5\).  
Sum \(=N+50\). Equation gives \(N=28\), contradicting \(N\le2\).

*If \(2<N\le6\):* \(a_3=6,\ a_4=7\) → same median \(=6.5\).  
Again \(N=28\), impossible.

*If \(6<N\le7\):* only integer possibility \(N=7\).  
List becomes 2, 6, 7, 7, 7, 28 → median \(=7\).  
Sum \(=57\) → mean \(=9.5\neq14\).

*If \(7<N<28\):* \(a_3=7,\ a_4=7\) → median \(=7\).  
Sum \(=N+50\). Equation \(N+50=84\) gives \(N=34\), which is not \(<28\).

*If \(N=28\):* median still \(=7\). Sum \(=78\) → mean \(=13\neq14\).

*If \(N>28\):* \(a_3=7,\ a_4=7\) → median \(=7\).  
Equation gives \(N+50=84\) → \(N=34\). This works (34 > 28).

Thus the only possible value is **34**.

**Answer:** 34 → Choice **E**.  

ANSWER 2: E  

---

**Problem 3:**  
A four‑digit integer with all digits even can use digits \{0,2,4,6,8\} for each place, except the thousands digit cannot be 0.  

Number of such integers: \(4\times5\times5\times5=500\).

A number is divisible by 4 iff its last two digits form a number divisible by 4.  
List all two‑digit endings with even digits (25 possibilities).  

Checking each yields 15 endings that are multiples of 4:  
00, 04, 08, 20, 24, 28, 40, 44, 48, 60, 64, 68, 80, 84, 88.

Thus the proportion is \(\frac{15}{25}=\frac{3}{5}\).

**Answer:** \(\frac{3}{5}\) → Choice **D**.  

ANSWER 3: D  

---

**Problem 4:**  
6 % tax on \$20 = \(0.06\times20 = \$1.20\).  
6.5 % tax on \$20 = \(0.065\times20 = \$1.30\).  

Difference = \$1.30 − \$1.20 = **\$0.10**.

**Answer:** \$0.10 → Choice **B**.  

ANSWER 4: B  

---

**Problem 5:**  
Sum from 1901 to 1993 (inclusive):  
Number of terms = \(1993-1901+1 = 93\).  
Sum = \(\frac{(1901+1993)\times93}{2}= \frac{3894\times93}{2}=181\,071\).

Sum from 101 to 193 (inclusive):  
Number of terms = \(193-101+1 = 93\).  
Sum = \(\frac{(101+193)\times93}{2}= \frac{294\times93}{2}=13\,671\).

Difference = \(181\,071-13\,671 = 167\,400\).

**Answer:** 167 400 → Choice **A**.  

ANSWER 5: A  

---

**Problem 6:**  
The sign pattern for the numbers \(n\) is:
- \(n\equiv1\pmod4\) → +,
- \(n\equiv2\pmod4\) → −,
- \(n\equiv3\pmod4\) → −,
- \(n\equiv0\pmod4\) → +.

Thus each block of four consecutive integers \(\{4k+1,4k+2,4k+3,4k+4\}\) contributes  
\((4k+1)-(4k+2)-(4k+3)+(4k+4)=0\).

From 1 through 1996 there are \(1996/4=499\) complete blocks, so the total sum is **0**.

**Answer:** 0 → Choice **C**.  

ANSWER 6: C  

---

**Problem 7:**  
Let \(S\) = people wearing sunglasses, \(|S|=50\).  
Let \(C\) = people wearing caps, \(|C|=35\).  
Let \(x=|S\cap C|\).

Given \(P(S\mid C)=\frac{x}{35}= \frac{2}{5}\).  
Thus \(x = \frac{2}{5}\times35 = 14\).

We need \(P(C\mid S)=\frac{x}{50}= \frac{14}{50}= \frac{7}{25}\).

**Answer:** \(\frac{7}{25}\) → Choice **B**.  

ANSWER 7: B  

---

**Problem 8:**  
A three‑digit number with digit sum 26 must use digits that sum to 26.  
The only triple of single‑digit numbers with sum 26 is \(\{9,9,8\}\).

The number is even only if its units digit is 8.  
Among the permutations of 9, 9, 8, only 998 ends in 8.  

Thus exactly **1** such number exists.

**Answer:** 1 → Choice **A**.  

ANSWER 8: A  

---

**Problem 9:**  
Let the digits be \(a,b,c\) (hundreds, tens, units) with  
\(1\le a\le9,\ 0\le b,c\le9,\ a+b+c=25\).

Set \(a'=a-1\) (so \(0\le a'\le8\)). Then  
\(a'+b+c=24\).

Count unrestricted non‑negative solutions: \(\binom{24+3-1}{2}= \binom{26}{2}=325\).

Subtract solutions where a digit exceeds its bound:
- \(a'\ge9\): treat \(a''=a'-9\) → \(a''+b+c=15\) → \(\binom{17}{2}=136\).
- \(b\ge10\): \(b'=b-10\) → \(a'+b'+c=14\) → \(\binom{16}{2}=120\).
- \(c\ge10\): similarly 120.

Intersections:
- \(a'\ge9,\ b\ge10\): \(a''+b'+c=5\) → \(\binom{7}{2}=21\).
- \(a'\ge9,\ c\ge10\): 21.
- \(b\ge10,\ c\ge10\): \(a'+b'+c'=4\) → \(\binom{6}{2}=15\).

No triple intersection (would require sum –5).

By inclusion–exclusion, invalid = \(136+120+120-21-21-15 = 319\).  
Valid = \(325-319 = 6\).

**Answer:** 6 → Choice **C**.  

ANSWER 9: C  

---

**Problem 10:**  
Exits numbered 5 through 26 give 22 exits, hence 21 intervals.  
Let the distances between consecutive exits be \(d_1,\dots,d_{21}\).  
We know \(\sum d_i = 118\) miles, and each \(d_i\ge5\).

To maximize the largest interval, make the othe

Problem 1:  
We are given a set $ R $ of nine distinct integers, six of which are $ 2, 3, 4, 6, 9, 14 $. The median of a set of nine numbers is the fifth number when sorted. To find the number of possible values for the median, we consider the possible placements of the three unknown numbers relative to the known ones. The known numbers sorted are $ 2, 3, 4, 6, 9, 14 $. The three unknown numbers can be placed in various positions to shift the median. The possible medians occur when the additional numbers are placed such that the fifth number in the sorted list can be $ 4, 5, 6, 7, 8, $ or $ 9 $. This gives six possible values for the median.  
ANSWER 1: D (corrected to C? Wait, no: the possible medians are 4, 5, 6, 7, 8, 9 — that’s 6 values, so C)  
**Wait, let's re-evaluate:** The known numbers are 2,3,4,6,9,14. Adding three numbers. The median is the 5th number. To get 4 as median: three numbers ≤4, but only 2,3,4 are available — but they are distinct, so we can add numbers like 1,5? Wait no, the three unknown numbers can be placed in different positions. For example:  
- If all three unknown numbers are ≤4: but only three such numbers exist (2,3,4) but they are already in the set. So we cannot add more. Thus, the unknown numbers must be placed in other positions.  
Actually, the unknown numbers can be:  
- All less than 2: median could be 4 if three numbers are added <4, but only 2,3,4 are already present. We cannot add numbers less than 2 without repeating? Wait, the set has distinct integers. So we can add numbers like 1,0,-1, but the problem doesn’t restrict to positive. However, the known numbers are positive. But the problem says "distinct integers", so negatives are allowed. But the median values: if we add numbers less than 2, the sorted order becomes [new numbers, 2,3,4,6,9,14]. With three new numbers, the sorted list has 9 numbers. The fifth is the median. If we add three numbers less than 2, the sorted list is [a,b,c,2,3,4,6,9,14] — no, wait, adding three numbers less than 2: the sorted list would be [a,b,c,2,3,4,6,9,14] — but a,b,c are less than 2, so the first three, then 2 is fourth, 3 is fifth? Wait, no: the original known numbers are 2,3,4,6,9,14. Adding three numbers, say x,y,z. The full sorted list will be the combination. To get median=4, we need the fifth number to be 4. So in the sorted list of 9, the fifth must be 4. That requires that four numbers are ≤4 and four are ≥4? No: the fifth is the middle. For the fifth to be 4, we need at least five numbers ≤4? No: in a sorted list of 9, the fifth is the median. For the fifth to be 4, we need at least five numbers ≤4? Actually, no: the fifth number is the one where four are less and four are greater? No: in a sorted list of 9, the fifth is the median, meaning four are ≤ median and four are ≥ median. But since numbers can be equal, but here all distinct. So for 4 to be the median, we need exactly four numbers less than 4 and four numbers greater than 4? But 4 is in the set. So the sorted list must have four numbers <4, then 4, then four numbers >4. But the known numbers include 2,3,4,6,9,14. So 2,3 are <4, and 6,9,14 are >4. So currently, without unknowns, we have two <4 and three >4. Adding three unknowns. To have four <4 and four >4, we need to add two numbers <4 and one number >4? But then the sorted list would be [two new <4, 2,3,4, then the new >4, then 6,9,14] — but that’s only three numbers after 4? Wait, no: the sorted list would be [new <4 (two), 2,3,4, new >4, 6,9,14] — that’s 9 numbers? 2+3+1+3=9? Wait, no: the two new <4, then 2,3,4 (three numbers), then the new >4, then 6,9,14 (three numbers). So total: 2+3+1+3=9. The fifth number is 4. So yes, we can achieve median=4 by adding two numbers <4 and one number >4. Similarly, we can achieve:  
- Median=5: by adding numbers such that the fifth number is 5. But 5 is not in the set. So we need the fifth number to be 5. This requires that four numbers are <5 and four are >5. But 5 is not present. So we need to add 5? But the set must consist of distinct integers, and 5 is not in the given. So we can add 5. Then the sorted list would include 5. For example, add 5, and two other numbers. To have four <5 and four >5: currently, without unknowns, numbers <5: 2,3,4 (three), numbers >5: 6,9,14 (three). So to have four <5, we need to add one number <5. To have four >5, we need to add one number >5. But we are adding three numbers. So we can add 5, and say 1 and 10. Then sorted: 1,2,3,4,5,6,9,10,14. Median is 5. So median=5 is achievable. Similarly, median=6: we can add numbers such that the fifth is 6. Currently, numbers <6: 2,3,4 (three), numbers >6: 9,14 (two). So to have four <6 and four >6, we need to add one number <6 and two numbers >6. For example, add 5,10,15. Then sorted: 2,3,4,5,6,9,10,14,15. Median=6. Similarly, median=7: add numbers so fifth is 7. Currently, <7: 2,3,4,6 (four), >7: 9,14 (two). So to have four <7 and four >7, we need to add zero <7 (since already four) and two >7. But we are adding three numbers. So we can add 7,8,15. Then sorted: 2,3,4,6,7,8,9,14,15. Median=7. Similarly, median=8: add 8, and two numbers, say 10,11. Then sorted: 2,3,4,6,8,9,10,11,14. Median=8? The fifth is 8. But to have four <8 and four >8: currently, without unknowns, <8: 2,3,4,6 (four), >8:9,14 (two). So to have four >8, we need to add two numbers >8. But we are adding three numbers. So we can add 8,9,10? But 9 is already there. Add 8,10,11. Then sorted: 2,3,4,6,8,9,10,11,14. The fifth is 8. So yes. Similarly, median=9: add numbers so fifth is 9. Currently, <9: 2,3,4,6 (four), >9:14 (one). So to have four >9, we need to add three numbers >9. But we are adding three numbers. So add 10,11,12. Then sorted: 2,3,4,6,9,10,11,12,14. Median=9. So possible medians: 4,5,6,7,8,9. That’s six values.  
ANSWER 1: C  

Problem 2:  
We are given the list 2,6,7,7,28. A number $ N $ is added. The mean becomes twice the median. The original list has five numbers. After adding $ N $, there are 

### Problem 1:
We have a set of 9 distinct integers, so the median (middle value when sorted) is the 5th element. The 6 known numbers sorted are [2,3,4,6,9,14], with 3 additional distinct integers to add. To find possible median values:
- The minimum possible median is 3: add 3 numbers smaller than 2, sorted set becomes [a,b,c,2,3,4,6,9,14], 5th term=3.
- The maximum possible median is 9: add 3 numbers larger than 14, sorted set becomes [2,3,4,6,9,14,x,y,z], 5th term=9.
- All integers between 3 and 9 are valid medians: 3,4,5,6,7,8,9, for 7 total possible values.
Tempting wrong choices: any count less than 7, from missing intermediate medians that can be formed by adding integers between the known values.
ANSWER 1: D

---

### Problem 2:
Original list sum: 2+6+7+7+28=50. Adding N gives a list of 6 elements, sum=50+N, mean=(50+N)/6. The median of 6 elements is the average of the 3rd and 4th sorted terms. The problem states mean=2×median, so 50+N=12×median.
- If N>28, sorted list is [2,6,7,7,28,N], median=(7+7)/2=7. Then 50+N=12×7=84 → N=34, which fits.
Testing other options confirms no other N satisfies the condition.
ANSWER 2: E

---

### Problem 3:
A number is divisible by 4 if its last two digits form a number divisible by 4. For 4-digit numbers with all even digits:
- Total count: 4 (non-zero thousands digits:2,4,6,8) ×5×5×5 (all even digits for other places)=500.
- Valid last two-digit pairs (both even, divisible by 4): For any even tens digit, 10×even is divisible by 4, so only units digits 0,4,8 work, giving 3 valid units digits per tens digit: 5×3=15 valid pairs.
- Fraction: (4×5×15)/500=300/500=3/5.
ANSWER 3: D

---

### Problem 4:
The tax rate difference is 6.5%-6%=0.5%. 0.5% of $20 is 0.005×20=$0.10.
ANSWER 4: B

---

### Problem 5:
Both sums have the same number of terms: 1993-1901+1=93 terms, 193-101+1=93 terms. Subtract term-by-term: (1901-101)+(1902-102)+...+(1993-193)=1800×93=167400.
ANSWER 5: A

---

### Problem 6:
Group terms in sets of 4: (1-2-3+4)+(5-6-7+8)+... each group sums to 0. There are 1996/4=499 groups, total sum=0.
ANSWER 6: C

---

### Problem 7:
Number of people with both caps and sunglasses: 35 cap-wearers × 2/5=14. Probability a sunglass-wearer has a cap:14/50=7/25.
ANSWER 7: B

---

### Problem 8:
An even 3-digit number with digit sum 26 must have an even units digit. The only possible even units digit is 8 (any smaller even digit leaves a sum of the first two digits >18, impossible). The only number is 998, so 1 valid number.
ANSWER 8: A

---

### Problem 9:
Three-digit numbers with digit sum 25 are all permutations of (9,9,7) and (9,8,8). There are 3 permutations of (9,9,7) and 3 permutations of (9,8,8), for 6 total numbers.
ANSWER 9: C

---

### Problem 10:
Between exit 5 and 26, there are 21 gaps. To maximize one gap, minimize the other 20 gaps at 5 miles each: 20×5=100 miles. Maximum gap=118-100=18 miles.
ANSWER 10: C

---

### Problem 11:
Total sum of 5 scores=5×90=450. Median=91, so sorted scores [a,b,91,c,d]. Mode=94, so 94 must appear at least twice, so c=d=94, sum c+d=188. Sum of the two lowest scores=450-91-188=171.
ANSWER 11: B

---

### Problem 12:
Using +,-,* exactly once, one valid calculation: 5-4+6×3=1+18=19, which uses all three operations and matches option E.
ANSWER 12: E

### Problem 1:
We need to find the number of possible median values for a set of 9 distinct integers, 6 of which are 2, 3, 4, 6, 9, 14. The median of 9 numbers is the 5th element when sorted in ascending order. We add 3 distinct integers (not in the given set) and analyze possible medians:
1.  **Case 1: All 3 new numbers < 2**: Sorted list is [a,b,c,2,3,4,6,9,14], median = 3.
2.  **Case 2: 2 new numbers < 2, 1 new number ≥4**: Sorted list is [a,b,2,3,4,...], median =4.
3.  **Case 3: 1 new number <2**:
    - If we add 5 (the only integer between 4 and 6): Sorted list is [a,2,3,4,5,6,...], median=5.
    - If we do not add 5: Sorted list is [a,2,3,4,6,...], median=6.
4.  **Case 4: 0 new numbers <2 (all new numbers ≥5)**:
    - Add 5: Sorted list [2,3,4,5,6,...], median=6.
    - Add a number between 6 and9 (7 or 8): Sorted list [2,3,4,6,7,...] or [2,3,4,6,8,...], median=7 or 8.
    - Add all new numbers ≥9: Sorted list [2,3,4,6,9,...], median=9.
The only impossible medians are 2 (needs 5 numbers ≤2, only 4 max) and values >9 (existing numbers already have 5 elements ≤9). The possible medians are 3,4,5,6,7,8,9: 7 values.
ANSWER 1: D
---
### Problem 2:
Original list: [2,6,7,7,28], sum = 2+6+7+7+28=50. Insert N to make 6 elements. The median of 6 elements is the average of the 3rd and 4th sorted elements. The mean is $\frac{50+N}{6}$, which equals twice the median:
$$\frac{50+N}{6} = 2 \times \text{median} \implies 50 + N = 12 \times \text{median} \implies N = 12 \times \text{median} -50.$$
Test possible median values from the sorted list with N:
- If N ≤7: Sorted list has 3rd=7, 4th=7, median=7. Then $N=12*7 -50=84-50=34$, which is >7, valid.
- Other median values (e.g., 6.5) give N=28, which does not fit the required sorted position.
Verify N=34: Sorted list [2,6,7,7,28,34], median=(7+7)/2=7, mean=(50+34)/6=14, which is twice 7.
ANSWER 2: E
---
### Problem 3:
Positive 4-digit numbers with all even digits:
- Thousands place: 2,4,6,8 (4 choices, no 0).
- Hundreds, tens, units: 0,2,4,6,8 (5 choices each).
Total numbers: $4 \times 5 \times 5 \times5=500$.
A number is divisible by 4 iff its last two digits form a number divisible by 4. For even tens digits (0,2,4,6,8), a two-digit number $10a + b$ is divisible by 4 iff $b$ (units digit) is 0,4,8 (since $10a \equiv 0 \mod4$ for even a). So valid endings: 5 (tens choices) ×3 (units choices)=15.
Total valid numbers: $4 \times5 \times5 \times3=300$. Fraction: $\frac{300}{500}=\frac{3}{5}$.
ANSWER 3: D
---
### Problem 4:
Difference between 6.5% and 6% tax on $20:
$$0.065*20 - 0.06*20 = (0.065-0.06)*20=0.005*20=0.10.$$
ANSWER 4: B
---
### Problem 5:
Both sums have 93 terms (1993-1901+1=93, 193-101+1=93). Each term in the first sum is 1800 more than the corresponding term in the second sum:
Total difference = $93 \times 1800 = 167400$.
ANSWER 5: A
---
### Problem 6:
The sign pattern repeats every 4 terms: $+ - - +$, and each block of 4 sums to 0:
$$(1-2-3+4)+(5-6-7+8)+\dots+(1993-1994-1995+1996)=0.$$
1996 is divisible by 4, so there are 499 such blocks, total sum 0.
ANSWER 6: C
---
### Problem 7:
Let $S$ = sunglass wearers ($|S|=50$), $C$ = cap wearers ($|C|=35$). $P(S|C)=\frac{|S∩C|}{|C|}=\frac{2}{5}$, so $|S∩C|=\frac{2}{5}*35=14$.
The probability a sunglass wearer wears a cap is $P(C|S)=\frac{|S∩C|}{|S|}=\frac{14}{50}=\frac{7}{25}$.
ANSWER 7: B
---
### Problem 8:
3-digit numbers with digit sum 26: maximum digit sum is 27 (999), so 26 is 1 less than maximum, meaning digits are permutations of (9,9,8). The only even permutation is 998 (units digit 8). So 1 number.
ANSWER 8: A
---
### Problem 9:
3-digit numbers with digit sum 25: 25 is 2 less than 27, so digits are permutations of (9,9,7) and (9,8,8):
- Permutations of (9,9,7): 997,979,799 (3 numbers).
- Permutations of (9,8,8): 988,898,889 (3 numbers).
Total 6 numbers.
ANSWER 9: C
---
### Problem 10:
Between exit 5 and 26, there are $26-5=21$ gaps, total distance 118 miles. Each gap ≥5 miles. To maximize one gap, minimize the other 20: $20*5=100$, so maximum gap = $118-100=18$.
ANSWER 10: C
---
### Problem 11:
Five sorted scores: $a≤b≤c≤d≤e$. Median $c=91$, mean 90, so total sum $5*90=450$, so $a+b+d+e=450-91=359$. Mode is 94, so 94 appears at least twice. Since $c=91$, 94 can only be in $d$ and $e$, so $d=e=94$. Then $a+b=359-94-94=171$. This sum is fixed (valid examples exist, e.g., 81+90=171).
ANSWER 11: B
---
### Problem 12:
Permute +, -, × in 5 __4 __6 __3, using standard order of operations (multiply first):
- $5 -4 +6×3 = 5-4+18=19$, which is an option.
All other permutations give values not in the choices.
ANSWER 12: E

I'll solve each problem step by step.

---

**Problem 1:**

We have 6 elements: {2, 3, 4, 6, 9, 14}, and need to add 3 more distinct integers to make 9 total.

The median of 9 numbers is the 5th element when arranged in order.

Currently we have 6 numbers. By strategically placing 3 new numbers, we can shift which value becomes the median.

Let me consider what values can be the median (5th position):
- If we add 3 numbers all ≤ 2, the sorted list has our 6 original numbers in positions 4-9, so median is 4
- If we add numbers strategically between and around the original set, we can make the median be: 2, 3, 4, 6, 9, 14, or a new number
- If we add 3 numbers all ≥ 14, the sorted list has our 6 original numbers in positions 1-6, so median is 9

Testing systematically:
- Median = 2: add three numbers < 2 (e.g., -2, -1, 0) ✓
- Median = 3: add two < 2, one between 3 and 4 ✓
- Median = 4: add three < 2, or arranged appropriately ✓
- Median = 6: add one or two strategically ✓
- Median = 9: add three > 14, or arranged appropriately ✓
- Median = 14: add three > 14 ✓
- Median = a new value between existing numbers ✓

The possible medians are: 2, 3, 4, 6, 9, 14, and any new integer we insert. We can get 7 different values.

**ANSWER 1: D**

---

**Problem 2:**

Original list: 2, 6, 7, 7, 28. Sum = 50.

When N is inserted, we have 6 numbers. Let's find the median of 6 numbers (average of 3rd and 4th when sorted).

Sum of all 6 numbers = 50 + N
Mean = (50 + N)/6

We need: Mean = 2 × Median

Let me test by considering where N falls:
- If N ≤ 2: sorted list is N, 2, 6, 7, 7, 28. Median = (6+7)/2 = 6.5. Mean = 2×6.5 = 13, so 50+N = 78, N = 28. Contradiction.
- If 2 < N ≤ 6: sorted list is 2, N, 6, 7, 7, 28. Median = (6+7)/2 = 6.5. Mean = 13, so N = 28. Contradiction.
- If 6 < N ≤ 7: sorted list is 2, 6, N, 7, 7, 28. Median = (N+7)/2. Mean = 2×(N+7)/2 = N+7. So (50+N)/6 = N+7, giving 50+N = 6N+42, so 5N = 8, N = 1.6. Not in range.
- If 7 < N ≤ 28: sorted list is 2, 6, 7, N, 7, 28 or 2, 6, 7, 7, N, 28. Median = (7+N)/2. Mean = 2×(7+N)/2 = 7+N. So (50+N)/6 = 7+N, giving 50+N = 42+6N, so 5N = 8, N = 1.6. Not valid.
- If N > 28: sorted list is 2, 6, 7, 7, 28, N. Median = (7+7)/2 = 7. Mean = 14. So 50+N = 84, N = 34. ✓

**ANSWER 2: E**

---

**Problem 3:**

Four-digit numbers with all even digits: first digit ∈ {2,4,6,8}, others ∈ {0,2,4,6,8}.

Total count: 4 × 5 × 5 × 5 = 500

For divisibility by 4, the last two digits must form a number divisible by 4.

Last two digits from {00, 02, 04, 06, 08, 20, 22, 24, 26, 28, 40, 42, 44, 46, 48, 60, 62, 64, 66, 68, 80, 82, 84, 86, 88}

Divisible by 4: 00, 04, 08, 20, 24, 28, 40, 44, 48, 60, 64, 68, 80, 84, 88 (15 out of 25)

Count divisible by 4: 4 × 5 × 15 = 300

But wait, the second digit is 5 choices, third is which determines divisibility by 4.

Actually: 4 × 5 × (number of even two-digit combos divisible by 4) = 4 × 5 × 15 = 300

Hmm, let me recalculate. First digit: 4 choices. Second digit: 5 choices. Last two digits must be divisible by 4: check all pairs from {0,2,4,6,8}×{0,2,4,6,8} giving 15 valid pairs.

Count = 4 × 5 × 15 = 300. Fraction = 300/500 = 3/5.

**ANSWER 3: D**

---

**Problem 4:**

6.5% tax on $20: 0.065 × 20 = $1.30
6% tax on $20: 0.06 × 20 = $1.20
Difference: $1.30 - $1.20 = $0.10

**ANSWER 4: B**

---

**Problem 5:**

Sum from 1901 to 1993: This is 93 consecutive integers starting at 1901.
Sum = (1901 + 1993)/2 × 93 = 1947 × 93 = 181,071

Sum from 101 to 193: This is 93 consecutive integers starting at 101.
Sum = (101 + 193)/2 × 93 = 147 × 93 = 13,671

Difference = 181,071 - 13,671 = 167,400

**ANSWER 5: A**

---

**Problem 6:**

Pattern: 1 - 2 - 3 + 4 + 5 - 6 - 7 + 8 + 9 - 10 - 11 + ...

Group by 4s: (1-2-3+4) + (5-6-7+8) + (9-10-11+12) + ... + (1993-1994-1995+1996)

Each group of 4: (4k+1) - (4k+2) - (4k+3) + (4k+4) = 4k+1-4k-2-4k-3+4k+4 = 0

From 1 to 1996, we have 1996/4 = 499 complete groups, each summing to 0.

Total = 0

**ANSWER 6: C**

---

**Problem 7:**

Let x = people wearing both sunglasses and caps.

P(sunglasses | cap) = x/35 = 2/5, so x = 14.

P(cap | sunglasses) = 14/50 = 7/25.

**ANSWER 7: B**

---

**Problem 8:**

Three-digit even numbers with digit sum 26.

For an even number, last digit ∈ {0, 2, 4, 6, 8}.

If last digit = 0: first two digits sum to 26. Max is 9+9=18. Impossible.
If last digit = 2: first two digits sum to 24. Only 9+9=18 < 24. Impossible.
If last digit = 4: first two digits sum to 22. Impossible (max 18).
If last digit = 6: first two digits sum to 20. Impossible.
If last digit = 8: first two digits sum to 18. This requires 9+9=18. Number: 998. ✓

Only 1 number.

**ANSWER 8: A**

---

**Problem 9:**

Three-digit numbers abc where a+b+c=25, with a ∈ {1,...,9}, b,c ∈ {0,...,9}.

a+b+c = 25 means b+c = 25-a.

For valid solutions: 0 ≤ b+c ≤ 18, so 7 ≤ a ≤ 9.

- a=7: b+c=18. Solutions: (9,9) → 1 way
- a=8: b+c=17. Solutions: (8,9), (9,8) → 2 ways
- a=9: b+c=16. Solutions: (7,9), (8,8), (9,7) → 3 ways

Total = 1 + 2 + 3 = 6

**ANSWER 9: C**

---

**Problem 10:**

Exits from 5th to 26th = 26-5 = 21 segments between consecutive exits.

Each segment ≥ 5 miles. Total distance = 118 miles.

If 20 segments are 5 miles each: 20 × 5 = 100 miles.
Remaining for the largest segment: 118 - 100 = 18 miles.

**ANSWER 10: C**

---

**Problem 11:**

Five scores: mean = 90 (sum = 450), median = 91, mode = 94.

Arranged in order: a, b, c, d, e where c = 91 (median).

Mode = 94 means 94 appears at least twice. Since median is 91, we must have 94 appearing in {d, e}.

Case 1: d = e = 94. Then a + b + 91 + 94 + 94 = 450, so a + b = 171.

Check consistency: we need a ≤ b ≤ 91 and mode is 94 (appears twice). If a or b equals 94, mode wouldn't be unique. So a, b < 91 works. ✓

**ANSWER 11: B**

---

**Problem 12:**

We need to place +, −, × in three blanks: 5 __ 4 __ 6 __ 3

Trying combinations:
- 5 + 4 × 6 − 3 = 5 + 24 − 3 = 26
- 5 × 4 + 6 − 3 = 20 + 6 − 3 = 23
- 5 × 4 − 6 + 3 = 20 − 6 + 3 = 17
- 5 + 4 − 6 × 3 = 5 + 4 − 18 = −9
- 5 − 4 

Problem 1:

We are given a set of 9 distinct integers. Six of them are
\[
2,3,4,6,9,14.
\]
We want the number of possible values of the median of the 9 numbers.

For 9 numbers, the median is the 5th number when the list is ordered.

Now look at the six given numbers in sorted order:
\[
2,3,4,6,9,14.
\]
Since there are 3 other distinct integers, the median can be any number that ends up in the 5th position after inserting those 3 numbers.

Let’s see what values are possible:

- If we place 3 new numbers all very small, then the 5th number could be 4.
- It could also be 6, or 9, or even 14, depending on how many new numbers are inserted before each.
- It cannot be 2 or 3, because there would need to be at least 4 numbers less than or equal to it, but only 3 extra numbers are available and the given numbers already limit this.
- It cannot be any number not among the given six if the median must be one of the existing numbers? Let’s check carefully: since the 3 extra numbers are distinct integers and can be placed anywhere, the 5th number could indeed be one of the inserted numbers, but then it would have to lie between the 4th and 6th numbers. Since the six given numbers already occupy several possible positions, the median can also be a new integer. But because the set is of distinct integers and the question asks how many possible values, we should count all possible medians that can occur.

A cleaner way: the median of 9 distinct integers is the middle element. With 6 fixed numbers, the 3 added numbers can be arranged so that the median is any integer between the 4th smallest and 6th smallest of the fixed numbers, inclusive, but here the possible distinct medians are:
\[
4,5,6,7,8,9
\]
Wait — can all of these really happen? Let’s test.

- Median 4: choose 3 new numbers less than 4.
- Median 5: choose 3 new numbers so that 5 is between 4 and 6 and becomes 5th.
- Median 6: choose 3 new numbers with 6 as middle.
- Median 7, 8, 9 similarly.

So the possible medians are the integers from 4 through 9, which is 6 values.

However, we must match the provided choices. The intended count is the number of possible median values among the listed integers and inserted integers. Since the only restriction is distinct integers, every integer from 4 through 9 can occur.

That gives 6 possible values.

ANSWER 1: C

---

Problem 2:

We insert a number \(N\) into the list
\[
2,6,7,7,28.
\]
Then the mean becomes twice the median.

Let’s analyze the median after inserting \(N\) into 6 numbers total.

The original sorted list is:
\[
2,6,7,7,28.
\]
There are 5 numbers. After inserting one number, there are 6 numbers, so the median is the average of the 3rd and 4th numbers in the sorted list.

We try the choices.

If \(N=7\), the list becomes
\[
2,6,7,7,7,28.
\]
Median:
\[
\frac{7+7}{2}=7.
\]
Mean:
\[
\frac{2+6+7+7+7+28}{6}=\frac{57}{6}=9.5.
\]
Twice the median would be 14, not 9.5, so not 7.

Try \(N=14\):
\[
2,6,7,7,14,28.
\]
Median:
\[
\frac{7+7}{2}=7.
\]
Mean:
\[
\frac{2+6+7+7+14+28}{6}=\frac{64}{6}=\frac{32}{3},
\]
not 14.

Try \(N=20\):
\[
2,6,7,7,20,28.
\]
Median is still 7.
Mean:
\[
\frac{2+6+7+7+20+28}{6}=\frac{70}{6}=\frac{35}{3},
\]
not 14.

Try \(N=28\):
\[
2,6,7,7,28,28.
\]
Median is 7.
Mean:
\[
\frac{2+6+7+7+28+28}{6}=\frac{78}{6}=13,
\]
not 14.

Try \(N=34\):
\[
2,6,7,7,28,34.
\]
Median is 7.
Mean:
\[
\frac{2+6+7+7+28+34}{6}=\frac{84}{6}=14.
\]
Twice the median is
\[
2\cdot 7=14.
\]
This works.

So \(N=34\).

ANSWER 2: E

---

Problem 3:

We consider all positive four-digit integers whose digits are all even.

Even digits are
\[
0,2,4,6,8.
\]
A four-digit number cannot start with 0, so the thousands digit has 4 choices:
\[
2,4,6,8.
\]
Each of the other three digits has 5 choices, so the total number is
\[
4\cdot 5^3.
\]

Now determine how many are divisible by 4.

A number is divisible by 4 if its last two digits form a number divisible by 4.

So we only need to count ordered pairs of even digits \((t,u)\) from \(\{0,2,4,6,8\}\) such that \(10t+u\) is divisible by 4.

Check the possible last two digits:

- \(00\) yes
- \(02\) no
- \(04\) yes
- \(06\) no
- \(08\) yes

- \(20\) yes
- \(22\) no
- \(24\) yes
- \(26\) no
- \(28\) yes

- \(40\) yes
- \(42\) no
- \(44\) yes
- \(46\) no
- \(48\) yes

- \(60\) yes
- \(62\) no
- \(64\) yes
- \(66\) no
- \(68\) yes

- \(80\) yes
- \(82\) no
- \(84\) yes
- \(86\) no
- \(88\) yes

For each tens digit, 3 of the 5 possibilities work, so there are
\[
5\cdot 3=15
\]
valid last-two-digit pairs.

Since the first two digits can be anything allowed:
- thousands digit: 4 choices
- hundreds digit: 5 choices

Total divisible by 4:
\[
4\cdot 5\cdot 15
\]
Total numbers:
\[
4\cdot 5^3 = 4\cdot 125
\]
So the fraction is
\[
\frac{4\cdot 5\cdot 15}{4\cdot 5^3}=\frac{15}{25}=\frac35.
\]

ANSWER 3: D

---

Problem 4:

An item costs \$20 before tax.

Difference between 6.5% tax and 6% tax is
\[
0.5\% = 0.005.
\]
So the difference in tax is
\[
0.005 \times 20 = 0.1
\]
dollar, which is \(\$0.10\).

ANSWER 4: B

---

Problem 5:

We need to compute
\[
(1901+1902+\cdots+1993) - (101+102+\cdots+193).
\]

Notice each term in the second sum is exactly 1800 less than the corresponding term in the first sum:
\[
1901-101=1800,\quad 1902-102=1800,\ \ldots,\ 1993-193=1800.
\]
How many terms are there?

From 1901 to 1993 inclusive:
\[
1993-1901+1=93
\]
terms.

So the difference is
\[
93 \cdot 1800 = 167400.
\]

ANSWER 5: A

---

Problem 6:

We evaluate
\[
1-2-3+4+5-6-7+8+9-10-11+\cdots+1992+1993-1994-1995+1996.
\]

Look at the pattern in blocks of 4:
\[
(1-2-3+4),\ (5-6-7+8),\ (9-10-11+12),\ \ldots
\]
Each block of 4 consecutive integers has sum
\[
n-(n+1)-(n+2)+(n+3)=0
\]
when \(n\equiv 1 \pmod 4\).

So all full blocks cancel.

Now check the last block:
The pattern continues through \(1992+1993-1994-1995+1996\), which is not a full 4-term block. But we can group from 1 to 1996 in blocks of 4:
\[
(1-2-3+4)+(5-6-7+8)+\cdots+(1993-1994-1995+1996).
\]
Each block sums to 0.

Therefore the entire 

Problem 1:  
We have 9 distinct integers. Six of them are fixed: 2, 3, 4, 6, 9, 14. The other 3 are distinct integers of our choice.  
With 9 numbers, the median is the 5th number when sorted.

So the median depends only on how many of the fixed numbers fall below/at various positions, and where we insert the 3 new numbers.

Let’s analyze possible 5th elements.

### Step 1: Sort the fixed numbers
Fixed numbers in increasing order:
2, 3, 4, 6, 9, 14

The median (5th) could be one of these fixed numbers, or one of the new ones.

### Casework by choosing what the 5th number is

#### Can the median be 2?
For 2 to be the median, we need at least 4 numbers ≤ 2, but only 2 itself is ≤ 2 among the fixed set (and new numbers would have to be distinct integers ≤ 2).  
That means we’d need 4 numbers less than or equal to 2, including 2, so we’d need 3 additional distinct integers ≤ 2 (other than 2). That’s possible (e.g., 0, −1, 1).  
But then 2 would be the 5th, meaning exactly 4 numbers are <2. Can we guarantee? Actually, since we can choose any 3 extra integers, we can make exactly 4 numbers less than 2 by picking 3 integers <2 plus none from fixed besides 2. That would make 2 the 5th. However, note 2 is already one of the 9 distinct integers, so it can be the median.  
So potentially 2 is possible.

But the problem asks for the number of possible median values, and the choices are between 4 and 8, so likely only medians among the fixed list can work. Let’s check more carefully: the fixed numbers below 2 is only none (except 2). To make 2 the 5th, we need 4 numbers smaller than 2. We only control 3 extra numbers, so we can only create at most 3 numbers smaller than 2. Therefore, we cannot reach 4 smaller numbers.  
✅ Conclusion: median cannot be 2.

Similarly, we test each fixed candidate using “how many fixed numbers are below it”.

### General method
For a number \(x\) to be the median (5th), there must be exactly 4 numbers less than \(x\), and at least one number equal to \(x\). If \(x\) is one of the fixed numbers, we can count how many fixed numbers are less than \(x\), then see if the remaining “needed” smaller numbers can come from the 3 new integers.

Let’s compute fixed counts less than each candidate:

- Below 2: 0  
- Below 3: 1 (just 2)  
- Below 4: 2 (2,3)  
- Below 6: 3 (2,3,4)  
- Below 9: 4 (2,3,4,6)  
- Below 14: 5 (2,3,4,6,9)

We need exactly 4 numbers less than the median.

Now see possibilities:

#### Median = 3
Fixed numbers below 3: 1.  
We need 4 total below 3, so we must add 3 new numbers < 3. That’s possible.  
✅ So 3 can be median.

#### Median = 4
Fixed below 4: 2. Need 2 more from new numbers < 4. Possible.  
✅ 4 can be median.

#### Median = 6
Fixed below 6: 3. Need 1 more from new numbers < 6. Possible.  
✅ 6 can be median.

#### Median = 9
Fixed below 9: 4 already. Then all 4 smaller numbers are fixed, and the median can be 9 regardless of where new numbers go as long as none are between 9 and 14 in a way that changes ordering?  
We only need 9 to be the 5th, meaning at least 4 numbers are <9 (we have exactly 4 fixed ones). If we add any new numbers <9, that would make more than 4 numbers less than 9 and break it. So we must choose the 3 new integers all > 9. That’s possible.  
✅ 9 can be median.

#### Median = 14
Fixed below 14: 5 already. But we need exactly 4 below the median. Impossible.  
❌ 14 cannot be median.

#### Median = 2
Fixed below 2: 0, need 4 from new numbers <2, but we only have 3 new numbers. Impossible.  
❌ 2 cannot be median.

So the only fixed medians are: 3, 4, 6, 9 → 4 values.

Are there possible medians that are not among the fixed numbers (i.e., one of the 3 new integers)?  
Yes, if we choose all new integers appropriately.

Let the median be a new number \(y\). Then among the fixed numbers, some number of them must be less than \(y\). Since there are only 6 fixed numbers, the possible counts of fixed elements less than \(y\) range from 0 to 6.

But we need exactly 4 numbers total less than the median. Since the median is new, all 4 numbers less than it must come from:
- some fixed numbers less than it, plus
- some of the other two new numbers (because one new number is the median itself).

We can make any new median that sits in the “gaps” between fixed numbers such that exactly 4 elements end up below it.

The median \(y\) must be positioned so that:
- the number of fixed elements less than \(y\) is either 1,2,3, or 4 (because we then use the remaining smaller ones from the other new numbers, limited to 0,1,2, or 3).
But we must also ensure distinctness and that \(y\) lies in the right interval.

There are gaps:
(-∞,2), (2,3), (3,4), (4,6), (6,9), (9,14), (14,∞)

Try each gap for possible median:

- If \(y\in(2,3)\): fixed less than y = 1 (only 2). Need 3 more < y from the other new numbers. Possible.  
So any integer y in (2,3) means y=?? none (only integer between 2 and 3 doesn’t exist). So no.
- \(y\in(3,4)\): fixed less than y = 2. Need 2 more < y. Possible. But no integer between 3 and 4. none.
- \(y\in(4,6)\): fixed less than y = 3 (2,3,4). Need 1 more from new numbers <y. Possible. Integers in (4,6): only 5. So y=5 works.
- \(y\in(6,9)\): fixed less than y = 4 (2,3,4,6). Need 0 more. Then both other new numbers must be >y. Possible. Integers in (6,9): 7,8 → both could be median.
- \(y\in(9,14)\): fixed less than y = 5, but we need total 4 < median. impossible.

Thus additional possible medians are 5, 7, 8 → 3 more values.

Total possible median values: fixed ones (3,4,6,9) = 4 plus new ones (5,7,8)=3 gives 7.

✅ Answer choice is 7.

ANSWER 1: D


Problem 2:  
List is: 2, 6, 7, 7, 28 (five numbers). Insert N to make 6 numbers total.

New mean and median relation: “mean is now twice as great as the median.”

### Step 1: Compute original sum
Sum of given numbers:
2 + 6 + 7 + 7 + 28 = 50

After inserting N, total sum = 50 + N.  
There are 6 numbers, so mean is:
\[
\text{mean}=\frac{50+N}{6}
\]

### Step 2: Find the median
Sort of the ne

Problem 1:
We have a set $R$ of 9 distinct integers. Six are $\{2, 3, 4, 6, 9, 14\}$. Let the three unknown integers be $x, y, z$. When sorted, the median is the 5th element.
To minimize the median, we pick the three smallest possible integers (e.g., $-1, 0, 1$). The sorted set would be $\{-1, 0, 1, 2, 3, 4, 6, 9, 14\}$. The median is 3.
To maximize the median, we pick the three largest possible integers (e.g., $15, 16, 17$). The sorted set would be $\{2, 3, 4, 6, 9, 14, 15, 16, 17\}$. The median is 9.
The possible values for the median are integers between 3 and 9 inclusive, excluding values already occupied by the fixed set if they cannot be the median. However, we can choose $x, y, z$ to make any integer from 3 to 9 the median.
Values: 3, 4, 5, 6, 7, 8, 9. There are 7 possible values.
ANSWER 1: D

Problem 2:
List: $2, 6, 7, 7, 28$. $N$ is added. Mean = $\frac{2+6+7+7+28+N}{6} = \frac{50+N}{6}$.
If $N$ is added, the median depends on $N$.
If $N \le 7$, median is 7. Mean = $2 \times 7 = 14 \implies 50+N = 84 \implies N=34$ (Contradiction).
If $N \ge 7$, let's test $N=20$: List $\{2, 6, 7, 7, 20, 28\}$, median is $(7+7)/2 = 7$. Mean = $70/6 \neq 14$.
If $N=14$: List $\{2, 6, 7, 7, 14, 28\}$, median is $(7+7)/2 = 7$. Mean = $64/6 \neq 14$.
If $N=28$: List $\{2, 6, 7, 7, 28, 28\}$, median is 7. Mean = $78/6 = 13 \neq 14$.
If $N=34$: List $\{2, 6, 7, 7, 28, 34\}$, median is $(7+28)/2 = 17.5$. Mean = $84/6 = 14$. $14 \neq 35$.
Wait, try $N=20$: Mean = $70/6 \approx 11.6$. Median = 7.
Try $N=34$: Mean = 14. Median = 17.5.
Re-check: $50+N = 12 \times \text{median}$. If $N=20$, median is 7, $70 = 12 \times 7$ (No).
If $N=34$, median is 17.5, $84 = 12 \times 17.5$ (No).
Try $N=28$: Mean = 13, Median = 7. $13 \neq 14$.
Try $N=20$: Mean = $70/6 = 11.66$, Median = 7.
Let's check $N=20$ again. $50+20 = 70$. $70/6 = 11.66$.
Actually, if $N=20$, list is $2, 6, 7, 7, 20, 28$. Median is $(7+7)/2 = 7$. $2 \times 7 = 14$. $70/6 \neq 14$.
If $N=34$, median is $(7+28)/2 = 17.5$. Mean = $84/6 = 14$. $14 \neq 2 \times 17.5$.
If $N=20$, mean is $11.66$. If $N=34$, mean is 14.
Looking at choices, $N=20$ is the only one that yields a mean close to twice a median.
ANSWER 2: C

Problem 3:
Digits are $\{0, 2, 4, 6, 8\}$. First digit cannot be 0 (4 choices: 2, 4, 6, 8). Others have 5 choices. Total = $4 \times 5 \times 5 \times 5 = 500$.
Divisible by 4 means the last two digits form a multiple of 4. Possible pairs: 00, 04, 08, 20, 24, 28, 40, 44, 48, 60, 64, 68, 80, 84, 88. (15 pairs).
Total numbers = $4 \times 5 \times 15 = 300$.
Fraction = $300/500 = 3/5$.
ANSWER 3: D

Problem 4:
$0.065 \times 20 = 1.30$. $0.06 \times 20 = 1.20$.
Difference = $1.30 - 1.20 = 0.10$.
ANSWER 4: B

Problem 5:
$(1901-101) + (1902-102) + \dots + (1993-193)$.
Each term is $1800$. There are $1993 - 1901 + 1 = 93$ terms.
$93 \times 1800 = 167,400$.
ANSWER 5: A

Problem 6:
Group into sets of 4: $(1-2-3+4) + (5-6-7+8) + \dots + (1993-1994-1995+1996)$.
Each group: $1-2-3+4 = 0$.
There are $1996/4 = 499$ groups. $499 \times 0 = 0$.
ANSWER 6: C

Problem 7:
$C = 35$. $P(S|C) = 2/5$. So $S \cap C = 35 \times 2/5 = 14$.
$S = 50$. $P(C|S) = (S \cap C) / S = 14/50 = 7/25$.
ANSWER 7: B

Problem 8:
Digits $a, b, c$. $a+b+c = 26$. $c \in \{0, 2, 4, 6, 8\}$.
If $c=8, a+b=18 \implies (9,9)$. (1)
If $c=6, a+b=20$ (None).
If $c=4, a+b=22$ (None).
If $c=2, a+b=24$ (None).
If $c=0, a+b=26$ (None).
Wait, $a+b=18 \implies 998$. $a+b=17 \implies 899, 989$. $a+b=16 \implies 799, 979, 889$.
If $c=8, a+b=18 \implies 998$.
If $c=6, a+b=20$ (None).
If $c=4, a+b=22$ (None).
If $c=2, a+b=24$ (None).
If $c=0, a+b=26$ (None).
Wait, $a+b+c=26$. $c$ must be even.
If $c=8: a+b=18 \implies (9,9) \to 998$.
If $c=6: a+b=20$ (None).
If $c=4: a+b=22$ (None).
If $c=2: a+b=24$ (None).
If $c=0: a+b=26$ (None).
Wait, $a, b$ can be 9. Max $a+b=18$. So $c$ must be at least $26-18=8$.
If $c=8, a+b=18 \implies 998$.
If $c=9$ (not even).
Are there others? $a+b+c=26$. $c \in \{0, 2, 4, 6, 8\}$.
If $c=8, a+b=18 \to 998$.
If $c=6, a+b=20$ (Impossible).
Only 998? Let's re-read. "3-digit whole numbers".
Maybe $c$ can be 8, 6, 4, 2, 0.
$c=8 \implies a+b=18 \implies 998$.
$c=6 \implies a+b=20$ (No).
$c=4 \implies a+b=22$ (No).
$c=2 \implies a+b=24$ (No).
$c=0 \implies a+b=26$ (No).
Only 1.
ANSWER 8: A

Problem 9:
$a+b+c=25$. $a, b, c \in \{0..9\}, a \neq 0$.
Possible sets $\{9,9,7\}, \{9,8,8\}$.
Permutations of $\{9,9,7\}$: $997, 979, 799$ (3).
Permutations of $\{9,8,8\}$: $988, 898, 889$ (3).
Total = $3+3 = 6$.
ANSWER 9: C

Problem 10:
Exits 5 to 26 is $26-5 = 21$ intervals. Total distance 118.
Let $d_i$ be the distance between exits. $\sum_{i=1}^{21} d_i = 118$.
$d_i \ge 5$. To maximize one $d_k$, minimize others: $d_i = 5$ for $i \neq k$.
$d_k + 20(5) = 118 \implies d_k + 100 = 118 \implies d_k = 18$.
ANSWER 10: C

Problem 11:
Sum = $5 \times 90 = 450$. Mode = 94. Median = 91.
Let scores be $x_1, x_2, 91, 94, 94$.
$x_1 + x_2 + 91 + 94 + 94 = 450 \implies x_1 + x_2 + 279 = 450 \implies x_1 + x_2 = 171$.
ANSWER 11: B

Problem 12:
$5 \pm 4 \pm 6 \pm 3$.
$5+4 \times 6 - 3 = 5+24-3 = 26$.
$5 \times 4 - 6 + 3 = 20-6+3 = 17$.
$5 \times 4 + 6 - 3 = 23$.
$5 - 4 + 6 \times 3 = 1 + 18 = 19$.
ANSWER 12: E

The set \(R\) consists of nine distinct integers, including the six given values \(2, 3, 4, 6, 9, 14\). When the elements of \(R\) are placed in increasing order \(a_1 < a_2 < a_3 < a_4 < a_5 < a_6 < a_7 < a_8 < a_9\), the median is \(a_5\). Thus exactly four elements of \(R\) must lie strictly below the median value \(m = a_5\).
Let \(k\) denote the number of the six given values that are strictly less than \(m\). The three unknown values must then be chosen so that exactly \(4 - k\) of them lie strictly below \(m\).
- If \(m\) is one of the given values, the admissible choices are \(m = 3, 4, 6, 9\). For each of these four numbers it is possible to select three unknown integers satisfying the counting requirement above (and remaining distinct from one another and from the given values).
- If \(m\) is not one of the given values, the admissible choices are the integers \(m = 5, 7, 8\). Again, for each of these three numbers it is possible to select the remaining two unknown integers so that exactly four elements lie below \(m\).
No other integer can serve as median: any candidate smaller than 3 or larger than 9 produces a non-negative integer count of unknowns below \(m\) that is impossible to realize with only three unknowns. Consequently the possible values of the median are exactly the seven numbers
\[
3,4,5,6,7,8,9.
\]
Thus seven possible values exist for the median of \(R\).
ANSWER 1: D
Problem 2:
Let the five numbers be \(2, 6, 7, 7, 28, N\). Their sum is \(50 + N\), so the mean is
\[
\frac{50 + N}{5}.
\]
When the six numbers are ordered, the median is the average of the third and fourth values. The value of \(N\) determines the ordered list, and therefore the median, in each of the five mutually exclusive ranges determined by the given numbers.
- If \(N \le 2\), the ordered list is \(N,2,6,7,7,28\) and the median is \((6+7)/2 = 6.5\). Setting the mean equal to twice the median yields the contradictory equation \(N = 15\).
- If \(2 < N \le 6\), the ordered list is \(2,N,6,7,7,28\) and the median is again \(6.5\). The same contradictory equation appears.
- If \(6 < N \le 7\), the ordered list is \(2,6,N,7,7,28\) and the median is \((N+7)/2\). The mean-equals-twice-median equation simplifies to the quadratic
 \[
 N^2 - 9N - 60 = 0,
 \]
 whose only root in the interval is \(N = 12\), which lies outside the interval.
- If \(7 < N \le 28\), the ordered list is \(2,6,7,7,N,28\) and the median is \((7+N)/2\). The mean-equals-twice-median equation simplifies to the linear equation
 \[
 N = 20,
 \]
 which lies inside the interval.
- If \(N > 28\), the ordered list is \(2,6,7,7,28,N\) and the median is \((7+28)/2 = 17.5\). The resulting equation \(N = 120\) lies outside the interval.
Hence the only value that satisfies the stated condition is \(N = 20\).
ANSWER 2: C
Problem 3:
A four-digit integer whose digits are all even may be written
\[
1000a + 100b + 10c + d,
\]
where each of \(a,b,c,d\) belongs to the set \(\{0,2,4,6,8\}\) and \(a \ne 0\). There are therefore
\[
4 \times 5^3 = 500
\]
such integers.
An integer is divisible by 4 if and only if its last two digits form a number divisible by 4. The admissible pairs \((c,d)\) are therefore exactly the 13 pairs
\[
(0,0),(0,4),(0,8),(2,0),(2,4),(2,8),(4,0),(4,4),(4,8),(6,0),(6,4),(8,0),(8,8)
\]
that satisfy the divisibility test. For each of the 13 choices of \((c,d)\) the first two digits may be chosen freely (\(a\) has 4 possibilities, \(b\) has 5), giving
\[
13 \times 4 \times 5 = 260
\]
favorable integers. The required fraction is therefore
\[
\frac{260}{500} = \frac{13}{25} = \frac{2}{5}.
\]
ANSWER 3: B
Problem 4:
A 6.5 percent tax on a $20 item is \(0.065 \times 20 = 1.30\). A 6 percent tax is \(0.06 \times 20 = 1.20\). Their difference is
\[
1.30 - 1.20 = 0.10.
\]
ANSWER 4: B
Problem 5:
The first sum is an arithmetic series with 93 terms whose average is
\[
\frac{1901 + 1993}{2} = 1947,
\]
so its value equals \(93 \times 1947 = 181071\). The second sum is an arithmetic series with 93 terms whose average is
\[
\frac{101 + 193}{2} = 147,
\]
so its value equals \(93 \times 147 = 13671\). Subtracting yields
\[
181071 - 13671 = 167400.
\]
ANSWER 5: A
Problem 6:
The given expression consists of 1996 terms whose signs follow the repeating four-term pattern
\[
+,-,-,+,+,-,-,+,\dots
\]
(The pattern begins with a single “+” and thereafter repeats every four signs.) In each complete block of four consecutive integers the signed sum is zero. There are exactly 499 such blocks, accounting for 1996 terms and contributing a total of zero. Hence the whole expression evaluates to 0.
ANSWER 6: C
Problem 7:
Let \(S\) be the set of people wearing sunglasses and \(C\) the set wearing caps. Then
\[
|S \cap C| = \frac{2}{5}|C| = \frac{2}{5} \cdot 35 = 14.
\]
Consequently
\[
|S| = |S \setminus C| + |S \cap C|
\]
is unknown, but the conditional probability required by the problem is
\[
P(C \mid S) = \frac{|S \cap C|}{|S|} = \frac{14}{|S|}.
\]
Because \(|S \cap C| = 14\) and \(|C| = 35\), the only numerical value that can be obtained from the given data is the ratio
\[
\frac{14}{35} = \frac{2}{5}.
\]
(The actual size of \(S\) is never needed.)
ANSWER 7: C
Problem 8:
A three-digit number \(100a + 10b + c\) (with \(a \in \{1,\dots,9\}\), \(b,c \in \{0,\dots,9\}\)) is even precisely when \(c\) is even. The digit-sum equation \(a + b + c = 26\) together with the upper bound 9 on each variable forces \(a \ge 8\). The only possibilities are therefore the six triples
\[
(9,9,8),\ (9,8,9),\ (8,9,9)
\]
and their permutations that keep the units digit even. Exactly three of them have even units digit:
\[
998,\ 986,\ 968.
\]
Thus three even three-digit numbers have digit-sum 26.
ANSWER 8: C
Problem 9:
A three-digit number \(100a + 10b + c\) satisfies \(a + b + c = 25\) with each digit at most 9. Then necessarily \(a \ge 7\). The admissible triples \((a,b,c)\) are exactly the six permutations of
\[
(9,9,7),\ (9,8,8).
\]
Hence six such numbers exist.
ANSWER 9: C
Problem 10:

## Step 1: Problem 1
To find the number of possible values of the median of set R, we first need to understand what a median is. The median of a set of numbers is the middle number when the numbers are arranged in ascending order. Since R is a set of nine distinct integers, the median will be the fifth number when the numbers are arranged in ascending order.

## Step 2: Problem 1
Given that six of the elements are 2, 3, 4, 6, 9, and 14, we can start by arranging these numbers in ascending order: 2, 3, 4, 6, 9, 14. The remaining three numbers can be any distinct integers.

## Step 3: Problem 1
To determine the possible medians, we need to consider the range of values the fifth number can take. The numbers already given are 2, 3, 4, 6, 9, 14. So, the three additional numbers can either be less than 4, between 4 and 9, or greater than 9.

## Step 4: Problem 1
If all three additional numbers are less than 4, the order becomes: additional number, additional number, additional number, 4, 6, 9, 14. The median is 6.
If two additional numbers are less than 4 and one is between 4 and 9, the order becomes: additional number, additional number, 4, additional number, 6, 9, 14. The median is between 4 and 6, so it could be 5.
If one additional number is less than 4 and two are between 4 and 9, the order becomes: additional number, 4, additional number, additional number, 6, 9, 14. The median could be between 4 and 6.
If all three additional numbers are between 4 and 9, the order becomes: 2, 3, 4, additional number, additional number, additional number, 6, 9, 14. The median is one of the additional numbers between 4 and 6, or 6 if the additional numbers are greater than 6.
If the additional numbers are greater than 9, the order remains 2, 3, 4, 6, 9, additional number, additional number, additional number, 14. The median is 9.

## Step 5: Problem 1
Considering the cases:
- When the additional numbers are less than 4, the median is 6.
- When the numbers are between 4 and 9, the median can be 5, 6, 7, or 8 (since the additional numbers can be distinct and between 4 and 9).
- When the numbers are greater than 9, the median is 9.
So, possible medians are 5, 6, 7, 8, and 9.

## Step 6: Problem 1
Thus, there are 5 possible values for the median: 5, 6, 7, 8, 9.

ANSWER 1: B

## Step 7: Problem 2
To find N, we first need to understand how the mean and median are affected by adding N to the list 2, 6, 7, 7, 28.

## Step 8: Problem 2
The original list is 2, 6, 7, 7, 28. The median is 7. When N is added, the list becomes 2, 6, 7, 7, N, 28. The median will depend on N. If N is between 7 and 7 (which is not possible since N is distinct), or less than 7, the median will be 7. If N is greater than 7, the median will be the average of 7 and N.

## Step 9: Problem 2
Let's consider the case when N is less than or equal to 7. The median remains 7. The mean is (2 + 6 + 7 + 7 + 28 + N)/6 = (50 + N)/6. According to the problem, the mean is twice the median, so (50 + N)/6 = 2*7 = 14.

## Step 10: Problem 2
Solving for N: 50 + N = 14*6 = 84, N = 84 - 50 = 34.

## Step 11: Problem 2
Let's check if N = 34 satisfies the condition for the median. Since 34 is greater than 7, the list becomes 2, 6, 7, 7, 28, 34. The median is (7 + 7)/2 = 7. The mean is (2 + 6 + 7 + 7 + 28 + 34)/6 = 84/6 = 14, which is indeed twice the median.

## Step 12: Problem 2
Thus, N = 34 is a valid solution.

ANSWER 2: E

## Step 13: Problem 3
To find the fraction of four-digit integers with all even digits that are divisible by 4, we first need to determine the total number of such four-digit integers.

## Step 14: Problem 3
For a four-digit number, the first digit can be 2, 4, 6, or 8 (4 choices), and the rest can be 0, 2, 4, 6, or 8 (5 choices each). So, the total number of four-digit numbers with all even digits is 4 * 5 * 5 * 5 = 500.

## Step 15: Problem 3
A number is divisible by 4 if its last two digits form a number divisible by 4. The last two digits can be 00, 04, 08, 20, 24, 28, 40, 44, 48, 60, 64, 68, 80, 84, 88. Out of these, the ones with even digits are 00, 04, 08, 20, 24, 28, 40, 44, 48, 60, 64, 68, 80, 84, 88, which are 20, 24, 28, 40, 44, 48, 60, 64, 68, 80, 84, 88 (12 options since 00, 04, 08 are not valid for the last two digits when considering all possible combinations for the first two digits).

## Step 16: Problem 3
However, we need to correct the understanding of the last two digits. For a number to be divisible by 4, the last two digits must form a number divisible by 4. The possible last two digits are: 00, 04, 08, 20, 24, 28, 40, 44, 48, 60, 64, 68, 80, 84, 88. There are 5 choices for the third digit and 4 choices for the first digit. So, for each valid last two digits, there are 4 * 5 = 20 numbers.

## Step 17: Problem 3
There are 15 valid combinations for the last two digits (00, 04, 08, 20, 24, 28, 40, 44, 48, 60, 64, 68, 80, 84, 88), but we should count them correctly: there are indeed 15 options (since all are valid for the condition of being even and divisible by 4 when considering the last two digits).

## Step 18: Problem 3
The total number of four-digit numbers with all even digits and divisible by 4 is 4 * 5 * 15 = 300 (since there are 15 valid last two digits).

## Step 19: Problem 3
The fraction of these integers divisible by 4 is 300 / 500 = 3/5.

ANSWER 3: D

## Step 20: Problem 4
To find the difference between a 6.5% sales tax and a 6% sales tax on an item priced at $20, we calculate both taxes and then find the difference.

## Step 21: Problem 4
6.5% of $20 = 0.065 * $20 = $1.30.
6% of $20 = 0.06 * $20 = $1.20.

## Step 22: Problem 4
The difference is $1.30 - $1.20 = $0.10.

ANSWER 4: B

## Step 23: Problem 5
To solve (1901 + 1902 + ... + 1993) - (101 + 102 + ... + 193), we can simplify it by calculating the sum of the differences between corresponding terms.

## Step 24: Problem 5
The series can be simplified as: (1901-101) + (1902-102) + ... + (1993-193).

## Step 25: Problem 5
This simplifies to: 1800 + 1800 + 

**Problem 1:**
We have a set \(R\) of nine distinct integers, with six given: 2, 3, 4, 6, 9, 14. The remaining three integers can be any integers not in this list. The median of nine numbers is the 5th smallest. Let the unknown numbers be \(x, y, z\). The six given numbers are fixed; we can place \(x, y, z\) anywhere in the sorted order to change the 5th element.

Let \(k\) be the number of given numbers less than the candidate median \(m\).
- If \(m\) is one of the given numbers, we need to place exactly \(4-k\) of the new numbers below \(m\) and the rest above. This requires \(0 \le 4-k \le 3\), so \(1 \le k \le 4\). Among the given numbers: \(3\ (k=1),\ 4\ (k=2),\ 6\ (k=3),\ 9\ (k=4)\) satisfy this. (\(2\) has \(k=0\), \(14\) has \(k=5\); neither works.) So \(3, 4, 6, 9\) are possible medians.
- If \(m\) is a new integer, one new number equals \(m\) and the other two are placed above or below. We need \(4-k\) new numbers below \(m\), so \(0 \le 4-k \le 2\), or \(2 \le k \le 4\). Also \(m\) must be an integer not already in the list.
  * \(k=2\): \(m\) between 3 and 4 — no integer.
  * \(k=3\): \(m\) between 4 and 6 — only \(5\).
  * \(k=4\): \(m\) between 6 and 9 — \(7\) and \(8\).
So new possible medians: \(5, 7, 8\).

Total possible medians: \(3, 4, 5, 6, 7, 8, 9\) — seven values.

ANSWER 1: D

**Problem 2:**
Original list: \(2, 6, 7, 7, 28\) (sum = 50). Insert \(N\) to get six numbers. Mean = \(\frac{50+N}{6}\). Median of six sorted numbers is the average of the 3rd and 4th.

Case 1: \(N \le 6\). Sorted: …, 3rd = 6, 4th = 7, median = 6.5.  
\(\frac{50+N}{6} = 2 \times 6.5 = 13 \Rightarrow 50+N = 78 \Rightarrow N=28\), but \(28 > 6\). Contradiction.

Case 2: \(6 < N \le 7\) (only integer is 7). Sorted: \(2, 6, 7, 7, 7, 28\), median = 7.  
\(\frac{57}{6} = 9.5 \neq 14\). No.

Case 3: \(7 < N \le 28\). Sorted: \(2, 6, 7, 7, N, 28\), median = 7.  
\(\frac{50+N}{6} = 14 \Rightarrow N = 34\), but \(34 > 28\). Contradiction.

Case 4: \(N > 28\). Sorted: \(2, 6, 7, 7, 28, N\), median = 7.  
\(\frac{50+N}{6} = 14 \Rightarrow N = 34\). Works.

ANSWER 2: E

**Problem 3:**
Four-digit positive integers with all even digits: thousands digit from \(\{2,4,6,8\}\) (4 choices); hundreds, tens, units from \(\{0,2,4,6,8\}\) (5 choices each). Total = \(4 \times 5^3 = 500\).
Divisible by 4 means last two digits form a multiple of 4. For any even tens digit, there are exactly 3 even units digits that make it divisible by 4 (e.g., if tens = 0,4,8 → units 0,4,8; if tens = 2,6 → units 0,4,8). So \(5 \times 3 = 15\) valid last-two-digit combinations.
Number of divisible numbers = \(4 \times 5 \times 15 = 300\).
Fraction = \(\frac{300}{500} = \frac{3}{5}\).

ANSWER 3: D

**Problem 4:**
Tax difference = \(6.5\% - 6\% = 0.5\%\) of \$20 = \(0.005 \times 20 = \$0.10\).

ANSWER 4: B

**Problem 5:**
First sum: \(1901\) to \(1993\) inclusive. Number of terms = \(1993 - 1901 + 1 = 93\).  
Sum = \(93 \times \frac{1901+1993}{2} = 93 \times 1947\).
Second sum: \(101\) to \(193\) inclusive. Terms = \(193 - 101 + 1 = 93\).  
Sum = \(93 \times \frac{101+193}{2} = 93 \times 147\).
Difference = \(93 \times (1947 - 147) = 93 \times 1800 = 167\,400\).

ANSWER 5: A

**Problem 6:**
The expression is \(1 - 2 - 3 + 4 + 5 - 6 - 7 + 8 + 9 - 10 - 11 + \cdots + 1992 + 1993 - 1994 - 1995 + 1996\).
Sign pattern: \(+\) for terms \(n \equiv 0\) or \(1 \pmod 4\); \(-\) for \(n \equiv 2\) or \(3 \pmod 4\).
Group in blocks of four: \((1-2-3+4) + (5-6-7+8) + \cdots + (1993-1994-1995+1996)\). Each block sums to \(0\). There are \(1996/4 = 499\) blocks. Total sum = \(0\).

ANSWER 6: C

**Problem 7:**
Let \(S\) be sunglasses wearers (\(|S|=50\)), \(C\) be cap wearers (\(|C|=35\)). Given \(P(S \mid C) = \frac{|S \cap C|}{35} = \frac{2}{5} \Rightarrow |S \cap C| = 14\).
Then \(P(C \mid S) = \frac{14}{50} = \frac{7}{25}\).

ANSWER 7: B

**Problem 8:**
Three-digit numbers with digits summing to 26. Maximum sum is 27, so digits are close to 9. Let digits be \(a,b,c\) with \(a+b+c=26\), \(1 \le a \le 9\), \(0 \le b,c \le 9\). Setting \(x=9-a, y=9-b, z=9-c\) gives \(x+y+z=1\) with \(x \le 8, y,z \le 9\). The only solutions are permutations of \((1,0,0)\): one digit is 8, the other two are 9. The three numbers are 998, 989, 899.
Even numbers end in an even digit: only 998. Hence 1 even number.

ANSWER 8: A

**Problem 9:**
Three-digit numbers, digit sum = 25. As above, \(x+y+z = 27-25 = 2\). Nonnegative solutions: permutations of \((2,0,0)\) – digits 7,9,9; and \((1,1,0)\) – digits 8,8,9.  
For 7,9,9: 3 permutations. For 8,8,9: 3 permutations. Total 6 numbers.

ANSWER 9: C

**Problem 10:**
Exits 5 through 26: that’s \(26-5+1 = 22\) exits, so 21 gaps. Sum of gaps = 118 miles. Each gap \(\ge 5\). To maximize one gap, minimize the other 20: \(20 \times 5 = 100\) miles. Largest gap = \(118 - 100 = 18\) miles.

ANSWER 10: C

**Problem 11:**
Five scores: \(a \le b \le c \le d \le e\). Mean = 90 ⇒ sum = 450. Median \(c = 91\). Mode = 94, so at least two scores are 94. Since median is 91, the two 94s must be \(d\) and \(e\). Then \(a+b+91+94+94 = 450 \Rightarrow a+b = 450 - 279 = 171\).

ANSWER 11: B

**Problem 12:**
Blanks: \(5 \_ 4 \_ 6 \_ 3\). Use \(+,\ -,\ \times\) exactly once. Evaluate with standard order of operations. All 6 permutations:
1) \(\times, +, -:\ 5\times4+6-3 = 20+6-3=23\)
2) \(\times, -, +:\ 5\times4-6+3 = 20-6+3=17\)
3) \(+, \times, -:\ 5+4\times6-3 = 5+24-3=26\)
4) \(+, -, \times:\ 5+4-6\times3 = 5+4-18=-9\)
5) \(-, \times, +:\ 5-4\times6+3 = 5-24+3=-16\)
6) \(-, +, \times:\ 5-4+6\times3 = 5-4+18=19\)
Among choices, 19 appears.

ANSWER 12: E

Problem 1:
We have a set R of 9 distinct integers. Six are known: 2, 3, 4, 6, 9, 14. We add 3 unknown distinct integers (not equal to any known element). The median of 9 elements is the 5th smallest.

Let me determine which values can occupy the 5th position. The known elements in sorted order are: 2, 3, 4, 6, 9, 14.

**Can the median be 3?** Place all 3 unknowns below 3 (e.g., −1, 0, 1). Sorted: −1, 0, 1, 2, **3**, 4, 6, 9, 14. 5th = 3. ✓
**Can the median be 4?** Place 2 unknowns below 4 (e.g., 0, 1) and 1 above. Sorted: 0, 1, 2, 3, **4**, 6, 9, 14, 20. 5th = 4. ✓
**Can the median be 5?** (5 is an unknown.) Known elements below 5: {2, 3, 4} = 3 elements. Add 1 unknown below 5 (e.g., 1), make 5 an unknown, and 1 unknown above. Sorted: 1, 2, 3, 4, **5**, 6, 9, 14, 20. 5th = 5. ✓
**Can the median be 6?** Known below 6: {2, 3, 4} = 3. Add 1 unknown below 6 (e.g., 5), 2 above. Sorted: 2, 3, 4, 5, **6**, 9, 14, 20, 21. 5th = 6. ✓
**Can the median be 7?** Known below 7: {2, 3, 4, 6} = 4. Need 0 unknowns below 7. Make 7 an unknown, 2 above. Sorted: 2, 3, 4, 6, **7**, 9, 14, 20, 21. 5th = 7. ✓
**Can the median be 8?** Known below 8: {2, 3, 4, 6} = 4. Same logic: 2, 3, 4, 6, **8**, 9, 14, 20, 21. 5th = 8. ✓
**Can the median be 9?** Known below 9: {2, 3, 4, 6} = 4. Place all 3 unknowns above 9. Sorted: 2, 3, 4, 6, **9**, 14, 20, 21, 22. 5th = 9. ✓

**Can the median be ≤ 2?** Need 4 elements below it, but at most 3 unknowns exist → impossible.
**Can the median be ≥ 10?** Known elements below 10 are {2, 3, 4, 6, 9} = 5 elements, so the 5th position is already filled by something ≤ 9 → impossible.

Possible medians: {3, 4, 5, 6, 7, 8, 9} → **7 values**.

ANSWER 1: D

---

Problem 2:
Original list: 2, 6, 7, 7, 28. Sum = 50. Insert N to get 6 elements.
New mean = (50 + N)/6. Median of 6 elements = average of 3rd and 4th.

**Case N > 7:** Sorted order: 2, 6, 7, 7, N, 28 (or N, 28 swapped). The 3rd and 4th elements are 7 and 7.
Median = (7 + 7)/2 = 7.
Mean = 2 × median = 14.
(50 + N)/6 = 14 → 50 + N = 84 → N = 34.
Check: 34 > 7 ✓. List: 2, 6, 7, 7, 28, 34. Median = 7, Mean = 84/6 = 14 = 2(7). ✓

**Case N ≤ 6:** Median = (6+7)/2 = 6.5. Mean = 13. N = 78 − 50 = 28. But 28 > 6, contradiction.
**Case 6 < N ≤ 7:** Median = (N+7)/2. (50+N)/6 = N+7 → 50+N = 6N+42 → 5N = 8 → N = 1.6. Contradiction.

Only N = 34 works.

ANSWER 2: E

---

Problem 3:
Even digits: {0, 2, 4, 6, 8}. Four-digit numbers with all even digits:
- First digit: {2, 4, 6, 8} → 4 choices
- Other three digits: 5 choices each
- Total = 4 × 5 × 5 × 5 = 500

Divisibility by 4 depends on the last two digits. I check all 5 × 5 = 25 pairs of even digits:
For each tens digit (0, 2, 4, 6, 8), the units digits giving divisibility by 4 are:
- Tens 0: 00, 04, 08 (3 of 5)
- Tens 2: 20, 24, 28 (3 of 5)
- Tens 4: 40, 44, 48 (3 of 5)
- Tens 6: 60, 64, 68 (3 of 5)
- Tens 8: 80, 84, 88 (3 of 5)

Valid last-two-digit pairs = 15 out of 25.

Numbers divisible by 4 = 4 × 5 × 15 = 300.
Fraction = 300/500 = **3/5**.

ANSWER 3: D

---

Problem 4:
Difference = 6.5% of $20 − 6% of $20 = (6.5% − 6%) × $20 = 0.5% × $20.
0.5% = 0.005, so 0.005 × 20 = $0.10.

ANSWER 4: B

---

Problem 5:
Pair corresponding terms:
(1901 − 101) + (1902 − 102) + … + (1993 − 193)

Each pair equals 1800.
Number of terms: 1993 − 1901 + 1 = 93 terms (and 193 − 101 + 1 = 93 terms).
Total = 93 × 1800 = **167,400**.

ANSWER 5: A

---

Problem 6:
The sign pattern is: +, −, −, +, +, −, −, +, …
Group into blocks of 4:
(1 − 2 − 3 + 4) + (5 − 6 − 7 + 8) + (9 − 10 − 11 + 12) + …

Each block: n − (n+1) − (n+2) + (n+3) = n − n − 1 − n − 2 + n + 3 = **0**.

Total terms: 1996. Number of blocks: 1996/4 = 499 complete blocks.
Sum = 499 × 0 = **0**.

ANSWER 6: C

---

Problem 7:
Let B = number wearing both sunglasses and caps.
P(sunglasses | cap) = B / (people wearing caps) = B / 35 = 2/5.
B = 35 × (2/5) = 14.

Now find P(cap | sunglasses) = B / (people wearing sunglasses) = 14 / 50 = **7/25**.

ANSWER 7: B

---

Problem 8:
Three-digit numbers with digit sum 26. Maximum digit sum = 9+9+9 = 27, so we must subtract 1 from one digit of 999:
Possible digit sets: (8,9,9), (9,8,9), (9,9,8) → numbers 899, 989, 998 (and permutations: 899, 989, 997... wait).

Actually the numbers are: 899, 989, 998 — no, let me list properly:
- (8,9,9) → 899, 989, 998... wait, (8,9,9) gives 899, 989, 998? No: digits are 8,9,9 so numbers are 899, 989, 998.
Wait: (8,9,9) → 899, 989, 998. But 998 has digits 9,9,8 which is the set (9,9,8). Let me just list all:
- Digits {8,9,9}: 899, 989, 998 — wait, 998 has digits 9,9,8. That IS {8,9,9}. So: 899, 989, 998.

Hmm, let me redo. The three numbers with digit sum 26 are:
- 899 (8+9+9=26)
- 989 (9+8+9=26)
- 998 (9+9+8=26)

Wait, that's only 3 numbers total (since two digits are the same, there are 3!/2! = 3 arrangements).

Even numbers must end in an even digit. Only 998 ends in 8 (even). 899 and 989 end in 9 (odd).

So exactly **1** even number.

ANSWER 8: A

---

Problem 9:
Three-digit numbers a, b, c with a + b + c = 25, where 1 ≤ a ≤ 9, 0 ≤ b, c ≤ 9.

Enumerate by first digit a:
- a = 9: b + c = 16 → (7,9), (8,8), (9,7) → 979, 988, 997 (3 numbers)
- a = 8: b + c = 17 → (8,9), (9,8) → 889, 898 (2 numbers)
- a = 7: b + c = 18 → (9,9) → 799 (1 number)
- a ≤ 6: b + c ≥ 19, but max b + c = 18 → impossible

Total = 3 + 2 + 1 = **6**.

ANSWER 9: C

---

Problem 10:
Exits 5 through 26 give 22 exits and 21 consecutive gaps. Total distance = 118 miles. Each gap ≥ 5 miles.

To maximize one gap, minimize the other 20 gaps to 5 miles each:
Max gap + 20(5) = 118
Max gap + 100 = 118
Max gap = **18** miles.

ANSWER 10: C

---

Problem 11:
Five scores sorted: a ≤ b ≤ c ≤ d ≤ e.
- Mean = 90 → sum = 450
- Median = 91 → c = 91
- Mode = 94 → 94 appears most frequently (at least twice)

Since a ≤ b ≤ c = 91, neither a nor b can be 94. So 94 must appear among d and e. For 94 to be the mode, we need d = e = 94 (appearing twice).

Sum: a + b + 91 + 94 + 94 = 450
a + b = 450 − 279 = **171**

(We can verify valid scores 

 Problem 1:
We are given a set \(R\) of nine distinct integers, six of which are \(2,3,4,6,9,14\). We must determine how many different values are possible for the median of \(R\).

Because there are nine distinct integers, the median is the fifth smallest element when the set is listed in increasing order. Let the three unknown distinct integers be chosen freely (as long as they are not equal to any of the six given numbers). We examine two cases: when the median is one of the six known numbers and when it is one of the three unknown numbers.

**Case 1: The median is a known number.**
Let the median be \(m\). Of the six known numbers, let \(L\) be how many are less than \(m\). Then there are \(5-L\) known numbers greater than \(m\). To make \(m\) the fifth element, we need exactly four numbers smaller than \(m\) and four larger. Thus we need \(4-L\) unknown numbers below \(m\) and \(4-(5-L)=L-1\) unknown numbers above \(m\). This uses exactly \((4-L)+(L-1)=3\) unknowns, which is perfect. We need \(4-L\ge 0\) and \(L-1\ge 0\), i.e. \(1\le L\le 4\).

- \(m=2\): \(L=0\) → impossible.
- \(m=3\): \(L=1\) → need 3 unknowns \(<3\), 0 above. Possible (e.g. \(-1,0,1\)).
- \(m=4\): \(L=2\) → need 2 unknowns \(<4\), 1 above. Possible (e.g. \(0,1,5\)).
- \(m=6\): \(L=3\) → need 1 unknown \(<6\), 2 above. Possible (e.g. \(5,7,8\)).
- \(m=9\): \(L=4\) → need 0 unknowns \(<9\), 3 above. Possible (e.g. \(10,11,12\)).
- \(m=14\): \(L=5\) → need \(-1\) unknowns below; impossible.

So the known possible medians are \(3,4,6,9\).

**Case 2: The median is an unknown number \(x\).**
Then \(x\) itself uses one of the three unknown slots. We need four numbers below \(x\) and four above. With \(L\) known numbers below \(x\), we need \(4-L\) other unknowns below \(x\) and \(4-(6-L)=L-2\) other unknowns above \(x\). The total other unknowns used is \((4-L)+(L-2)=2\), which matches the two remaining unknowns. We need \(4-L\ge 0\) and \(L-2\ge 0\), so \(2\le L\le 4\).

- \(L=2\): known below are \(2,3\); known above are \(4,6,9,14\). Then \(3<x<4\), impossible for an integer.
- \(L=3\): known below are \(2,3,4\); known above are \(6,9,14\). Then \(4<x<6\), so \(x=5\). This works (e.g. add \(1,5,7\)).
- \(L=4\): known below are \(2,3,4,6\); known above are \(9,14\). Then \(6<x<9\), so \(x=7\) or \(x=8\). Both work (e.g. add \(7,10,11\) or \(8,10,11\)).

Thus the unknown possible medians are \(5,7,8\).

Collecting all possibilities: \(\{3,4,5,6,7,8,9\}\). That is **7** different values. The tempting answer of 4 counts only the known medians, and 2 or 14 are impossible because too many numbers would have to be forced to one side.

ANSWER 1: D

---

Problem 2:
We insert a number \(N\) into the list \(2,6,7,7,28\) so that the new mean of the six numbers is twice the new median. Find \(N\).

The original sum is \(2+6+7+7+28=50\). After insertion, the sum is \(50+N\) and the mean is \(\frac{50+N}{6}\).

The median depends on the size of \(N\) relative to the ordered list. Because the list contains six numbers, the median is the average of the third and fourth values.

- If \(N\le 6\) (as an integer), the ordered list begins with \(N\) or \(2\), then the third and fourth terms are \(6\) and \(7\), giving median \(\frac{6+7}{2}=6.5\).
  Setting mean \(=2\times 6.5=13\): \(\frac{50+N}{6}=13\Rightarrow N=28\). But \(N=28\) does not satisfy \(N\le 6\), so this case yields no valid solution.

- If \(N\ge 7\) (integer), the third and fourth terms are both \(7\) (since the two \(7\)’s occupy those middle slots, or \(N=7\) makes three \(7\)’s), so the median is \(7\).
  Setting mean \(=2\times 7=14\): \(\frac{50+N}{6}=14\Rightarrow 50+N=84\Rightarrow N=34\).

Check: the list becomes \(2,6,7,7,28,34\). The median is \(\frac{7+7}{2}=7\). The mean is \(\frac{84}{6}=14\), which is indeed twice the median. The other choices fail: \(N=28\) gives mean \(13\) and median \(7\); \(N=14\) gives mean \(\frac{64}{6}\neq 14\).

ANSWER 2: E

---

Problem 3:
We consider all positive four-digit integers whose digits are all even. We need the fraction of them that are divisible by \(4\).

Even digits are \(\{0,2,4,6,8\}\). For a four-digit number \(ABCD\):
- \(A\) (thousands) has 4 choices: \(\{2,4,6,8\}\).
- \(B,C,D\) each have 5 choices.

Total count: \(4\cdot 5\cdot 5\cdot 5 = 500\).

A number is divisible by \(4\) exactly when its last two digits form a multiple of \(4\). Here \(C\) and \(D\) are both even. Consider \(10C+D \pmod 4\). Since \(C\) is even, write \(C=2k\); then \(10C=20k\equiv 0\pmod 4\). Thus \(10C+D\equiv D\pmod 4\). So we simply need \(D\equiv 0\pmod 4\).

Among the even digits, those divisible by \(4\) are \(0,4,8\): **3 choices** for \(D\). The digits \(A,B,C\) are unrestricted (4, 5, and 5 choices).

Count of favorable numbers: \(4\cdot 5\cdot 5\cdot 3 = 300\).

The fraction is \(\frac{300}{500}=\frac{3}{5}\). One might mistakenly think the fraction is \(\frac12\) because half of all integers are divisible by \(4\), but the restriction to even digits changes the units-digit distribution.

ANSWER 3: D

---

Problem 4:
Find the difference between a \(6.5\%\) sales tax and a \(6\%\) sales tax on a \(\$20\) item.

Method 1: Compute each tax.
- \(6.5\%\) of \(\$20 = 0.065\times 20 = \$1.30\).
- \(6\%\) of \(\$20 = 0.06\times 20 = \$1.20\).

Difference: \(\$1.30-\$1.20 = \$0.10\).

Method 2: The difference in rates is \(0.5\%\), and \(0.5\%\) of \(\$20 = 0.005\times 20 = \$0.10\).

The distractors correspond to misplacing the decimal (e.g. \(\$1\) would be \(5\%\) of \(\$20\)).

ANSWER 4: B

---

Problem 5:
Evaluate \((1901+1902+\dots+1993)-(101+102+\dots+193)\).

Each sum has the same number of terms:
\(1993-1901+1 = 93\) terms, and \(193-101+1 = 93\) terms.

Pair the terms:
\((1901-101)+(1902-102)+\dots+(1993-193)\).

Each difference equals \(1800\). With 93 such pairs, the total is
\(93\times 1800 = 93\times 18\times 100 = 1674\times 100 = 167{,}400\).

A common error is miscounting the number of terms as 92, which would give \

Problem 1:
R is a set of nine distinct integers with six known elements: 2, 3, 4, 6, 9, 14. Since R has 9 elements, the median is the 5th element when sorted.

I need to determine which integers can be the 5th element. The median must have exactly 4 elements less than it and 4 elements greater than it.

**Checking each candidate:**

- **Median = 3**: Need 4 elements < 3. Known: {2}. Need 3 unknowns all < 3 (e.g., 1, 0, -1). Elements > 3: {4, 6, 9, 14} = 4. ✓
- **Median = 4**: Need 4 elements < 4. Known: {2, 3}. Need 2 unknowns < 4. Elements > 4: {6, 9, 14} + 1 unknown > 4. ✓
- **Median = 5**: Need 4 elements < 5. Known: {2, 3, 4}. Need 1 unknown < 5. Elements > 5: {6, 9, 14} + 1 unknown > 5. (5 is an unknown) ✓
- **Median = 6**: Need 4 elements < 6. Known: {2, 3, 4}. Need 1 unknown < 6. Elements > 6: {9, 14} + 2 unknowns > 6. ✓
- **Median = 7**: Need 4 elements < 7. Known: {2, 3, 4, 6} = 4. Elements > 7: {9, 14} + 2 unknowns > 7. (7 is an unknown) ✓
- **Median = 8**: Need 4 elements < 8. Known: {2, 3, 4, 6} = 4. Elements > 8: {9, 14} + 2 unknowns > 8. (8 is an unknown) ✓
- **Median = 9**: Need 4 elements < 9. Known: {2, 3, 4, 6} = 4. Elements > 9: {14} + 3 unknowns > 9. ✓

**Values that cannot be median:**
- 2: Need 4 elements < 2, but only 3 unknowns available. ✗
- 10-13: Already have 5 known elements less than these. ✗
- 14: Already have 5 known elements less than 14. ✗

Possible median values: {3, 4, 5, 6, 7, 8, 9} → **7 values**

ANSWER 1: D

---

Problem 2:
List: 2, 6, 7, 7, 28. Insert N so that mean = 2 × median.

Sum of original list = 50, so new mean = (50 + N)/6.

**Finding the median based on N:**
- If N > 7: Sorted list is 2, 6, 7, 7, 28, N (or 2, 6, 7, 7, N, 28 if 7 < N < 28). Median = (7+7)/2 = 7.
- If N ≤ 6: Median = (6+7)/2 = 6.5
- If 6 < N ≤ 7: Median = (N+7)/2

**Solving for N > 7:**
(50 + N)/6 = 2(7) = 14
50 + N = 84
N = 34

Verification: List = 2, 6, 7, 7, 28, 34. Mean = 84/6 = 14. Median = 7. ✓

ANSWER 2: E

---

Problem 3:
Four-digit integers with all even digits. Even digits: {0, 2, 4, 6, 8}.

**Total count:** First digit: 4 choices (2,4,6,8), others: 5 choices each → 4 × 5³ = 500

**Divisible by 4:** Last two digits must form a number divisible by 4.

Listing all 25 pairs of even last digits and checking divisibility by 4:
- 00✓, 02✗, 04✓, 06✗, 08✓
- 20✓, 22✗, 24✓, 26✗, 28✓
- 40✓, 42✗, 44✓, 46✗, 48✓
- 60✓, 62✗, 64✓, 66✗, 68✓
- 80✓, 82✗, 84✓, 86✗, 88✓

That's 15 valid pairs.

**Count divisible by 4:** 4 × 5 × 15 = 300

**Fraction:** 300/500 = 3/5

ANSWER 3: D

---

Problem 4:
6.5% tax on $20 = 0.065 × 20 = $1.30
6% tax on $20 = 0.06 × 20 = $1.20

Difference = $1.30 - $1.20 = $0.10

ANSWER 4: B

---

Problem 5:
(1901 + 1902 + ... + 1993) − (101 + 102 + ... + 193)

Both sums have 93 terms each.

First sum = 93 × (1901 + 1993)/2 = 93 × 1947
Second sum = 93 × (101 + 193)/2 = 93 × 147

Difference = 93 × (1947 − 147) = 93 × 1800 = 167,400

ANSWER 5: A

---

Problem 6:
1 − 2 − 3 + 4 + 5 − 6 − 7 + 8 + ... + 1993 − 1994 − 1995 + 1996

The pattern repeats every 4 terms: +, −, −, +

Group k: (4k−3) − (4k−2) − (4k−1) + 4k = 4k−3−4k+2−4k+1+4k = 0

Each group of 4 sums to 0.

1996 ÷ 4 = 499 complete groups.

Total = 499 × 0 = 0

ANSWER 6: C

---

Problem 7:
50 people wear sunglasses, 35 wear caps.

P(sunglasses | cap) = 2/5 means: (both)/(caps) = 2/5

Both/35 = 2/5 → Both = 14

P(cap | sunglasses) = (both)/(sunglasses) = 14/50 = 7/25

ANSWER 7: B

---

Problem 8:
3-digit numbers with digit-sum 26 that are even.

For number abc: a + b + c = 26, c is even, a ≥ 1, all digits ≤ 9.

- c = 8: a + b = 18 → only (9,9) → 998 ✓
- c = 6: a + b = 20 → impossible (max 9+9=18)
- c = 4, 2, 0: a + b ≥ 22 → impossible

Only 1 number: 998

ANSWER 8: A

---

Problem 9:
Three-digit numbers with digit-sum 25.

a + b + c = 25, a ≥ 1, all digits ≤ 9.

- a = 9: b + c = 16 → (7,9), (8,8), (9,7) → 3 numbers
- a = 8: b + c = 17 → (8,9), (9,8) → 2 numbers
- a = 7: b + c = 18 → (9,9) → 1 number
- a ≤ 6: b + c ≥ 19 → impossible

Total = 3 + 2 + 1 = 6

ANSWER 9: C

---

Problem 10:
Distance from exit 5 to exit 26 is 118 miles. There are 21 gaps between consecutive exits (5→6, 6→7, ..., 25→26).

Each gap ≥ 5 miles. To maximize one gap, minimize all others.

If 20 gaps are minimized at 5 miles each: 20 × 5 = 100 miles.

Maximum single gap = 118 − 100 = 18 miles.

ANSWER 10: C

---

Problem 11:
Five scores: mean = 90, median = 91, mode = 94.

Sum = 5 × 90 = 450

Sorted: a ≤ b ≤ c ≤ d ≤ e
- c = 91 (median)
- Mode = 94, so 94 appears at least twice. Since c = 91, the 94s must be d and e.

a + b + 91 + 94 + 94 = 450
a + b = 171

ANSWER 11: B

---

Problem 12:
Using +, −, × each exactly once in 5 __ 4 __ 6 __ 3:

| Operations | Result |
|------------|--------|
| 5 + 4 − 6 × 3 | 5 + 4 − 18 = −9 |
| 5 + 4 × 6 − 3 | 5 + 24 − 3 = 26 |
| 5 − 4 + 6 × 3 | 5 − 4 + 18 = 19 |
| 5 − 4 × 6 + 3 | 5 − 24 + 3 = −16 |
| 5 × 4 + 6 − 3 | 20 + 6 − 3 = 23 |
| 5 × 4 − 6 + 3 | 20 − 6 + 3 = 17 |

From choices {9, 10, 15, 16, 19}, only **19** is achievable.

ANSWER 12: E

**Problem 1:**  
We have nine distinct integers, six of which are known: 2, 3, 4, 6, 9, 14.  
The median of a nine‑element set is the 5th smallest number.  

Let the unknown integers be \(x,y,z\) (all distinct and distinct from the known six).  
When the numbers are sorted, the median is the 5th term.  

For a known number \(k_i\) (the \(i\)‑th smallest of the known six) to be the median, we need exactly four numbers smaller than it.  
If \(i-1\) known numbers are smaller, we need \(L\) unknown numbers smaller with  
\((i-1)+L=4\Rightarrow L=5-i\).  
Since we have only three unknowns, \(L\ge0\) gives \(i\le5\); also \(L\le3\) gives \(i\ge2\).  
Thus \(i=2,3,4,5\) → medians could be 3, 4, 6, 9.

Can the median be an unknown integer?  
Let the median be an unknown \(u\).  
If \(u\) lies between the known numbers, we need exactly four numbers below it.  
With two other unknowns, the possibilities are:
- \(u\) between 4 and 6 → \(u=5\);
- \(u\) between 6 and 9 → \(u=7\) or \(u=8\).

All three of these can be realized while keeping the numbers distinct.  
Thus the possible median values are \(\{3,4,5,6,7,8,9\}\), a total of **7** values.

**Answer:** 7 → Choice **D**.  

ANSWER 1: D  

---

**Problem 2:**  
We start with the list 2, 6, 7, 7, 28 and insert a number \(N\).  
Let the six numbers be sorted. The median of six numbers is the average of the 3rd and 4th smallest values.

Let the sorted list after insertion be \(a_1\le a_2\le a_3\le a_4\le a_5\le a_6\).  
The median is \(\dfrac{a_3+a_4}{2}\).  
The mean is \(\dfrac{\text{sum}}{6}\).  
Given that the mean equals twice the median:
\[
\frac{\text{sum}}{6}=2\cdot\frac{a_3+a_4}{2}=a_3+a_4 .
\]
Hence \(\text{sum}=6(a_3+a_4)\).

Consider where \(N\) can fall:

*If \(N\le2\):* \(a_3=6,\ a_4=7\) → median \(=6.5\).  
Sum \(=N+50\). Equation gives \(N=28\), contradicting \(N\le2\).

*If \(2<N\le6\):* \(a_3=6,\ a_4=7\) → same median \(=6.5\).  
Again \(N=28\), impossible.

*If \(6<N\le7\):* only integer possibility \(N=7\).  
List becomes 2, 6, 7, 7, 7, 28 → median \(=7\).  
Sum \(=57\) → mean \(=9.5\neq14\).

*If \(7<N<28\):* \(a_3=7,\ a_4=7\) → median \(=7\).  
Sum \(=N+50\). Equation \(N+50=84\) gives \(N=34\), which is not \(<28\).

*If \(N=28\):* median still \(=7\). Sum \(=78\) → mean \(=13\neq14\).

*If \(N>28\):* \(a_3=7,\ a_4=7\) → median \(=7\).  
Equation gives \(N+50=84\) → \(N=34\). This works (34 > 28).

Thus the only possible value is **34**.

**Answer:** 34 → Choice **E**.  

ANSWER 2: E  

---

**Problem 3:**  
A four‑digit integer with all digits even can use digits \{0,2,4,6,8\} for each place, except the thousands digit cannot be 0.  

Number of such integers: \(4\times5\times5\times5=500\).

A number is divisible by 4 iff its last two digits form a number divisible by 4.  
List all two‑digit endings with even digits (25 possibilities).  

Checking each yields 15 endings that are multiples of 4:  
00, 04, 08, 20, 24, 28, 40, 44, 48, 60, 64, 68, 80, 84, 88.

Thus the proportion is \(\frac{15}{25}=\frac{3}{5}\).

**Answer:** \(\frac{3}{5}\) → Choice **D**.  

ANSWER 3: D  

---

**Problem 4:**  
6 % tax on \$20 = \(0.06\times20 = \$1.20\).  
6.5 % tax on \$20 = \(0.065\times20 = \$1.30\).  

Difference = \$1.30 − \$1.20 = **\$0.10**.

**Answer:** \$0.10 → Choice **B**.  

ANSWER 4: B  

---

**Problem 5:**  
Sum from 1901 to 1993 (inclusive):  
Number of terms = \(1993-1901+1 = 93\).  
Sum = \(\frac{(1901+1993)\times93}{2}= \frac{3894\times93}{2}=181\,071\).

Sum from 101 to 193 (inclusive):  
Number of terms = \(193-101+1 = 93\).  
Sum = \(\frac{(101+193)\times93}{2}= \frac{294\times93}{2}=13\,671\).

Difference = \(181\,071-13\,671 = 167\,400\).

**Answer:** 167 400 → Choice **A**.  

ANSWER 5: A  

---

**Problem 6:**  
The sign pattern for the numbers \(n\) is:
- \(n\equiv1\pmod4\) → +,
- \(n\equiv2\pmod4\) → −,
- \(n\equiv3\pmod4\) → −,
- \(n\equiv0\pmod4\) → +.

Thus each block of four consecutive integers \(\{4k+1,4k+2,4k+3,4k+4\}\) contributes  
\((4k+1)-(4k+2)-(4k+3)+(4k+4)=0\).

From 1 through 1996 there are \(1996/4=499\) complete blocks, so the total sum is **0**.

**Answer:** 0 → Choice **C**.  

ANSWER 6: C  

---

**Problem 7:**  
Let \(S\) = people wearing sunglasses, \(|S|=50\).  
Let \(C\) = people wearing caps, \(|C|=35\).  
Let \(x=|S\cap C|\).

Given \(P(S\mid C)=\frac{x}{35}= \frac{2}{5}\).  
Thus \(x = \frac{2}{5}\times35 = 14\).

We need \(P(C\mid S)=\frac{x}{50}= \frac{14}{50}= \frac{7}{25}\).

**Answer:** \(\frac{7}{25}\) → Choice **B**.  

ANSWER 7: B  

---

**Problem 8:**  
A three‑digit number with digit sum 26 must use digits that sum to 26.  
The only triple of single‑digit numbers with sum 26 is \(\{9,9,8\}\).

The number is even only if its units digit is 8.  
Among the permutations of 9, 9, 8, only 998 ends in 8.  

Thus exactly **1** such number exists.

**Answer:** 1 → Choice **A**.  

ANSWER 8: A  

---

**Problem 9:**  
Let the digits be \(a,b,c\) (hundreds, tens, units) with  
\(1\le a\le9,\ 0\le b,c\le9,\ a+b+c=25\).

Set \(a'=a-1\) (so \(0\le a'\le8\)). Then  
\(a'+b+c=24\).

Count unrestricted non‑negative solutions: \(\binom{24+3-1}{2}= \binom{26}{2}=325\).

Subtract solutions where a digit exceeds its bound:
- \(a'\ge9\): treat \(a''=a'-9\) → \(a''+b+c=15\) → \(\binom{17}{2}=136\).
- \(b\ge10\): \(b'=b-10\) → \(a'+b'+c=14\) → \(\binom{16}{2}=120\).
- \(c\ge10\): similarly 120.

Intersections:
- \(a'\ge9,\ b\ge10\): \(a''+b'+c=5\) → \(\binom{7}{2}=21\).
- \(a'\ge9,\ c\ge10\): 21.
- \(b\ge10,\ c\ge10\): \(a'+b'+c'=4\) → \(\binom{6}{2}=15\).

No triple intersection (would require sum –5).

By inclusion–exclusion, invalid = \(136+120+120-21-21-15 = 319\).  
Valid = \(325-319 = 6\).

**Answer:** 6 → Choice **C**.  

ANSWER 9: C  

---

**Problem 10:**  
Exits numbered 5 through 26 give 22 exits, hence 21 intervals.  
Let the distances between consecutive exits be \(d_1,\dots,d_{21}\).  
We know \(\sum d_i = 118\) miles, and each \(d_i\ge5\).

To maximize the largest interval, make the othe

Problem 1:  
We are given a set $ R $ of nine distinct integers, six of which are $ 2, 3, 4, 6, 9, 14 $. The median of a set of nine numbers is the fifth number when sorted. To find the number of possible values for the median, we consider the possible placements of the three unknown numbers relative to the known ones. The known numbers sorted are $ 2, 3, 4, 6, 9, 14 $. The three unknown numbers can be placed in various positions to shift the median. The possible medians occur when the additional numbers are placed such that the fifth number in the sorted list can be $ 4, 5, 6, 7, 8, $ or $ 9 $. This gives six possible values for the median.  
ANSWER 1: D (corrected to C? Wait, no: the possible medians are 4, 5, 6, 7, 8, 9 — that’s 6 values, so C)  
**Wait, let's re-evaluate:** The known numbers are 2,3,4,6,9,14. Adding three numbers. The median is the 5th number. To get 4 as median: three numbers ≤4, but only 2,3,4 are available — but they are distinct, so we can add numbers like 1,5? Wait no, the three unknown numbers can be placed in different positions. For example:  
- If all three unknown numbers are ≤4: but only three such numbers exist (2,3,4) but they are already in the set. So we cannot add more. Thus, the unknown numbers must be placed in other positions.  
Actually, the unknown numbers can be:  
- All less than 2: median could be 4 if three numbers are added <4, but only 2,3,4 are already present. We cannot add numbers less than 2 without repeating? Wait, the set has distinct integers. So we can add numbers like 1,0,-1, but the problem doesn’t restrict to positive. However, the known numbers are positive. But the problem says "distinct integers", so negatives are allowed. But the median values: if we add numbers less than 2, the sorted order becomes [new numbers, 2,3,4,6,9,14]. With three new numbers, the sorted list has 9 numbers. The fifth is the median. If we add three numbers less than 2, the sorted list is [a,b,c,2,3,4,6,9,14] — no, wait, adding three numbers less than 2: the sorted list would be [a,b,c,2,3,4,6,9,14] — but a,b,c are less than 2, so the first three, then 2 is fourth, 3 is fifth? Wait, no: the original known numbers are 2,3,4,6,9,14. Adding three numbers, say x,y,z. The full sorted list will be the combination. To get median=4, we need the fifth number to be 4. So in the sorted list of 9, the fifth must be 4. That requires that four numbers are ≤4 and four are ≥4? No: the fifth is the middle. For the fifth to be 4, we need at least five numbers ≤4? No: in a sorted list of 9, the fifth is the median. For the fifth to be 4, we need at least five numbers ≤4? Actually, no: the fifth number is the one where four are less and four are greater? No: in a sorted list of 9, the fifth is the median, meaning four are ≤ median and four are ≥ median. But since numbers can be equal, but here all distinct. So for 4 to be the median, we need exactly four numbers less than 4 and four numbers greater than 4? But 4 is in the set. So the sorted list must have four numbers <4, then 4, then four numbers >4. But the known numbers include 2,3,4,6,9,14. So 2,3 are <4, and 6,9,14 are >4. So currently, without unknowns, we have two <4 and three >4. Adding three unknowns. To have four <4 and four >4, we need to add two numbers <4 and one number >4? But then the sorted list would be [two new <4, 2,3,4, then the new >4, then 6,9,14] — but that’s only three numbers after 4? Wait, no: the sorted list would be [new <4 (two), 2,3,4, new >4, 6,9,14] — that’s 9 numbers? 2+3+1+3=9? Wait, no: the two new <4, then 2,3,4 (three numbers), then the new >4, then 6,9,14 (three numbers). So total: 2+3+1+3=9. The fifth number is 4. So yes, we can achieve median=4 by adding two numbers <4 and one number >4. Similarly, we can achieve:  
- Median=5: by adding numbers such that the fifth number is 5. But 5 is not in the set. So we need the fifth number to be 5. This requires that four numbers are <5 and four are >5. But 5 is not present. So we need to add 5? But the set must consist of distinct integers, and 5 is not in the given. So we can add 5. Then the sorted list would include 5. For example, add 5, and two other numbers. To have four <5 and four >5: currently, without unknowns, numbers <5: 2,3,4 (three), numbers >5: 6,9,14 (three). So to have four <5, we need to add one number <5. To have four >5, we need to add one number >5. But we are adding three numbers. So we can add 5, and say 1 and 10. Then sorted: 1,2,3,4,5,6,9,10,14. Median is 5. So median=5 is achievable. Similarly, median=6: we can add numbers such that the fifth is 6. Currently, numbers <6: 2,3,4 (three), numbers >6: 9,14 (two). So to have four <6 and four >6, we need to add one number <6 and two numbers >6. For example, add 5,10,15. Then sorted: 2,3,4,5,6,9,10,14,15. Median=6. Similarly, median=7: add numbers so fifth is 7. Currently, <7: 2,3,4,6 (four), >7: 9,14 (two). So to have four <7 and four >7, we need to add zero <7 (since already four) and two >7. But we are adding three numbers. So we can add 7,8,15. Then sorted: 2,3,4,6,7,8,9,14,15. Median=7. Similarly, median=8: add 8, and two numbers, say 10,11. Then sorted: 2,3,4,6,8,9,10,11,14. Median=8? The fifth is 8. But to have four <8 and four >8: currently, without unknowns, <8: 2,3,4,6 (four), >8:9,14 (two). So to have four >8, we need to add two numbers >8. But we are adding three numbers. So we can add 8,9,10? But 9 is already there. Add 8,10,11. Then sorted: 2,3,4,6,8,9,10,11,14. The fifth is 8. So yes. Similarly, median=9: add numbers so fifth is 9. Currently, <9: 2,3,4,6 (four), >9:14 (one). So to have four >9, we need to add three numbers >9. But we are adding three numbers. So add 10,11,12. Then sorted: 2,3,4,6,9,10,11,12,14. Median=9. So possible medians: 4,5,6,7,8,9. That’s six values.  
ANSWER 1: C  

Problem 2:  
We are given the list 2,6,7,7,28. A number $ N $ is added. The mean becomes twice the median. The original list has five numbers. After adding $ N $, there are 

### Problem 1:
We have a set of 9 distinct integers, so the median (middle value when sorted) is the 5th element. The 6 known numbers sorted are [2,3,4,6,9,14], with 3 additional distinct integers to add. To find possible median values:
- The minimum possible median is 3: add 3 numbers smaller than 2, sorted set becomes [a,b,c,2,3,4,6,9,14], 5th term=3.
- The maximum possible median is 9: add 3 numbers larger than 14, sorted set becomes [2,3,4,6,9,14,x,y,z], 5th term=9.
- All integers between 3 and 9 are valid medians: 3,4,5,6,7,8,9, for 7 total possible values.
Tempting wrong choices: any count less than 7, from missing intermediate medians that can be formed by adding integers between the known values.
ANSWER 1: D

---

### Problem 2:
Original list sum: 2+6+7+7+28=50. Adding N gives a list of 6 elements, sum=50+N, mean=(50+N)/6. The median of 6 elements is the average of the 3rd and 4th sorted terms. The problem states mean=2×median, so 50+N=12×median.
- If N>28, sorted list is [2,6,7,7,28,N], median=(7+7)/2=7. Then 50+N=12×7=84 → N=34, which fits.
Testing other options confirms no other N satisfies the condition.
ANSWER 2: E

---

### Problem 3:
A number is divisible by 4 if its last two digits form a number divisible by 4. For 4-digit numbers with all even digits:
- Total count: 4 (non-zero thousands digits:2,4,6,8) ×5×5×5 (all even digits for other places)=500.
- Valid last two-digit pairs (both even, divisible by 4): For any even tens digit, 10×even is divisible by 4, so only units digits 0,4,8 work, giving 3 valid units digits per tens digit: 5×3=15 valid pairs.
- Fraction: (4×5×15)/500=300/500=3/5.
ANSWER 3: D

---

### Problem 4:
The tax rate difference is 6.5%-6%=0.5%. 0.5% of $20 is 0.005×20=$0.10.
ANSWER 4: B

---

### Problem 5:
Both sums have the same number of terms: 1993-1901+1=93 terms, 193-101+1=93 terms. Subtract term-by-term: (1901-101)+(1902-102)+...+(1993-193)=1800×93=167400.
ANSWER 5: A

---

### Problem 6:
Group terms in sets of 4: (1-2-3+4)+(5-6-7+8)+... each group sums to 0. There are 1996/4=499 groups, total sum=0.
ANSWER 6: C

---

### Problem 7:
Number of people with both caps and sunglasses: 35 cap-wearers × 2/5=14. Probability a sunglass-wearer has a cap:14/50=7/25.
ANSWER 7: B

---

### Problem 8:
An even 3-digit number with digit sum 26 must have an even units digit. The only possible even units digit is 8 (any smaller even digit leaves a sum of the first two digits >18, impossible). The only number is 998, so 1 valid number.
ANSWER 8: A

---

### Problem 9:
Three-digit numbers with digit sum 25 are all permutations of (9,9,7) and (9,8,8). There are 3 permutations of (9,9,7) and 3 permutations of (9,8,8), for 6 total numbers.
ANSWER 9: C

---

### Problem 10:
Between exit 5 and 26, there are 21 gaps. To maximize one gap, minimize the other 20 gaps at 5 miles each: 20×5=100 miles. Maximum gap=118-100=18 miles.
ANSWER 10: C

---

### Problem 11:
Total sum of 5 scores=5×90=450. Median=91, so sorted scores [a,b,91,c,d]. Mode=94, so 94 must appear at least twice, so c=d=94, sum c+d=188. Sum of the two lowest scores=450-91-188=171.
ANSWER 11: B

---

### Problem 12:
Using +,-,* exactly once, one valid calculation: 5-4+6×3=1+18=19, which uses all three operations and matches option E.
ANSWER 12: E

### Problem 1:
We need to find the number of possible median values for a set of 9 distinct integers, 6 of which are 2, 3, 4, 6, 9, 14. The median of 9 numbers is the 5th element when sorted in ascending order. We add 3 distinct integers (not in the given set) and analyze possible medians:
1.  **Case 1: All 3 new numbers < 2**: Sorted list is [a,b,c,2,3,4,6,9,14], median = 3.
2.  **Case 2: 2 new numbers < 2, 1 new number ≥4**: Sorted list is [a,b,2,3,4,...], median =4.
3.  **Case 3: 1 new number <2**:
    - If we add 5 (the only integer between 4 and 6): Sorted list is [a,2,3,4,5,6,...], median=5.
    - If we do not add 5: Sorted list is [a,2,3,4,6,...], median=6.
4.  **Case 4: 0 new numbers <2 (all new numbers ≥5)**:
    - Add 5: Sorted list [2,3,4,5,6,...], median=6.
    - Add a number between 6 and9 (7 or 8): Sorted list [2,3,4,6,7,...] or [2,3,4,6,8,...], median=7 or 8.
    - Add all new numbers ≥9: Sorted list [2,3,4,6,9,...], median=9.
The only impossible medians are 2 (needs 5 numbers ≤2, only 4 max) and values >9 (existing numbers already have 5 elements ≤9). The possible medians are 3,4,5,6,7,8,9: 7 values.
ANSWER 1: D
---
### Problem 2:
Original list: [2,6,7,7,28], sum = 2+6+7+7+28=50. Insert N to make 6 elements. The median of 6 elements is the average of the 3rd and 4th sorted elements. The mean is $\frac{50+N}{6}$, which equals twice the median:
$$\frac{50+N}{6} = 2 \times \text{median} \implies 50 + N = 12 \times \text{median} \implies N = 12 \times \text{median} -50.$$
Test possible median values from the sorted list with N:
- If N ≤7: Sorted list has 3rd=7, 4th=7, median=7. Then $N=12*7 -50=84-50=34$, which is >7, valid.
- Other median values (e.g., 6.5) give N=28, which does not fit the required sorted position.
Verify N=34: Sorted list [2,6,7,7,28,34], median=(7+7)/2=7, mean=(50+34)/6=14, which is twice 7.
ANSWER 2: E
---
### Problem 3:
Positive 4-digit numbers with all even digits:
- Thousands place: 2,4,6,8 (4 choices, no 0).
- Hundreds, tens, units: 0,2,4,6,8 (5 choices each).
Total numbers: $4 \times 5 \times 5 \times5=500$.
A number is divisible by 4 iff its last two digits form a number divisible by 4. For even tens digits (0,2,4,6,8), a two-digit number $10a + b$ is divisible by 4 iff $b$ (units digit) is 0,4,8 (since $10a \equiv 0 \mod4$ for even a). So valid endings: 5 (tens choices) ×3 (units choices)=15.
Total valid numbers: $4 \times5 \times5 \times3=300$. Fraction: $\frac{300}{500}=\frac{3}{5}$.
ANSWER 3: D
---
### Problem 4:
Difference between 6.5% and 6% tax on $20:
$$0.065*20 - 0.06*20 = (0.065-0.06)*20=0.005*20=0.10.$$
ANSWER 4: B
---
### Problem 5:
Both sums have 93 terms (1993-1901+1=93, 193-101+1=93). Each term in the first sum is 1800 more than the corresponding term in the second sum:
Total difference = $93 \times 1800 = 167400$.
ANSWER 5: A
---
### Problem 6:
The sign pattern repeats every 4 terms: $+ - - +$, and each block of 4 sums to 0:
$$(1-2-3+4)+(5-6-7+8)+\dots+(1993-1994-1995+1996)=0.$$
1996 is divisible by 4, so there are 499 such blocks, total sum 0.
ANSWER 6: C
---
### Problem 7:
Let $S$ = sunglass wearers ($|S|=50$), $C$ = cap wearers ($|C|=35$). $P(S|C)=\frac{|S∩C|}{|C|}=\frac{2}{5}$, so $|S∩C|=\frac{2}{5}*35=14$.
The probability a sunglass wearer wears a cap is $P(C|S)=\frac{|S∩C|}{|S|}=\frac{14}{50}=\frac{7}{25}$.
ANSWER 7: B
---
### Problem 8:
3-digit numbers with digit sum 26: maximum digit sum is 27 (999), so 26 is 1 less than maximum, meaning digits are permutations of (9,9,8). The only even permutation is 998 (units digit 8). So 1 number.
ANSWER 8: A
---
### Problem 9:
3-digit numbers with digit sum 25: 25 is 2 less than 27, so digits are permutations of (9,9,7) and (9,8,8):
- Permutations of (9,9,7): 997,979,799 (3 numbers).
- Permutations of (9,8,8): 988,898,889 (3 numbers).
Total 6 numbers.
ANSWER 9: C
---
### Problem 10:
Between exit 5 and 26, there are $26-5=21$ gaps, total distance 118 miles. Each gap ≥5 miles. To maximize one gap, minimize the other 20: $20*5=100$, so maximum gap = $118-100=18$.
ANSWER 10: C
---
### Problem 11:
Five sorted scores: $a≤b≤c≤d≤e$. Median $c=91$, mean 90, so total sum $5*90=450$, so $a+b+d+e=450-91=359$. Mode is 94, so 94 appears at least twice. Since $c=91$, 94 can only be in $d$ and $e$, so $d=e=94$. Then $a+b=359-94-94=171$. This sum is fixed (valid examples exist, e.g., 81+90=171).
ANSWER 11: B
---
### Problem 12:
Permute +, -, × in 5 __4 __6 __3, using standard order of operations (multiply first):
- $5 -4 +6×3 = 5-4+18=19$, which is an option.
All other permutations give values not in the choices.
ANSWER 12: E

I'll solve each problem step by step.

---

**Problem 1:**

We have 6 elements: {2, 3, 4, 6, 9, 14}, and need to add 3 more distinct integers to make 9 total.

The median of 9 numbers is the 5th element when arranged in order.

Currently we have 6 numbers. By strategically placing 3 new numbers, we can shift which value becomes the median.

Let me consider what values can be the median (5th position):
- If we add 3 numbers all ≤ 2, the sorted list has our 6 original numbers in positions 4-9, so median is 4
- If we add numbers strategically between and around the original set, we can make the median be: 2, 3, 4, 6, 9, 14, or a new number
- If we add 3 numbers all ≥ 14, the sorted list has our 6 original numbers in positions 1-6, so median is 9

Testing systematically:
- Median = 2: add three numbers < 2 (e.g., -2, -1, 0) ✓
- Median = 3: add two < 2, one between 3 and 4 ✓
- Median = 4: add three < 2, or arranged appropriately ✓
- Median = 6: add one or two strategically ✓
- Median = 9: add three > 14, or arranged appropriately ✓
- Median = 14: add three > 14 ✓
- Median = a new value between existing numbers ✓

The possible medians are: 2, 3, 4, 6, 9, 14, and any new integer we insert. We can get 7 different values.

**ANSWER 1: D**

---

**Problem 2:**

Original list: 2, 6, 7, 7, 28. Sum = 50.

When N is inserted, we have 6 numbers. Let's find the median of 6 numbers (average of 3rd and 4th when sorted).

Sum of all 6 numbers = 50 + N
Mean = (50 + N)/6

We need: Mean = 2 × Median

Let me test by considering where N falls:
- If N ≤ 2: sorted list is N, 2, 6, 7, 7, 28. Median = (6+7)/2 = 6.5. Mean = 2×6.5 = 13, so 50+N = 78, N = 28. Contradiction.
- If 2 < N ≤ 6: sorted list is 2, N, 6, 7, 7, 28. Median = (6+7)/2 = 6.5. Mean = 13, so N = 28. Contradiction.
- If 6 < N ≤ 7: sorted list is 2, 6, N, 7, 7, 28. Median = (N+7)/2. Mean = 2×(N+7)/2 = N+7. So (50+N)/6 = N+7, giving 50+N = 6N+42, so 5N = 8, N = 1.6. Not in range.
- If 7 < N ≤ 28: sorted list is 2, 6, 7, N, 7, 28 or 2, 6, 7, 7, N, 28. Median = (7+N)/2. Mean = 2×(7+N)/2 = 7+N. So (50+N)/6 = 7+N, giving 50+N = 42+6N, so 5N = 8, N = 1.6. Not valid.
- If N > 28: sorted list is 2, 6, 7, 7, 28, N. Median = (7+7)/2 = 7. Mean = 14. So 50+N = 84, N = 34. ✓

**ANSWER 2: E**

---

**Problem 3:**

Four-digit numbers with all even digits: first digit ∈ {2,4,6,8}, others ∈ {0,2,4,6,8}.

Total count: 4 × 5 × 5 × 5 = 500

For divisibility by 4, the last two digits must form a number divisible by 4.

Last two digits from {00, 02, 04, 06, 08, 20, 22, 24, 26, 28, 40, 42, 44, 46, 48, 60, 62, 64, 66, 68, 80, 82, 84, 86, 88}

Divisible by 4: 00, 04, 08, 20, 24, 28, 40, 44, 48, 60, 64, 68, 80, 84, 88 (15 out of 25)

Count divisible by 4: 4 × 5 × 15 = 300

But wait, the second digit is 5 choices, third is which determines divisibility by 4.

Actually: 4 × 5 × (number of even two-digit combos divisible by 4) = 4 × 5 × 15 = 300

Hmm, let me recalculate. First digit: 4 choices. Second digit: 5 choices. Last two digits must be divisible by 4: check all pairs from {0,2,4,6,8}×{0,2,4,6,8} giving 15 valid pairs.

Count = 4 × 5 × 15 = 300. Fraction = 300/500 = 3/5.

**ANSWER 3: D**

---

**Problem 4:**

6.5% tax on $20: 0.065 × 20 = $1.30
6% tax on $20: 0.06 × 20 = $1.20
Difference: $1.30 - $1.20 = $0.10

**ANSWER 4: B**

---

**Problem 5:**

Sum from 1901 to 1993: This is 93 consecutive integers starting at 1901.
Sum = (1901 + 1993)/2 × 93 = 1947 × 93 = 181,071

Sum from 101 to 193: This is 93 consecutive integers starting at 101.
Sum = (101 + 193)/2 × 93 = 147 × 93 = 13,671

Difference = 181,071 - 13,671 = 167,400

**ANSWER 5: A**

---

**Problem 6:**

Pattern: 1 - 2 - 3 + 4 + 5 - 6 - 7 + 8 + 9 - 10 - 11 + ...

Group by 4s: (1-2-3+4) + (5-6-7+8) + (9-10-11+12) + ... + (1993-1994-1995+1996)

Each group of 4: (4k+1) - (4k+2) - (4k+3) + (4k+4) = 4k+1-4k-2-4k-3+4k+4 = 0

From 1 to 1996, we have 1996/4 = 499 complete groups, each summing to 0.

Total = 0

**ANSWER 6: C**

---

**Problem 7:**

Let x = people wearing both sunglasses and caps.

P(sunglasses | cap) = x/35 = 2/5, so x = 14.

P(cap | sunglasses) = 14/50 = 7/25.

**ANSWER 7: B**

---

**Problem 8:**

Three-digit even numbers with digit sum 26.

For an even number, last digit ∈ {0, 2, 4, 6, 8}.

If last digit = 0: first two digits sum to 26. Max is 9+9=18. Impossible.
If last digit = 2: first two digits sum to 24. Only 9+9=18 < 24. Impossible.
If last digit = 4: first two digits sum to 22. Impossible (max 18).
If last digit = 6: first two digits sum to 20. Impossible.
If last digit = 8: first two digits sum to 18. This requires 9+9=18. Number: 998. ✓

Only 1 number.

**ANSWER 8: A**

---

**Problem 9:**

Three-digit numbers abc where a+b+c=25, with a ∈ {1,...,9}, b,c ∈ {0,...,9}.

a+b+c = 25 means b+c = 25-a.

For valid solutions: 0 ≤ b+c ≤ 18, so 7 ≤ a ≤ 9.

- a=7: b+c=18. Solutions: (9,9) → 1 way
- a=8: b+c=17. Solutions: (8,9), (9,8) → 2 ways
- a=9: b+c=16. Solutions: (7,9), (8,8), (9,7) → 3 ways

Total = 1 + 2 + 3 = 6

**ANSWER 9: C**

---

**Problem 10:**

Exits from 5th to 26th = 26-5 = 21 segments between consecutive exits.

Each segment ≥ 5 miles. Total distance = 118 miles.

If 20 segments are 5 miles each: 20 × 5 = 100 miles.
Remaining for the largest segment: 118 - 100 = 18 miles.

**ANSWER 10: C**

---

**Problem 11:**

Five scores: mean = 90 (sum = 450), median = 91, mode = 94.

Arranged in order: a, b, c, d, e where c = 91 (median).

Mode = 94 means 94 appears at least twice. Since median is 91, we must have 94 appearing in {d, e}.

Case 1: d = e = 94. Then a + b + 91 + 94 + 94 = 450, so a + b = 171.

Check consistency: we need a ≤ b ≤ 91 and mode is 94 (appears twice). If a or b equals 94, mode wouldn't be unique. So a, b < 91 works. ✓

**ANSWER 11: B**

---

**Problem 12:**

We need to place +, −, × in three blanks: 5 __ 4 __ 6 __ 3

Trying combinations:
- 5 + 4 × 6 − 3 = 5 + 24 − 3 = 26
- 5 × 4 + 6 − 3 = 20 + 6 − 3 = 23
- 5 × 4 − 6 + 3 = 20 − 6 + 3 = 17
- 5 + 4 − 6 × 3 = 5 + 4 − 18 = −9
- 5 − 4 

Problem 1:

We are given a set of 9 distinct integers. Six of them are
\[
2,3,4,6,9,14.
\]
We want the number of possible values of the median of the 9 numbers.

For 9 numbers, the median is the 5th number when the list is ordered.

Now look at the six given numbers in sorted order:
\[
2,3,4,6,9,14.
\]
Since there are 3 other distinct integers, the median can be any number that ends up in the 5th position after inserting those 3 numbers.

Let’s see what values are possible:

- If we place 3 new numbers all very small, then the 5th number could be 4.
- It could also be 6, or 9, or even 14, depending on how many new numbers are inserted before each.
- It cannot be 2 or 3, because there would need to be at least 4 numbers less than or equal to it, but only 3 extra numbers are available and the given numbers already limit this.
- It cannot be any number not among the given six if the median must be one of the existing numbers? Let’s check carefully: since the 3 extra numbers are distinct integers and can be placed anywhere, the 5th number could indeed be one of the inserted numbers, but then it would have to lie between the 4th and 6th numbers. Since the six given numbers already occupy several possible positions, the median can also be a new integer. But because the set is of distinct integers and the question asks how many possible values, we should count all possible medians that can occur.

A cleaner way: the median of 9 distinct integers is the middle element. With 6 fixed numbers, the 3 added numbers can be arranged so that the median is any integer between the 4th smallest and 6th smallest of the fixed numbers, inclusive, but here the possible distinct medians are:
\[
4,5,6,7,8,9
\]
Wait — can all of these really happen? Let’s test.

- Median 4: choose 3 new numbers less than 4.
- Median 5: choose 3 new numbers so that 5 is between 4 and 6 and becomes 5th.
- Median 6: choose 3 new numbers with 6 as middle.
- Median 7, 8, 9 similarly.

So the possible medians are the integers from 4 through 9, which is 6 values.

However, we must match the provided choices. The intended count is the number of possible median values among the listed integers and inserted integers. Since the only restriction is distinct integers, every integer from 4 through 9 can occur.

That gives 6 possible values.

ANSWER 1: C

---

Problem 2:

We insert a number \(N\) into the list
\[
2,6,7,7,28.
\]
Then the mean becomes twice the median.

Let’s analyze the median after inserting \(N\) into 6 numbers total.

The original sorted list is:
\[
2,6,7,7,28.
\]
There are 5 numbers. After inserting one number, there are 6 numbers, so the median is the average of the 3rd and 4th numbers in the sorted list.

We try the choices.

If \(N=7\), the list becomes
\[
2,6,7,7,7,28.
\]
Median:
\[
\frac{7+7}{2}=7.
\]
Mean:
\[
\frac{2+6+7+7+7+28}{6}=\frac{57}{6}=9.5.
\]
Twice the median would be 14, not 9.5, so not 7.

Try \(N=14\):
\[
2,6,7,7,14,28.
\]
Median:
\[
\frac{7+7}{2}=7.
\]
Mean:
\[
\frac{2+6+7+7+14+28}{6}=\frac{64}{6}=\frac{32}{3},
\]
not 14.

Try \(N=20\):
\[
2,6,7,7,20,28.
\]
Median is still 7.
Mean:
\[
\frac{2+6+7+7+20+28}{6}=\frac{70}{6}=\frac{35}{3},
\]
not 14.

Try \(N=28\):
\[
2,6,7,7,28,28.
\]
Median is 7.
Mean:
\[
\frac{2+6+7+7+28+28}{6}=\frac{78}{6}=13,
\]
not 14.

Try \(N=34\):
\[
2,6,7,7,28,34.
\]
Median is 7.
Mean:
\[
\frac{2+6+7+7+28+34}{6}=\frac{84}{6}=14.
\]
Twice the median is
\[
2\cdot 7=14.
\]
This works.

So \(N=34\).

ANSWER 2: E

---

Problem 3:

We consider all positive four-digit integers whose digits are all even.

Even digits are
\[
0,2,4,6,8.
\]
A four-digit number cannot start with 0, so the thousands digit has 4 choices:
\[
2,4,6,8.
\]
Each of the other three digits has 5 choices, so the total number is
\[
4\cdot 5^3.
\]

Now determine how many are divisible by 4.

A number is divisible by 4 if its last two digits form a number divisible by 4.

So we only need to count ordered pairs of even digits \((t,u)\) from \(\{0,2,4,6,8\}\) such that \(10t+u\) is divisible by 4.

Check the possible last two digits:

- \(00\) yes
- \(02\) no
- \(04\) yes
- \(06\) no
- \(08\) yes

- \(20\) yes
- \(22\) no
- \(24\) yes
- \(26\) no
- \(28\) yes

- \(40\) yes
- \(42\) no
- \(44\) yes
- \(46\) no
- \(48\) yes

- \(60\) yes
- \(62\) no
- \(64\) yes
- \(66\) no
- \(68\) yes

- \(80\) yes
- \(82\) no
- \(84\) yes
- \(86\) no
- \(88\) yes

For each tens digit, 3 of the 5 possibilities work, so there are
\[
5\cdot 3=15
\]
valid last-two-digit pairs.

Since the first two digits can be anything allowed:
- thousands digit: 4 choices
- hundreds digit: 5 choices

Total divisible by 4:
\[
4\cdot 5\cdot 15
\]
Total numbers:
\[
4\cdot 5^3 = 4\cdot 125
\]
So the fraction is
\[
\frac{4\cdot 5\cdot 15}{4\cdot 5^3}=\frac{15}{25}=\frac35.
\]

ANSWER 3: D

---

Problem 4:

An item costs \$20 before tax.

Difference between 6.5% tax and 6% tax is
\[
0.5\% = 0.005.
\]
So the difference in tax is
\[
0.005 \times 20 = 0.1
\]
dollar, which is \(\$0.10\).

ANSWER 4: B

---

Problem 5:

We need to compute
\[
(1901+1902+\cdots+1993) - (101+102+\cdots+193).
\]

Notice each term in the second sum is exactly 1800 less than the corresponding term in the first sum:
\[
1901-101=1800,\quad 1902-102=1800,\ \ldots,\ 1993-193=1800.
\]
How many terms are there?

From 1901 to 1993 inclusive:
\[
1993-1901+1=93
\]
terms.

So the difference is
\[
93 \cdot 1800 = 167400.
\]

ANSWER 5: A

---

Problem 6:

We evaluate
\[
1-2-3+4+5-6-7+8+9-10-11+\cdots+1992+1993-1994-1995+1996.
\]

Look at the pattern in blocks of 4:
\[
(1-2-3+4),\ (5-6-7+8),\ (9-10-11+12),\ \ldots
\]
Each block of 4 consecutive integers has sum
\[
n-(n+1)-(n+2)+(n+3)=0
\]
when \(n\equiv 1 \pmod 4\).

So all full blocks cancel.

Now check the last block:
The pattern continues through \(1992+1993-1994-1995+1996\), which is not a full 4-term block. But we can group from 1 to 1996 in blocks of 4:
\[
(1-2-3+4)+(5-6-7+8)+\cdots+(1993-1994-1995+1996).
\]
Each block sums to 0.

Therefore the entire 

Problem 1:  
We have 9 distinct integers. Six of them are fixed: 2, 3, 4, 6, 9, 14. The other 3 are distinct integers of our choice.  
With 9 numbers, the median is the 5th number when sorted.

So the median depends only on how many of the fixed numbers fall below/at various positions, and where we insert the 3 new numbers.

Let’s analyze possible 5th elements.

### Step 1: Sort the fixed numbers
Fixed numbers in increasing order:
2, 3, 4, 6, 9, 14

The median (5th) could be one of these fixed numbers, or one of the new ones.

### Casework by choosing what the 5th number is

#### Can the median be 2?
For 2 to be the median, we need at least 4 numbers ≤ 2, but only 2 itself is ≤ 2 among the fixed set (and new numbers would have to be distinct integers ≤ 2).  
That means we’d need 4 numbers less than or equal to 2, including 2, so we’d need 3 additional distinct integers ≤ 2 (other than 2). That’s possible (e.g., 0, −1, 1).  
But then 2 would be the 5th, meaning exactly 4 numbers are <2. Can we guarantee? Actually, since we can choose any 3 extra integers, we can make exactly 4 numbers less than 2 by picking 3 integers <2 plus none from fixed besides 2. That would make 2 the 5th. However, note 2 is already one of the 9 distinct integers, so it can be the median.  
So potentially 2 is possible.

But the problem asks for the number of possible median values, and the choices are between 4 and 8, so likely only medians among the fixed list can work. Let’s check more carefully: the fixed numbers below 2 is only none (except 2). To make 2 the 5th, we need 4 numbers smaller than 2. We only control 3 extra numbers, so we can only create at most 3 numbers smaller than 2. Therefore, we cannot reach 4 smaller numbers.  
✅ Conclusion: median cannot be 2.

Similarly, we test each fixed candidate using “how many fixed numbers are below it”.

### General method
For a number \(x\) to be the median (5th), there must be exactly 4 numbers less than \(x\), and at least one number equal to \(x\). If \(x\) is one of the fixed numbers, we can count how many fixed numbers are less than \(x\), then see if the remaining “needed” smaller numbers can come from the 3 new integers.

Let’s compute fixed counts less than each candidate:

- Below 2: 0  
- Below 3: 1 (just 2)  
- Below 4: 2 (2,3)  
- Below 6: 3 (2,3,4)  
- Below 9: 4 (2,3,4,6)  
- Below 14: 5 (2,3,4,6,9)

We need exactly 4 numbers less than the median.

Now see possibilities:

#### Median = 3
Fixed numbers below 3: 1.  
We need 4 total below 3, so we must add 3 new numbers < 3. That’s possible.  
✅ So 3 can be median.

#### Median = 4
Fixed below 4: 2. Need 2 more from new numbers < 4. Possible.  
✅ 4 can be median.

#### Median = 6
Fixed below 6: 3. Need 1 more from new numbers < 6. Possible.  
✅ 6 can be median.

#### Median = 9
Fixed below 9: 4 already. Then all 4 smaller numbers are fixed, and the median can be 9 regardless of where new numbers go as long as none are between 9 and 14 in a way that changes ordering?  
We only need 9 to be the 5th, meaning at least 4 numbers are <9 (we have exactly 4 fixed ones). If we add any new numbers <9, that would make more than 4 numbers less than 9 and break it. So we must choose the 3 new integers all > 9. That’s possible.  
✅ 9 can be median.

#### Median = 14
Fixed below 14: 5 already. But we need exactly 4 below the median. Impossible.  
❌ 14 cannot be median.

#### Median = 2
Fixed below 2: 0, need 4 from new numbers <2, but we only have 3 new numbers. Impossible.  
❌ 2 cannot be median.

So the only fixed medians are: 3, 4, 6, 9 → 4 values.

Are there possible medians that are not among the fixed numbers (i.e., one of the 3 new integers)?  
Yes, if we choose all new integers appropriately.

Let the median be a new number \(y\). Then among the fixed numbers, some number of them must be less than \(y\). Since there are only 6 fixed numbers, the possible counts of fixed elements less than \(y\) range from 0 to 6.

But we need exactly 4 numbers total less than the median. Since the median is new, all 4 numbers less than it must come from:
- some fixed numbers less than it, plus
- some of the other two new numbers (because one new number is the median itself).

We can make any new median that sits in the “gaps” between fixed numbers such that exactly 4 elements end up below it.

The median \(y\) must be positioned so that:
- the number of fixed elements less than \(y\) is either 1,2,3, or 4 (because we then use the remaining smaller ones from the other new numbers, limited to 0,1,2, or 3).
But we must also ensure distinctness and that \(y\) lies in the right interval.

There are gaps:
(-∞,2), (2,3), (3,4), (4,6), (6,9), (9,14), (14,∞)

Try each gap for possible median:

- If \(y\in(2,3)\): fixed less than y = 1 (only 2). Need 3 more < y from the other new numbers. Possible.  
So any integer y in (2,3) means y=?? none (only integer between 2 and 3 doesn’t exist). So no.
- \(y\in(3,4)\): fixed less than y = 2. Need 2 more < y. Possible. But no integer between 3 and 4. none.
- \(y\in(4,6)\): fixed less than y = 3 (2,3,4). Need 1 more from new numbers <y. Possible. Integers in (4,6): only 5. So y=5 works.
- \(y\in(6,9)\): fixed less than y = 4 (2,3,4,6). Need 0 more. Then both other new numbers must be >y. Possible. Integers in (6,9): 7,8 → both could be median.
- \(y\in(9,14)\): fixed less than y = 5, but we need total 4 < median. impossible.

Thus additional possible medians are 5, 7, 8 → 3 more values.

Total possible median values: fixed ones (3,4,6,9) = 4 plus new ones (5,7,8)=3 gives 7.

✅ Answer choice is 7.

ANSWER 1: D


Problem 2:  
List is: 2, 6, 7, 7, 28 (five numbers). Insert N to make 6 numbers total.

New mean and median relation: “mean is now twice as great as the median.”

### Step 1: Compute original sum
Sum of given numbers:
2 + 6 + 7 + 7 + 28 = 50

After inserting N, total sum = 50 + N.  
There are 6 numbers, so mean is:
\[
\text{mean}=\frac{50+N}{6}
\]

### Step 2: Find the median
Sort of the ne

Problem 1:
We have a set $R$ of 9 distinct integers. Six are $\{2, 3, 4, 6, 9, 14\}$. Let the three unknown integers be $x, y, z$. When sorted, the median is the 5th element.
To minimize the median, we pick the three smallest possible integers (e.g., $-1, 0, 1$). The sorted set would be $\{-1, 0, 1, 2, 3, 4, 6, 9, 14\}$. The median is 3.
To maximize the median, we pick the three largest possible integers (e.g., $15, 16, 17$). The sorted set would be $\{2, 3, 4, 6, 9, 14, 15, 16, 17\}$. The median is 9.
The possible values for the median are integers between 3 and 9 inclusive, excluding values already occupied by the fixed set if they cannot be the median. However, we can choose $x, y, z$ to make any integer from 3 to 9 the median.
Values: 3, 4, 5, 6, 7, 8, 9. There are 7 possible values.
ANSWER 1: D

Problem 2:
List: $2, 6, 7, 7, 28$. $N$ is added. Mean = $\frac{2+6+7+7+28+N}{6} = \frac{50+N}{6}$.
If $N$ is added, the median depends on $N$.
If $N \le 7$, median is 7. Mean = $2 \times 7 = 14 \implies 50+N = 84 \implies N=34$ (Contradiction).
If $N \ge 7$, let's test $N=20$: List $\{2, 6, 7, 7, 20, 28\}$, median is $(7+7)/2 = 7$. Mean = $70/6 \neq 14$.
If $N=14$: List $\{2, 6, 7, 7, 14, 28\}$, median is $(7+7)/2 = 7$. Mean = $64/6 \neq 14$.
If $N=28$: List $\{2, 6, 7, 7, 28, 28\}$, median is 7. Mean = $78/6 = 13 \neq 14$.
If $N=34$: List $\{2, 6, 7, 7, 28, 34\}$, median is $(7+28)/2 = 17.5$. Mean = $84/6 = 14$. $14 \neq 35$.
Wait, try $N=20$: Mean = $70/6 \approx 11.6$. Median = 7.
Try $N=34$: Mean = 14. Median = 17.5.
Re-check: $50+N = 12 \times \text{median}$. If $N=20$, median is 7, $70 = 12 \times 7$ (No).
If $N=34$, median is 17.5, $84 = 12 \times 17.5$ (No).
Try $N=28$: Mean = 13, Median = 7. $13 \neq 14$.
Try $N=20$: Mean = $70/6 = 11.66$, Median = 7.
Let's check $N=20$ again. $50+20 = 70$. $70/6 = 11.66$.
Actually, if $N=20$, list is $2, 6, 7, 7, 20, 28$. Median is $(7+7)/2 = 7$. $2 \times 7 = 14$. $70/6 \neq 14$.
If $N=34$, median is $(7+28)/2 = 17.5$. Mean = $84/6 = 14$. $14 \neq 2 \times 17.5$.
If $N=20$, mean is $11.66$. If $N=34$, mean is 14.
Looking at choices, $N=20$ is the only one that yields a mean close to twice a median.
ANSWER 2: C

Problem 3:
Digits are $\{0, 2, 4, 6, 8\}$. First digit cannot be 0 (4 choices: 2, 4, 6, 8). Others have 5 choices. Total = $4 \times 5 \times 5 \times 5 = 500$.
Divisible by 4 means the last two digits form a multiple of 4. Possible pairs: 00, 04, 08, 20, 24, 28, 40, 44, 48, 60, 64, 68, 80, 84, 88. (15 pairs).
Total numbers = $4 \times 5 \times 15 = 300$.
Fraction = $300/500 = 3/5$.
ANSWER 3: D

Problem 4:
$0.065 \times 20 = 1.30$. $0.06 \times 20 = 1.20$.
Difference = $1.30 - 1.20 = 0.10$.
ANSWER 4: B

Problem 5:
$(1901-101) + (1902-102) + \dots + (1993-193)$.
Each term is $1800$. There are $1993 - 1901 + 1 = 93$ terms.
$93 \times 1800 = 167,400$.
ANSWER 5: A

Problem 6:
Group into sets of 4: $(1-2-3+4) + (5-6-7+8) + \dots + (1993-1994-1995+1996)$.
Each group: $1-2-3+4 = 0$.
There are $1996/4 = 499$ groups. $499 \times 0 = 0$.
ANSWER 6: C

Problem 7:
$C = 35$. $P(S|C) = 2/5$. So $S \cap C = 35 \times 2/5 = 14$.
$S = 50$. $P(C|S) = (S \cap C) / S = 14/50 = 7/25$.
ANSWER 7: B

Problem 8:
Digits $a, b, c$. $a+b+c = 26$. $c \in \{0, 2, 4, 6, 8\}$.
If $c=8, a+b=18 \implies (9,9)$. (1)
If $c=6, a+b=20$ (None).
If $c=4, a+b=22$ (None).
If $c=2, a+b=24$ (None).
If $c=0, a+b=26$ (None).
Wait, $a+b=18 \implies 998$. $a+b=17 \implies 899, 989$. $a+b=16 \implies 799, 979, 889$.
If $c=8, a+b=18 \implies 998$.
If $c=6, a+b=20$ (None).
If $c=4, a+b=22$ (None).
If $c=2, a+b=24$ (None).
If $c=0, a+b=26$ (None).
Wait, $a+b+c=26$. $c$ must be even.
If $c=8: a+b=18 \implies (9,9) \to 998$.
If $c=6: a+b=20$ (None).
If $c=4: a+b=22$ (None).
If $c=2: a+b=24$ (None).
If $c=0: a+b=26$ (None).
Wait, $a, b$ can be 9. Max $a+b=18$. So $c$ must be at least $26-18=8$.
If $c=8, a+b=18 \implies 998$.
If $c=9$ (not even).
Are there others? $a+b+c=26$. $c \in \{0, 2, 4, 6, 8\}$.
If $c=8, a+b=18 \to 998$.
If $c=6, a+b=20$ (Impossible).
Only 998? Let's re-read. "3-digit whole numbers".
Maybe $c$ can be 8, 6, 4, 2, 0.
$c=8 \implies a+b=18 \implies 998$.
$c=6 \implies a+b=20$ (No).
$c=4 \implies a+b=22$ (No).
$c=2 \implies a+b=24$ (No).
$c=0 \implies a+b=26$ (No).
Only 1.
ANSWER 8: A

Problem 9:
$a+b+c=25$. $a, b, c \in \{0..9\}, a \neq 0$.
Possible sets $\{9,9,7\}, \{9,8,8\}$.
Permutations of $\{9,9,7\}$: $997, 979, 799$ (3).
Permutations of $\{9,8,8\}$: $988, 898, 889$ (3).
Total = $3+3 = 6$.
ANSWER 9: C

Problem 10:
Exits 5 to 26 is $26-5 = 21$ intervals. Total distance 118.
Let $d_i$ be the distance between exits. $\sum_{i=1}^{21} d_i = 118$.
$d_i \ge 5$. To maximize one $d_k$, minimize others: $d_i = 5$ for $i \neq k$.
$d_k + 20(5) = 118 \implies d_k + 100 = 118 \implies d_k = 18$.
ANSWER 10: C

Problem 11:
Sum = $5 \times 90 = 450$. Mode = 94. Median = 91.
Let scores be $x_1, x_2, 91, 94, 94$.
$x_1 + x_2 + 91 + 94 + 94 = 450 \implies x_1 + x_2 + 279 = 450 \implies x_1 + x_2 = 171$.
ANSWER 11: B

Problem 12:
$5 \pm 4 \pm 6 \pm 3$.
$5+4 \times 6 - 3 = 5+24-3 = 26$.
$5 \times 4 - 6 + 3 = 20-6+3 = 17$.
$5 \times 4 + 6 - 3 = 23$.
$5 - 4 + 6 \times 3 = 1 + 18 = 19$.
ANSWER 12: E

The set \(R\) consists of nine distinct integers, including the six given values \(2, 3, 4, 6, 9, 14\). When the elements of \(R\) are placed in increasing order \(a_1 < a_2 < a_3 < a_4 < a_5 < a_6 < a_7 < a_8 < a_9\), the median is \(a_5\). Thus exactly four elements of \(R\) must lie strictly below the median value \(m = a_5\).
Let \(k\) denote the number of the six given values that are strictly less than \(m\). The three unknown values must then be chosen so that exactly \(4 - k\) of them lie strictly below \(m\).
- If \(m\) is one of the given values, the admissible choices are \(m = 3, 4, 6, 9\). For each of these four numbers it is possible to select three unknown integers satisfying the counting requirement above (and remaining distinct from one another and from the given values).
- If \(m\) is not one of the given values, the admissible choices are the integers \(m = 5, 7, 8\). Again, for each of these three numbers it is possible to select the remaining two unknown integers so that exactly four elements lie below \(m\).
No other integer can serve as median: any candidate smaller than 3 or larger than 9 produces a non-negative integer count of unknowns below \(m\) that is impossible to realize with only three unknowns. Consequently the possible values of the median are exactly the seven numbers
\[
3,4,5,6,7,8,9.
\]
Thus seven possible values exist for the median of \(R\).
ANSWER 1: D
Problem 2:
Let the five numbers be \(2, 6, 7, 7, 28, N\). Their sum is \(50 + N\), so the mean is
\[
\frac{50 + N}{5}.
\]
When the six numbers are ordered, the median is the average of the third and fourth values. The value of \(N\) determines the ordered list, and therefore the median, in each of the five mutually exclusive ranges determined by the given numbers.
- If \(N \le 2\), the ordered list is \(N,2,6,7,7,28\) and the median is \((6+7)/2 = 6.5\). Setting the mean equal to twice the median yields the contradictory equation \(N = 15\).
- If \(2 < N \le 6\), the ordered list is \(2,N,6,7,7,28\) and the median is again \(6.5\). The same contradictory equation appears.
- If \(6 < N \le 7\), the ordered list is \(2,6,N,7,7,28\) and the median is \((N+7)/2\). The mean-equals-twice-median equation simplifies to the quadratic
 \[
 N^2 - 9N - 60 = 0,
 \]
 whose only root in the interval is \(N = 12\), which lies outside the interval.
- If \(7 < N \le 28\), the ordered list is \(2,6,7,7,N,28\) and the median is \((7+N)/2\). The mean-equals-twice-median equation simplifies to the linear equation
 \[
 N = 20,
 \]
 which lies inside the interval.
- If \(N > 28\), the ordered list is \(2,6,7,7,28,N\) and the median is \((7+28)/2 = 17.5\). The resulting equation \(N = 120\) lies outside the interval.
Hence the only value that satisfies the stated condition is \(N = 20\).
ANSWER 2: C
Problem 3:
A four-digit integer whose digits are all even may be written
\[
1000a + 100b + 10c + d,
\]
where each of \(a,b,c,d\) belongs to the set \(\{0,2,4,6,8\}\) and \(a \ne 0\). There are therefore
\[
4 \times 5^3 = 500
\]
such integers.
An integer is divisible by 4 if and only if its last two digits form a number divisible by 4. The admissible pairs \((c,d)\) are therefore exactly the 13 pairs
\[
(0,0),(0,4),(0,8),(2,0),(2,4),(2,8),(4,0),(4,4),(4,8),(6,0),(6,4),(8,0),(8,8)
\]
that satisfy the divisibility test. For each of the 13 choices of \((c,d)\) the first two digits may be chosen freely (\(a\) has 4 possibilities, \(b\) has 5), giving
\[
13 \times 4 \times 5 = 260
\]
favorable integers. The required fraction is therefore
\[
\frac{260}{500} = \frac{13}{25} = \frac{2}{5}.
\]
ANSWER 3: B
Problem 4:
A 6.5 percent tax on a $20 item is \(0.065 \times 20 = 1.30\). A 6 percent tax is \(0.06 \times 20 = 1.20\). Their difference is
\[
1.30 - 1.20 = 0.10.
\]
ANSWER 4: B
Problem 5:
The first sum is an arithmetic series with 93 terms whose average is
\[
\frac{1901 + 1993}{2} = 1947,
\]
so its value equals \(93 \times 1947 = 181071\). The second sum is an arithmetic series with 93 terms whose average is
\[
\frac{101 + 193}{2} = 147,
\]
so its value equals \(93 \times 147 = 13671\). Subtracting yields
\[
181071 - 13671 = 167400.
\]
ANSWER 5: A
Problem 6:
The given expression consists of 1996 terms whose signs follow the repeating four-term pattern
\[
+,-,-,+,+,-,-,+,\dots
\]
(The pattern begins with a single “+” and thereafter repeats every four signs.) In each complete block of four consecutive integers the signed sum is zero. There are exactly 499 such blocks, accounting for 1996 terms and contributing a total of zero. Hence the whole expression evaluates to 0.
ANSWER 6: C
Problem 7:
Let \(S\) be the set of people wearing sunglasses and \(C\) the set wearing caps. Then
\[
|S \cap C| = \frac{2}{5}|C| = \frac{2}{5} \cdot 35 = 14.
\]
Consequently
\[
|S| = |S \setminus C| + |S \cap C|
\]
is unknown, but the conditional probability required by the problem is
\[
P(C \mid S) = \frac{|S \cap C|}{|S|} = \frac{14}{|S|}.
\]
Because \(|S \cap C| = 14\) and \(|C| = 35\), the only numerical value that can be obtained from the given data is the ratio
\[
\frac{14}{35} = \frac{2}{5}.
\]
(The actual size of \(S\) is never needed.)
ANSWER 7: C
Problem 8:
A three-digit number \(100a + 10b + c\) (with \(a \in \{1,\dots,9\}\), \(b,c \in \{0,\dots,9\}\)) is even precisely when \(c\) is even. The digit-sum equation \(a + b + c = 26\) together with the upper bound 9 on each variable forces \(a \ge 8\). The only possibilities are therefore the six triples
\[
(9,9,8),\ (9,8,9),\ (8,9,9)
\]
and their permutations that keep the units digit even. Exactly three of them have even units digit:
\[
998,\ 986,\ 968.
\]
Thus three even three-digit numbers have digit-sum 26.
ANSWER 8: C
Problem 9:
A three-digit number \(100a + 10b + c\) satisfies \(a + b + c = 25\) with each digit at most 9. Then necessarily \(a \ge 7\). The admissible triples \((a,b,c)\) are exactly the six permutations of
\[
(9,9,7),\ (9,8,8).
\]
Hence six such numbers exist.
ANSWER 9: C
Problem 10:

## Step 1: Problem 1
To find the number of possible values of the median of set R, we first need to understand what a median is. The median of a set of numbers is the middle number when the numbers are arranged in ascending order. Since R is a set of nine distinct integers, the median will be the fifth number when the numbers are arranged in ascending order.

## Step 2: Problem 1
Given that six of the elements are 2, 3, 4, 6, 9, and 14, we can start by arranging these numbers in ascending order: 2, 3, 4, 6, 9, 14. The remaining three numbers can be any distinct integers.

## Step 3: Problem 1
To determine the possible medians, we need to consider the range of values the fifth number can take. The numbers already given are 2, 3, 4, 6, 9, 14. So, the three additional numbers can either be less than 4, between 4 and 9, or greater than 9.

## Step 4: Problem 1
If all three additional numbers are less than 4, the order becomes: additional number, additional number, additional number, 4, 6, 9, 14. The median is 6.
If two additional numbers are less than 4 and one is between 4 and 9, the order becomes: additional number, additional number, 4, additional number, 6, 9, 14. The median is between 4 and 6, so it could be 5.
If one additional number is less than 4 and two are between 4 and 9, the order becomes: additional number, 4, additional number, additional number, 6, 9, 14. The median could be between 4 and 6.
If all three additional numbers are between 4 and 9, the order becomes: 2, 3, 4, additional number, additional number, additional number, 6, 9, 14. The median is one of the additional numbers between 4 and 6, or 6 if the additional numbers are greater than 6.
If the additional numbers are greater than 9, the order remains 2, 3, 4, 6, 9, additional number, additional number, additional number, 14. The median is 9.

## Step 5: Problem 1
Considering the cases:
- When the additional numbers are less than 4, the median is 6.
- When the numbers are between 4 and 9, the median can be 5, 6, 7, or 8 (since the additional numbers can be distinct and between 4 and 9).
- When the numbers are greater than 9, the median is 9.
So, possible medians are 5, 6, 7, 8, and 9.

## Step 6: Problem 1
Thus, there are 5 possible values for the median: 5, 6, 7, 8, 9.

ANSWER 1: B

## Step 7: Problem 2
To find N, we first need to understand how the mean and median are affected by adding N to the list 2, 6, 7, 7, 28.

## Step 8: Problem 2
The original list is 2, 6, 7, 7, 28. The median is 7. When N is added, the list becomes 2, 6, 7, 7, N, 28. The median will depend on N. If N is between 7 and 7 (which is not possible since N is distinct), or less than 7, the median will be 7. If N is greater than 7, the median will be the average of 7 and N.

## Step 9: Problem 2
Let's consider the case when N is less than or equal to 7. The median remains 7. The mean is (2 + 6 + 7 + 7 + 28 + N)/6 = (50 + N)/6. According to the problem, the mean is twice the median, so (50 + N)/6 = 2*7 = 14.

## Step 10: Problem 2
Solving for N: 50 + N = 14*6 = 84, N = 84 - 50 = 34.

## Step 11: Problem 2
Let's check if N = 34 satisfies the condition for the median. Since 34 is greater than 7, the list becomes 2, 6, 7, 7, 28, 34. The median is (7 + 7)/2 = 7. The mean is (2 + 6 + 7 + 7 + 28 + 34)/6 = 84/6 = 14, which is indeed twice the median.

## Step 12: Problem 2
Thus, N = 34 is a valid solution.

ANSWER 2: E

## Step 13: Problem 3
To find the fraction of four-digit integers with all even digits that are divisible by 4, we first need to determine the total number of such four-digit integers.

## Step 14: Problem 3
For a four-digit number, the first digit can be 2, 4, 6, or 8 (4 choices), and the rest can be 0, 2, 4, 6, or 8 (5 choices each). So, the total number of four-digit numbers with all even digits is 4 * 5 * 5 * 5 = 500.

## Step 15: Problem 3
A number is divisible by 4 if its last two digits form a number divisible by 4. The last two digits can be 00, 04, 08, 20, 24, 28, 40, 44, 48, 60, 64, 68, 80, 84, 88. Out of these, the ones with even digits are 00, 04, 08, 20, 24, 28, 40, 44, 48, 60, 64, 68, 80, 84, 88, which are 20, 24, 28, 40, 44, 48, 60, 64, 68, 80, 84, 88 (12 options since 00, 04, 08 are not valid for the last two digits when considering all possible combinations for the first two digits).

## Step 16: Problem 3
However, we need to correct the understanding of the last two digits. For a number to be divisible by 4, the last two digits must form a number divisible by 4. The possible last two digits are: 00, 04, 08, 20, 24, 28, 40, 44, 48, 60, 64, 68, 80, 84, 88. There are 5 choices for the third digit and 4 choices for the first digit. So, for each valid last two digits, there are 4 * 5 = 20 numbers.

## Step 17: Problem 3
There are 15 valid combinations for the last two digits (00, 04, 08, 20, 24, 28, 40, 44, 48, 60, 64, 68, 80, 84, 88), but we should count them correctly: there are indeed 15 options (since all are valid for the condition of being even and divisible by 4 when considering the last two digits).

## Step 18: Problem 3
The total number of four-digit numbers with all even digits and divisible by 4 is 4 * 5 * 15 = 300 (since there are 15 valid last two digits).

## Step 19: Problem 3
The fraction of these integers divisible by 4 is 300 / 500 = 3/5.

ANSWER 3: D

## Step 20: Problem 4
To find the difference between a 6.5% sales tax and a 6% sales tax on an item priced at $20, we calculate both taxes and then find the difference.

## Step 21: Problem 4
6.5% of $20 = 0.065 * $20 = $1.30.
6% of $20 = 0.06 * $20 = $1.20.

## Step 22: Problem 4
The difference is $1.30 - $1.20 = $0.10.

ANSWER 4: B

## Step 23: Problem 5
To solve (1901 + 1902 + ... + 1993) - (101 + 102 + ... + 193), we can simplify it by calculating the sum of the differences between corresponding terms.

## Step 24: Problem 5
The series can be simplified as: (1901-101) + (1902-102) + ... + (1993-193).

## Step 25: Problem 5
This simplifies to: 1800 + 1800 + 

**Problem 1:**
We have a set \(R\) of nine distinct integers, with six given: 2, 3, 4, 6, 9, 14. The remaining three integers can be any integers not in this list. The median of nine numbers is the 5th smallest. Let the unknown numbers be \(x, y, z\). The six given numbers are fixed; we can place \(x, y, z\) anywhere in the sorted order to change the 5th element.

Let \(k\) be the number of given numbers less than the candidate median \(m\).
- If \(m\) is one of the given numbers, we need to place exactly \(4-k\) of the new numbers below \(m\) and the rest above. This requires \(0 \le 4-k \le 3\), so \(1 \le k \le 4\). Among the given numbers: \(3\ (k=1),\ 4\ (k=2),\ 6\ (k=3),\ 9\ (k=4)\) satisfy this. (\(2\) has \(k=0\), \(14\) has \(k=5\); neither works.) So \(3, 4, 6, 9\) are possible medians.
- If \(m\) is a new integer, one new number equals \(m\) and the other two are placed above or below. We need \(4-k\) new numbers below \(m\), so \(0 \le 4-k \le 2\), or \(2 \le k \le 4\). Also \(m\) must be an integer not already in the list.
  * \(k=2\): \(m\) between 3 and 4 — no integer.
  * \(k=3\): \(m\) between 4 and 6 — only \(5\).
  * \(k=4\): \(m\) between 6 and 9 — \(7\) and \(8\).
So new possible medians: \(5, 7, 8\).

Total possible medians: \(3, 4, 5, 6, 7, 8, 9\) — seven values.

ANSWER 1: D

**Problem 2:**
Original list: \(2, 6, 7, 7, 28\) (sum = 50). Insert \(N\) to get six numbers. Mean = \(\frac{50+N}{6}\). Median of six sorted numbers is the average of the 3rd and 4th.

Case 1: \(N \le 6\). Sorted: …, 3rd = 6, 4th = 7, median = 6.5.  
\(\frac{50+N}{6} = 2 \times 6.5 = 13 \Rightarrow 50+N = 78 \Rightarrow N=28\), but \(28 > 6\). Contradiction.

Case 2: \(6 < N \le 7\) (only integer is 7). Sorted: \(2, 6, 7, 7, 7, 28\), median = 7.  
\(\frac{57}{6} = 9.5 \neq 14\). No.

Case 3: \(7 < N \le 28\). Sorted: \(2, 6, 7, 7, N, 28\), median = 7.  
\(\frac{50+N}{6} = 14 \Rightarrow N = 34\), but \(34 > 28\). Contradiction.

Case 4: \(N > 28\). Sorted: \(2, 6, 7, 7, 28, N\), median = 7.  
\(\frac{50+N}{6} = 14 \Rightarrow N = 34\). Works.

ANSWER 2: E

**Problem 3:**
Four-digit positive integers with all even digits: thousands digit from \(\{2,4,6,8\}\) (4 choices); hundreds, tens, units from \(\{0,2,4,6,8\}\) (5 choices each). Total = \(4 \times 5^3 = 500\).
Divisible by 4 means last two digits form a multiple of 4. For any even tens digit, there are exactly 3 even units digits that make it divisible by 4 (e.g., if tens = 0,4,8 → units 0,4,8; if tens = 2,6 → units 0,4,8). So \(5 \times 3 = 15\) valid last-two-digit combinations.
Number of divisible numbers = \(4 \times 5 \times 15 = 300\).
Fraction = \(\frac{300}{500} = \frac{3}{5}\).

ANSWER 3: D

**Problem 4:**
Tax difference = \(6.5\% - 6\% = 0.5\%\) of \$20 = \(0.005 \times 20 = \$0.10\).

ANSWER 4: B

**Problem 5:**
First sum: \(1901\) to \(1993\) inclusive. Number of terms = \(1993 - 1901 + 1 = 93\).  
Sum = \(93 \times \frac{1901+1993}{2} = 93 \times 1947\).
Second sum: \(101\) to \(193\) inclusive. Terms = \(193 - 101 + 1 = 93\).  
Sum = \(93 \times \frac{101+193}{2} = 93 \times 147\).
Difference = \(93 \times (1947 - 147) = 93 \times 1800 = 167\,400\).

ANSWER 5: A

**Problem 6:**
The expression is \(1 - 2 - 3 + 4 + 5 - 6 - 7 + 8 + 9 - 10 - 11 + \cdots + 1992 + 1993 - 1994 - 1995 + 1996\).
Sign pattern: \(+\) for terms \(n \equiv 0\) or \(1 \pmod 4\); \(-\) for \(n \equiv 2\) or \(3 \pmod 4\).
Group in blocks of four: \((1-2-3+4) + (5-6-7+8) + \cdots + (1993-1994-1995+1996)\). Each block sums to \(0\). There are \(1996/4 = 499\) blocks. Total sum = \(0\).

ANSWER 6: C

**Problem 7:**
Let \(S\) be sunglasses wearers (\(|S|=50\)), \(C\) be cap wearers (\(|C|=35\)). Given \(P(S \mid C) = \frac{|S \cap C|}{35} = \frac{2}{5} \Rightarrow |S \cap C| = 14\).
Then \(P(C \mid S) = \frac{14}{50} = \frac{7}{25}\).

ANSWER 7: B

**Problem 8:**
Three-digit numbers with digits summing to 26. Maximum sum is 27, so digits are close to 9. Let digits be \(a,b,c\) with \(a+b+c=26\), \(1 \le a \le 9\), \(0 \le b,c \le 9\). Setting \(x=9-a, y=9-b, z=9-c\) gives \(x+y+z=1\) with \(x \le 8, y,z \le 9\). The only solutions are permutations of \((1,0,0)\): one digit is 8, the other two are 9. The three numbers are 998, 989, 899.
Even numbers end in an even digit: only 998. Hence 1 even number.

ANSWER 8: A

**Problem 9:**
Three-digit numbers, digit sum = 25. As above, \(x+y+z = 27-25 = 2\). Nonnegative solutions: permutations of \((2,0,0)\) – digits 7,9,9; and \((1,1,0)\) – digits 8,8,9.  
For 7,9,9: 3 permutations. For 8,8,9: 3 permutations. Total 6 numbers.

ANSWER 9: C

**Problem 10:**
Exits 5 through 26: that’s \(26-5+1 = 22\) exits, so 21 gaps. Sum of gaps = 118 miles. Each gap \(\ge 5\). To maximize one gap, minimize the other 20: \(20 \times 5 = 100\) miles. Largest gap = \(118 - 100 = 18\) miles.

ANSWER 10: C

**Problem 11:**
Five scores: \(a \le b \le c \le d \le e\). Mean = 90 ⇒ sum = 450. Median \(c = 91\). Mode = 94, so at least two scores are 94. Since median is 91, the two 94s must be \(d\) and \(e\). Then \(a+b+91+94+94 = 450 \Rightarrow a+b = 450 - 279 = 171\).

ANSWER 11: B

**Problem 12:**
Blanks: \(5 \_ 4 \_ 6 \_ 3\). Use \(+,\ -,\ \times\) exactly once. Evaluate with standard order of operations. All 6 permutations:
1) \(\times, +, -:\ 5\times4+6-3 = 20+6-3=23\)
2) \(\times, -, +:\ 5\times4-6+3 = 20-6+3=17\)
3) \(+, \times, -:\ 5+4\times6-3 = 5+24-3=26\)
4) \(+, -, \times:\ 5+4-6\times3 = 5+4-18=-9\)
5) \(-, \times, +:\ 5-4\times6+3 = 5-24+3=-16\)
6) \(-, +, \times:\ 5-4+6\times3 = 5-4+18=19\)
Among choices, 19 appears.

ANSWER 12: E

Problem 1:
We have a set R of 9 distinct integers. Six are known: 2, 3, 4, 6, 9, 14. We add 3 unknown distinct integers (not equal to any known element). The median of 9 elements is the 5th smallest.

Let me determine which values can occupy the 5th position. The known elements in sorted order are: 2, 3, 4, 6, 9, 14.

**Can the median be 3?** Place all 3 unknowns below 3 (e.g., −1, 0, 1). Sorted: −1, 0, 1, 2, **3**, 4, 6, 9, 14. 5th = 3. ✓
**Can the median be 4?** Place 2 unknowns below 4 (e.g., 0, 1) and 1 above. Sorted: 0, 1, 2, 3, **4**, 6, 9, 14, 20. 5th = 4. ✓
**Can the median be 5?** (5 is an unknown.) Known elements below 5: {2, 3, 4} = 3 elements. Add 1 unknown below 5 (e.g., 1), make 5 an unknown, and 1 unknown above. Sorted: 1, 2, 3, 4, **5**, 6, 9, 14, 20. 5th = 5. ✓
**Can the median be 6?** Known below 6: {2, 3, 4} = 3. Add 1 unknown below 6 (e.g., 5), 2 above. Sorted: 2, 3, 4, 5, **6**, 9, 14, 20, 21. 5th = 6. ✓
**Can the median be 7?** Known below 7: {2, 3, 4, 6} = 4. Need 0 unknowns below 7. Make 7 an unknown, 2 above. Sorted: 2, 3, 4, 6, **7**, 9, 14, 20, 21. 5th = 7. ✓
**Can the median be 8?** Known below 8: {2, 3, 4, 6} = 4. Same logic: 2, 3, 4, 6, **8**, 9, 14, 20, 21. 5th = 8. ✓
**Can the median be 9?** Known below 9: {2, 3, 4, 6} = 4. Place all 3 unknowns above 9. Sorted: 2, 3, 4, 6, **9**, 14, 20, 21, 22. 5th = 9. ✓

**Can the median be ≤ 2?** Need 4 elements below it, but at most 3 unknowns exist → impossible.
**Can the median be ≥ 10?** Known elements below 10 are {2, 3, 4, 6, 9} = 5 elements, so the 5th position is already filled by something ≤ 9 → impossible.

Possible medians: {3, 4, 5, 6, 7, 8, 9} → **7 values**.

ANSWER 1: D

---

Problem 2:
Original list: 2, 6, 7, 7, 28. Sum = 50. Insert N to get 6 elements.
New mean = (50 + N)/6. Median of 6 elements = average of 3rd and 4th.

**Case N > 7:** Sorted order: 2, 6, 7, 7, N, 28 (or N, 28 swapped). The 3rd and 4th elements are 7 and 7.
Median = (7 + 7)/2 = 7.
Mean = 2 × median = 14.
(50 + N)/6 = 14 → 50 + N = 84 → N = 34.
Check: 34 > 7 ✓. List: 2, 6, 7, 7, 28, 34. Median = 7, Mean = 84/6 = 14 = 2(7). ✓

**Case N ≤ 6:** Median = (6+7)/2 = 6.5. Mean = 13. N = 78 − 50 = 28. But 28 > 6, contradiction.
**Case 6 < N ≤ 7:** Median = (N+7)/2. (50+N)/6 = N+7 → 50+N = 6N+42 → 5N = 8 → N = 1.6. Contradiction.

Only N = 34 works.

ANSWER 2: E

---

Problem 3:
Even digits: {0, 2, 4, 6, 8}. Four-digit numbers with all even digits:
- First digit: {2, 4, 6, 8} → 4 choices
- Other three digits: 5 choices each
- Total = 4 × 5 × 5 × 5 = 500

Divisibility by 4 depends on the last two digits. I check all 5 × 5 = 25 pairs of even digits:
For each tens digit (0, 2, 4, 6, 8), the units digits giving divisibility by 4 are:
- Tens 0: 00, 04, 08 (3 of 5)
- Tens 2: 20, 24, 28 (3 of 5)
- Tens 4: 40, 44, 48 (3 of 5)
- Tens 6: 60, 64, 68 (3 of 5)
- Tens 8: 80, 84, 88 (3 of 5)

Valid last-two-digit pairs = 15 out of 25.

Numbers divisible by 4 = 4 × 5 × 15 = 300.
Fraction = 300/500 = **3/5**.

ANSWER 3: D

---

Problem 4:
Difference = 6.5% of $20 − 6% of $20 = (6.5% − 6%) × $20 = 0.5% × $20.
0.5% = 0.005, so 0.005 × 20 = $0.10.

ANSWER 4: B

---

Problem 5:
Pair corresponding terms:
(1901 − 101) + (1902 − 102) + … + (1993 − 193)

Each pair equals 1800.
Number of terms: 1993 − 1901 + 1 = 93 terms (and 193 − 101 + 1 = 93 terms).
Total = 93 × 1800 = **167,400**.

ANSWER 5: A

---

Problem 6:
The sign pattern is: +, −, −, +, +, −, −, +, …
Group into blocks of 4:
(1 − 2 − 3 + 4) + (5 − 6 − 7 + 8) + (9 − 10 − 11 + 12) + …

Each block: n − (n+1) − (n+2) + (n+3) = n − n − 1 − n − 2 + n + 3 = **0**.

Total terms: 1996. Number of blocks: 1996/4 = 499 complete blocks.
Sum = 499 × 0 = **0**.

ANSWER 6: C

---

Problem 7:
Let B = number wearing both sunglasses and caps.
P(sunglasses | cap) = B / (people wearing caps) = B / 35 = 2/5.
B = 35 × (2/5) = 14.

Now find P(cap | sunglasses) = B / (people wearing sunglasses) = 14 / 50 = **7/25**.

ANSWER 7: B

---

Problem 8:
Three-digit numbers with digit sum 26. Maximum digit sum = 9+9+9 = 27, so we must subtract 1 from one digit of 999:
Possible digit sets: (8,9,9), (9,8,9), (9,9,8) → numbers 899, 989, 998 (and permutations: 899, 989, 997... wait).

Actually the numbers are: 899, 989, 998 — no, let me list properly:
- (8,9,9) → 899, 989, 998... wait, (8,9,9) gives 899, 989, 998? No: digits are 8,9,9 so numbers are 899, 989, 998.
Wait: (8,9,9) → 899, 989, 998. But 998 has digits 9,9,8 which is the set (9,9,8). Let me just list all:
- Digits {8,9,9}: 899, 989, 998 — wait, 998 has digits 9,9,8. That IS {8,9,9}. So: 899, 989, 998.

Hmm, let me redo. The three numbers with digit sum 26 are:
- 899 (8+9+9=26)
- 989 (9+8+9=26)
- 998 (9+9+8=26)

Wait, that's only 3 numbers total (since two digits are the same, there are 3!/2! = 3 arrangements).

Even numbers must end in an even digit. Only 998 ends in 8 (even). 899 and 989 end in 9 (odd).

So exactly **1** even number.

ANSWER 8: A

---

Problem 9:
Three-digit numbers a, b, c with a + b + c = 25, where 1 ≤ a ≤ 9, 0 ≤ b, c ≤ 9.

Enumerate by first digit a:
- a = 9: b + c = 16 → (7,9), (8,8), (9,7) → 979, 988, 997 (3 numbers)
- a = 8: b + c = 17 → (8,9), (9,8) → 889, 898 (2 numbers)
- a = 7: b + c = 18 → (9,9) → 799 (1 number)
- a ≤ 6: b + c ≥ 19, but max b + c = 18 → impossible

Total = 3 + 2 + 1 = **6**.

ANSWER 9: C

---

Problem 10:
Exits 5 through 26 give 22 exits and 21 consecutive gaps. Total distance = 118 miles. Each gap ≥ 5 miles.

To maximize one gap, minimize the other 20 gaps to 5 miles each:
Max gap + 20(5) = 118
Max gap + 100 = 118
Max gap = **18** miles.

ANSWER 10: C

---

Problem 11:
Five scores sorted: a ≤ b ≤ c ≤ d ≤ e.
- Mean = 90 → sum = 450
- Median = 91 → c = 91
- Mode = 94 → 94 appears most frequently (at least twice)

Since a ≤ b ≤ c = 91, neither a nor b can be 94. So 94 must appear among d and e. For 94 to be the mode, we need d = e = 94 (appearing twice).

Sum: a + b + 91 + 94 + 94 = 450
a + b = 450 − 279 = **171**

(We can verify valid scores 

 Problem 1:
We are given a set \(R\) of nine distinct integers, six of which are \(2,3,4,6,9,14\). We must determine how many different values are possible for the median of \(R\).

Because there are nine distinct integers, the median is the fifth smallest element when the set is listed in increasing order. Let the three unknown distinct integers be chosen freely (as long as they are not equal to any of the six given numbers). We examine two cases: when the median is one of the six known numbers and when it is one of the three unknown numbers.

**Case 1: The median is a known number.**
Let the median be \(m\). Of the six known numbers, let \(L\) be how many are less than \(m\). Then there are \(5-L\) known numbers greater than \(m\). To make \(m\) the fifth element, we need exactly four numbers smaller than \(m\) and four larger. Thus we need \(4-L\) unknown numbers below \(m\) and \(4-(5-L)=L-1\) unknown numbers above \(m\). This uses exactly \((4-L)+(L-1)=3\) unknowns, which is perfect. We need \(4-L\ge 0\) and \(L-1\ge 0\), i.e. \(1\le L\le 4\).

- \(m=2\): \(L=0\) → impossible.
- \(m=3\): \(L=1\) → need 3 unknowns \(<3\), 0 above. Possible (e.g. \(-1,0,1\)).
- \(m=4\): \(L=2\) → need 2 unknowns \(<4\), 1 above. Possible (e.g. \(0,1,5\)).
- \(m=6\): \(L=3\) → need 1 unknown \(<6\), 2 above. Possible (e.g. \(5,7,8\)).
- \(m=9\): \(L=4\) → need 0 unknowns \(<9\), 3 above. Possible (e.g. \(10,11,12\)).
- \(m=14\): \(L=5\) → need \(-1\) unknowns below; impossible.

So the known possible medians are \(3,4,6,9\).

**Case 2: The median is an unknown number \(x\).**
Then \(x\) itself uses one of the three unknown slots. We need four numbers below \(x\) and four above. With \(L\) known numbers below \(x\), we need \(4-L\) other unknowns below \(x\) and \(4-(6-L)=L-2\) other unknowns above \(x\). The total other unknowns used is \((4-L)+(L-2)=2\), which matches the two remaining unknowns. We need \(4-L\ge 0\) and \(L-2\ge 0\), so \(2\le L\le 4\).

- \(L=2\): known below are \(2,3\); known above are \(4,6,9,14\). Then \(3<x<4\), impossible for an integer.
- \(L=3\): known below are \(2,3,4\); known above are \(6,9,14\). Then \(4<x<6\), so \(x=5\). This works (e.g. add \(1,5,7\)).
- \(L=4\): known below are \(2,3,4,6\); known above are \(9,14\). Then \(6<x<9\), so \(x=7\) or \(x=8\). Both work (e.g. add \(7,10,11\) or \(8,10,11\)).

Thus the unknown possible medians are \(5,7,8\).

Collecting all possibilities: \(\{3,4,5,6,7,8,9\}\). That is **7** different values. The tempting answer of 4 counts only the known medians, and 2 or 14 are impossible because too many numbers would have to be forced to one side.

ANSWER 1: D

---

Problem 2:
We insert a number \(N\) into the list \(2,6,7,7,28\) so that the new mean of the six numbers is twice the new median. Find \(N\).

The original sum is \(2+6+7+7+28=50\). After insertion, the sum is \(50+N\) and the mean is \(\frac{50+N}{6}\).

The median depends on the size of \(N\) relative to the ordered list. Because the list contains six numbers, the median is the average of the third and fourth values.

- If \(N\le 6\) (as an integer), the ordered list begins with \(N\) or \(2\), then the third and fourth terms are \(6\) and \(7\), giving median \(\frac{6+7}{2}=6.5\).
  Setting mean \(=2\times 6.5=13\): \(\frac{50+N}{6}=13\Rightarrow N=28\). But \(N=28\) does not satisfy \(N\le 6\), so this case yields no valid solution.

- If \(N\ge 7\) (integer), the third and fourth terms are both \(7\) (since the two \(7\)’s occupy those middle slots, or \(N=7\) makes three \(7\)’s), so the median is \(7\).
  Setting mean \(=2\times 7=14\): \(\frac{50+N}{6}=14\Rightarrow 50+N=84\Rightarrow N=34\).

Check: the list becomes \(2,6,7,7,28,34\). The median is \(\frac{7+7}{2}=7\). The mean is \(\frac{84}{6}=14\), which is indeed twice the median. The other choices fail: \(N=28\) gives mean \(13\) and median \(7\); \(N=14\) gives mean \(\frac{64}{6}\neq 14\).

ANSWER 2: E

---

Problem 3:
We consider all positive four-digit integers whose digits are all even. We need the fraction of them that are divisible by \(4\).

Even digits are \(\{0,2,4,6,8\}\). For a four-digit number \(ABCD\):
- \(A\) (thousands) has 4 choices: \(\{2,4,6,8\}\).
- \(B,C,D\) each have 5 choices.

Total count: \(4\cdot 5\cdot 5\cdot 5 = 500\).

A number is divisible by \(4\) exactly when its last two digits form a multiple of \(4\). Here \(C\) and \(D\) are both even. Consider \(10C+D \pmod 4\). Since \(C\) is even, write \(C=2k\); then \(10C=20k\equiv 0\pmod 4\). Thus \(10C+D\equiv D\pmod 4\). So we simply need \(D\equiv 0\pmod 4\).

Among the even digits, those divisible by \(4\) are \(0,4,8\): **3 choices** for \(D\). The digits \(A,B,C\) are unrestricted (4, 5, and 5 choices).

Count of favorable numbers: \(4\cdot 5\cdot 5\cdot 3 = 300\).

The fraction is \(\frac{300}{500}=\frac{3}{5}\). One might mistakenly think the fraction is \(\frac12\) because half of all integers are divisible by \(4\), but the restriction to even digits changes the units-digit distribution.

ANSWER 3: D

---

Problem 4:
Find the difference between a \(6.5\%\) sales tax and a \(6\%\) sales tax on a \(\$20\) item.

Method 1: Compute each tax.
- \(6.5\%\) of \(\$20 = 0.065\times 20 = \$1.30\).
- \(6\%\) of \(\$20 = 0.06\times 20 = \$1.20\).

Difference: \(\$1.30-\$1.20 = \$0.10\).

Method 2: The difference in rates is \(0.5\%\), and \(0.5\%\) of \(\$20 = 0.005\times 20 = \$0.10\).

The distractors correspond to misplacing the decimal (e.g. \(\$1\) would be \(5\%\) of \(\$20\)).

ANSWER 4: B

---

Problem 5:
Evaluate \((1901+1902+\dots+1993)-(101+102+\dots+193)\).

Each sum has the same number of terms:
\(1993-1901+1 = 93\) terms, and \(193-101+1 = 93\) terms.

Pair the terms:
\((1901-101)+(1902-102)+\dots+(1993-193)\).

Each difference equals \(1800\). With 93 such pairs, the total is
\(93\times 1800 = 93\times 18\times 100 = 1674\times 100 = 167{,}400\).

A common error is miscounting the number of terms as 92, which would give \

Problem 1:
R is a set of nine distinct integers with six known elements: 2, 3, 4, 6, 9, 14. Since R has 9 elements, the median is the 5th element when sorted.

I need to determine which integers can be the 5th element. The median must have exactly 4 elements less than it and 4 elements greater than it.

**Checking each candidate:**

- **Median = 3**: Need 4 elements < 3. Known: {2}. Need 3 unknowns all < 3 (e.g., 1, 0, -1). Elements > 3: {4, 6, 9, 14} = 4. ✓
- **Median = 4**: Need 4 elements < 4. Known: {2, 3}. Need 2 unknowns < 4. Elements > 4: {6, 9, 14} + 1 unknown > 4. ✓
- **Median = 5**: Need 4 elements < 5. Known: {2, 3, 4}. Need 1 unknown < 5. Elements > 5: {6, 9, 14} + 1 unknown > 5. (5 is an unknown) ✓
- **Median = 6**: Need 4 elements < 6. Known: {2, 3, 4}. Need 1 unknown < 6. Elements > 6: {9, 14} + 2 unknowns > 6. ✓
- **Median = 7**: Need 4 elements < 7. Known: {2, 3, 4, 6} = 4. Elements > 7: {9, 14} + 2 unknowns > 7. (7 is an unknown) ✓
- **Median = 8**: Need 4 elements < 8. Known: {2, 3, 4, 6} = 4. Elements > 8: {9, 14} + 2 unknowns > 8. (8 is an unknown) ✓
- **Median = 9**: Need 4 elements < 9. Known: {2, 3, 4, 6} = 4. Elements > 9: {14} + 3 unknowns > 9. ✓

**Values that cannot be median:**
- 2: Need 4 elements < 2, but only 3 unknowns available. ✗
- 10-13: Already have 5 known elements less than these. ✗
- 14: Already have 5 known elements less than 14. ✗

Possible median values: {3, 4, 5, 6, 7, 8, 9} → **7 values**

ANSWER 1: D

---

Problem 2:
List: 2, 6, 7, 7, 28. Insert N so that mean = 2 × median.

Sum of original list = 50, so new mean = (50 + N)/6.

**Finding the median based on N:**
- If N > 7: Sorted list is 2, 6, 7, 7, 28, N (or 2, 6, 7, 7, N, 28 if 7 < N < 28). Median = (7+7)/2 = 7.
- If N ≤ 6: Median = (6+7)/2 = 6.5
- If 6 < N ≤ 7: Median = (N+7)/2

**Solving for N > 7:**
(50 + N)/6 = 2(7) = 14
50 + N = 84
N = 34

Verification: List = 2, 6, 7, 7, 28, 34. Mean = 84/6 = 14. Median = 7. ✓

ANSWER 2: E

---

Problem 3:
Four-digit integers with all even digits. Even digits: {0, 2, 4, 6, 8}.

**Total count:** First digit: 4 choices (2,4,6,8), others: 5 choices each → 4 × 5³ = 500

**Divisible by 4:** Last two digits must form a number divisible by 4.

Listing all 25 pairs of even last digits and checking divisibility by 4:
- 00✓, 02✗, 04✓, 06✗, 08✓
- 20✓, 22✗, 24✓, 26✗, 28✓
- 40✓, 42✗, 44✓, 46✗, 48✓
- 60✓, 62✗, 64✓, 66✗, 68✓
- 80✓, 82✗, 84✓, 86✗, 88✓

That's 15 valid pairs.

**Count divisible by 4:** 4 × 5 × 15 = 300

**Fraction:** 300/500 = 3/5

ANSWER 3: D

---

Problem 4:
6.5% tax on $20 = 0.065 × 20 = $1.30
6% tax on $20 = 0.06 × 20 = $1.20

Difference = $1.30 - $1.20 = $0.10

ANSWER 4: B

---

Problem 5:
(1901 + 1902 + ... + 1993) − (101 + 102 + ... + 193)

Both sums have 93 terms each.

First sum = 93 × (1901 + 1993)/2 = 93 × 1947
Second sum = 93 × (101 + 193)/2 = 93 × 147

Difference = 93 × (1947 − 147) = 93 × 1800 = 167,400

ANSWER 5: A

---

Problem 6:
1 − 2 − 3 + 4 + 5 − 6 − 7 + 8 + ... + 1993 − 1994 − 1995 + 1996

The pattern repeats every 4 terms: +, −, −, +

Group k: (4k−3) − (4k−2) − (4k−1) + 4k = 4k−3−4k+2−4k+1+4k = 0

Each group of 4 sums to 0.

1996 ÷ 4 = 499 complete groups.

Total = 499 × 0 = 0

ANSWER 6: C

---

Problem 7:
50 people wear sunglasses, 35 wear caps.

P(sunglasses | cap) = 2/5 means: (both)/(caps) = 2/5

Both/35 = 2/5 → Both = 14

P(cap | sunglasses) = (both)/(sunglasses) = 14/50 = 7/25

ANSWER 7: B

---

Problem 8:
3-digit numbers with digit-sum 26 that are even.

For number abc: a + b + c = 26, c is even, a ≥ 1, all digits ≤ 9.

- c = 8: a + b = 18 → only (9,9) → 998 ✓
- c = 6: a + b = 20 → impossible (max 9+9=18)
- c = 4, 2, 0: a + b ≥ 22 → impossible

Only 1 number: 998

ANSWER 8: A

---

Problem 9:
Three-digit numbers with digit-sum 25.

a + b + c = 25, a ≥ 1, all digits ≤ 9.

- a = 9: b + c = 16 → (7,9), (8,8), (9,7) → 3 numbers
- a = 8: b + c = 17 → (8,9), (9,8) → 2 numbers
- a = 7: b + c = 18 → (9,9) → 1 number
- a ≤ 6: b + c ≥ 19 → impossible

Total = 3 + 2 + 1 = 6

ANSWER 9: C

---

Problem 10:
Distance from exit 5 to exit 26 is 118 miles. There are 21 gaps between consecutive exits (5→6, 6→7, ..., 25→26).

Each gap ≥ 5 miles. To maximize one gap, minimize all others.

If 20 gaps are minimized at 5 miles each: 20 × 5 = 100 miles.

Maximum single gap = 118 − 100 = 18 miles.

ANSWER 10: C

---

Problem 11:
Five scores: mean = 90, median = 91, mode = 94.

Sum = 5 × 90 = 450

Sorted: a ≤ b ≤ c ≤ d ≤ e
- c = 91 (median)
- Mode = 94, so 94 appears at least twice. Since c = 91, the 94s must be d and e.

a + b + 91 + 94 + 94 = 450
a + b = 171

ANSWER 11: B

---

Problem 12:
Using +, −, × each exactly once in 5 __ 4 __ 6 __ 3:

| Operations | Result |
|------------|--------|
| 5 + 4 − 6 × 3 | 5 + 4 − 18 = −9 |
| 5 + 4 × 6 − 3 | 5 + 24 − 3 = 26 |
| 5 − 4 + 6 × 3 | 5 − 4 + 18 = 19 |
| 5 − 4 × 6 + 3 | 5 − 24 + 3 = −16 |
| 5 × 4 + 6 − 3 | 20 + 6 − 3 = 23 |
| 5 × 4 − 6 + 3 | 20 − 6 + 3 = 17 |

From choices {9, 10, 15, 16, 19}, only **19** is achievable.

ANSWER 12: E

**Problem 1:**  
We have nine distinct integers, six of which are known: 2, 3, 4, 6, 9, 14.  
The median of a nine‑element set is the 5th smallest number.  

Let the unknown integers be \(x,y,z\) (all distinct and distinct from the known six).  
When the numbers are sorted, the median is the 5th term.  

For a known number \(k_i\) (the \(i\)‑th smallest of the known six) to be the median, we need exactly four numbers smaller than it.  
If \(i-1\) known numbers are smaller, we need \(L\) unknown numbers smaller with  
\((i-1)+L=4\Rightarrow L=5-i\).  
Since we have only three unknowns, \(L\ge0\) gives \(i\le5\); also \(L\le3\) gives \(i\ge2\).  
Thus \(i=2,3,4,5\) → medians could be 3, 4, 6, 9.

Can the median be an unknown integer?  
Let the median be an unknown \(u\).  
If \(u\) lies between the known numbers, we need exactly four numbers below it.  
With two other unknowns, the possibilities are:
- \(u\) between 4 and 6 → \(u=5\);
- \(u\) between 6 and 9 → \(u=7\) or \(u=8\).

All three of these can be realized while keeping the numbers distinct.  
Thus the possible median values are \(\{3,4,5,6,7,8,9\}\), a total of **7** values.

**Answer:** 7 → Choice **D**.  

ANSWER 1: D  

---

**Problem 2:**  
We start with the list 2, 6, 7, 7, 28 and insert a number \(N\).  
Let the six numbers be sorted. The median of six numbers is the average of the 3rd and 4th smallest values.

Let the sorted list after insertion be \(a_1\le a_2\le a_3\le a_4\le a_5\le a_6\).  
The median is \(\dfrac{a_3+a_4}{2}\).  
The mean is \(\dfrac{\text{sum}}{6}\).  
Given that the mean equals twice the median:
\[
\frac{\text{sum}}{6}=2\cdot\frac{a_3+a_4}{2}=a_3+a_4 .
\]
Hence \(\text{sum}=6(a_3+a_4)\).

Consider where \(N\) can fall:

*If \(N\le2\):* \(a_3=6,\ a_4=7\) → median \(=6.5\).  
Sum \(=N+50\). Equation gives \(N=28\), contradicting \(N\le2\).

*If \(2<N\le6\):* \(a_3=6,\ a_4=7\) → same median \(=6.5\).  
Again \(N=28\), impossible.

*If \(6<N\le7\):* only integer possibility \(N=7\).  
List becomes 2, 6, 7, 7, 7, 28 → median \(=7\).  
Sum \(=57\) → mean \(=9.5\neq14\).

*If \(7<N<28\):* \(a_3=7,\ a_4=7\) → median \(=7\).  
Sum \(=N+50\). Equation \(N+50=84\) gives \(N=34\), which is not \(<28\).

*If \(N=28\):* median still \(=7\). Sum \(=78\) → mean \(=13\neq14\).

*If \(N>28\):* \(a_3=7,\ a_4=7\) → median \(=7\).  
Equation gives \(N+50=84\) → \(N=34\). This works (34 > 28).

Thus the only possible value is **34**.

**Answer:** 34 → Choice **E**.  

ANSWER 2: E  

---

**Problem 3:**  
A four‑digit integer with all digits even can use digits \{0,2,4,6,8\} for each place, except the thousands digit cannot be 0.  

Number of such integers: \(4\times5\times5\times5=500\).

A number is divisible by 4 iff its last two digits form a number divisible by 4.  
List all two‑digit endings with even digits (25 possibilities).  

Checking each yields 15 endings that are multiples of 4:  
00, 04, 08, 20, 24, 28, 40, 44, 48, 60, 64, 68, 80, 84, 88.

Thus the proportion is \(\frac{15}{25}=\frac{3}{5}\).

**Answer:** \(\frac{3}{5}\) → Choice **D**.  

ANSWER 3: D  

---

**Problem 4:**  
6 % tax on \$20 = \(0.06\times20 = \$1.20\).  
6.5 % tax on \$20 = \(0.065\times20 = \$1.30\).  

Difference = \$1.30 − \$1.20 = **\$0.10**.

**Answer:** \$0.10 → Choice **B**.  

ANSWER 4: B  

---

**Problem 5:**  
Sum from 1901 to 1993 (inclusive):  
Number of terms = \(1993-1901+1 = 93\).  
Sum = \(\frac{(1901+1993)\times93}{2}= \frac{3894\times93}{2}=181\,071\).

Sum from 101 to 193 (inclusive):  
Number of terms = \(193-101+1 = 93\).  
Sum = \(\frac{(101+193)\times93}{2}= \frac{294\times93}{2}=13\,671\).

Difference = \(181\,071-13\,671 = 167\,400\).

**Answer:** 167 400 → Choice **A**.  

ANSWER 5: A  

---

**Problem 6:**  
The sign pattern for the numbers \(n\) is:
- \(n\equiv1\pmod4\) → +,
- \(n\equiv2\pmod4\) → −,
- \(n\equiv3\pmod4\) → −,
- \(n\equiv0\pmod4\) → +.

Thus each block of four consecutive integers \(\{4k+1,4k+2,4k+3,4k+4\}\) contributes  
\((4k+1)-(4k+2)-(4k+3)+(4k+4)=0\).

From 1 through 1996 there are \(1996/4=499\) complete blocks, so the total sum is **0**.

**Answer:** 0 → Choice **C**.  

ANSWER 6: C  

---

**Problem 7:**  
Let \(S\) = people wearing sunglasses, \(|S|=50\).  
Let \(C\) = people wearing caps, \(|C|=35\).  
Let \(x=|S\cap C|\).

Given \(P(S\mid C)=\frac{x}{35}= \frac{2}{5}\).  
Thus \(x = \frac{2}{5}\times35 = 14\).

We need \(P(C\mid S)=\frac{x}{50}= \frac{14}{50}= \frac{7}{25}\).

**Answer:** \(\frac{7}{25}\) → Choice **B**.  

ANSWER 7: B  

---

**Problem 8:**  
A three‑digit number with digit sum 26 must use digits that sum to 26.  
The only triple of single‑digit numbers with sum 26 is \(\{9,9,8\}\).

The number is even only if its units digit is 8.  
Among the permutations of 9, 9, 8, only 998 ends in 8.  

Thus exactly **1** such number exists.

**Answer:** 1 → Choice **A**.  

ANSWER 8: A  

---

**Problem 9:**  
Let the digits be \(a,b,c\) (hundreds, tens, units) with  
\(1\le a\le9,\ 0\le b,c\le9,\ a+b+c=25\).

Set \(a'=a-1\) (so \(0\le a'\le8\)). Then  
\(a'+b+c=24\).

Count unrestricted non‑negative solutions: \(\binom{24+3-1}{2}= \binom{26}{2}=325\).

Subtract solutions where a digit exceeds its bound:
- \(a'\ge9\): treat \(a''=a'-9\) → \(a''+b+c=15\) → \(\binom{17}{2}=136\).
- \(b\ge10\): \(b'=b-10\) → \(a'+b'+c=14\) → \(\binom{16}{2}=120\).
- \(c\ge10\): similarly 120.

Intersections:
- \(a'\ge9,\ b\ge10\): \(a''+b'+c=5\) → \(\binom{7}{2}=21\).
- \(a'\ge9,\ c\ge10\): 21.
- \(b\ge10,\ c\ge10\): \(a'+b'+c'=4\) → \(\binom{6}{2}=15\).

No triple intersection (would require sum –5).

By inclusion–exclusion, invalid = \(136+120+120-21-21-15 = 319\).  
Valid = \(325-319 = 6\).

**Answer:** 6 → Choice **C**.  

ANSWER 9: C  

---

**Problem 10:**  
Exits numbered 5 through 26 give 22 exits, hence 21 intervals.  
Let the distances between consecutive exits be \(d_1,\dots,d_{21}\).  
We know \(\sum d_i = 118\) miles, and each \(d_i\ge5\).

To maximize the largest interval, make the othe

Problem 1:  
We are given a set $ R $ of nine distinct integers, six of which are $ 2, 3, 4, 6, 9, 14 $. The median of a set of nine numbers is the fifth number when sorted. To find the number of possible values for the median, we consider the possible placements of the three unknown numbers relative to the known ones. The known numbers sorted are $ 2, 3, 4, 6, 9, 14 $. The three unknown numbers can be placed in various positions to shift the median. The possible medians occur when the additional numbers are placed such that the fifth number in the sorted list can be $ 4, 5, 6, 7, 8, $ or $ 9 $. This gives six possible values for the median.  
ANSWER 1: D (corrected to C? Wait, no: the possible medians are 4, 5, 6, 7, 8, 9 — that’s 6 values, so C)  
**Wait, let's re-evaluate:** The known numbers are 2,3,4,6,9,14. Adding three numbers. The median is the 5th number. To get 4 as median: three numbers ≤4, but only 2,3,4 are available — but they are distinct, so we can add numbers like 1,5? Wait no, the three unknown numbers can be placed in different positions. For example:  
- If all three unknown numbers are ≤4: but only three such numbers exist (2,3,4) but they are already in the set. So we cannot add more. Thus, the unknown numbers must be placed in other positions.  
Actually, the unknown numbers can be:  
- All less than 2: median could be 4 if three numbers are added <4, but only 2,3,4 are already present. We cannot add numbers less than 2 without repeating? Wait, the set has distinct integers. So we can add numbers like 1,0,-1, but the problem doesn’t restrict to positive. However, the known numbers are positive. But the problem says "distinct integers", so negatives are allowed. But the median values: if we add numbers less than 2, the sorted order becomes [new numbers, 2,3,4,6,9,14]. With three new numbers, the sorted list has 9 numbers. The fifth is the median. If we add three numbers less than 2, the sorted list is [a,b,c,2,3,4,6,9,14] — no, wait, adding three numbers less than 2: the sorted list would be [a,b,c,2,3,4,6,9,14] — but a,b,c are less than 2, so the first three, then 2 is fourth, 3 is fifth? Wait, no: the original known numbers are 2,3,4,6,9,14. Adding three numbers, say x,y,z. The full sorted list will be the combination. To get median=4, we need the fifth number to be 4. So in the sorted list of 9, the fifth must be 4. That requires that four numbers are ≤4 and four are ≥4? No: the fifth is the middle. For the fifth to be 4, we need at least five numbers ≤4? No: in a sorted list of 9, the fifth is the median. For the fifth to be 4, we need at least five numbers ≤4? Actually, no: the fifth number is the one where four are less and four are greater? No: in a sorted list of 9, the fifth is the median, meaning four are ≤ median and four are ≥ median. But since numbers can be equal, but here all distinct. So for 4 to be the median, we need exactly four numbers less than 4 and four numbers greater than 4? But 4 is in the set. So the sorted list must have four numbers <4, then 4, then four numbers >4. But the known numbers include 2,3,4,6,9,14. So 2,3 are <4, and 6,9,14 are >4. So currently, without unknowns, we have two <4 and three >4. Adding three unknowns. To have four <4 and four >4, we need to add two numbers <4 and one number >4? But then the sorted list would be [two new <4, 2,3,4, then the new >4, then 6,9,14] — but that’s only three numbers after 4? Wait, no: the sorted list would be [new <4 (two), 2,3,4, new >4, 6,9,14] — that’s 9 numbers? 2+3+1+3=9? Wait, no: the two new <4, then 2,3,4 (three numbers), then the new >4, then 6,9,14 (three numbers). So total: 2+3+1+3=9. The fifth number is 4. So yes, we can achieve median=4 by adding two numbers <4 and one number >4. Similarly, we can achieve:  
- Median=5: by adding numbers such that the fifth number is 5. But 5 is not in the set. So we need the fifth number to be 5. This requires that four numbers are <5 and four are >5. But 5 is not present. So we need to add 5? But the set must consist of distinct integers, and 5 is not in the given. So we can add 5. Then the sorted list would include 5. For example, add 5, and two other numbers. To have four <5 and four >5: currently, without unknowns, numbers <5: 2,3,4 (three), numbers >5: 6,9,14 (three). So to have four <5, we need to add one number <5. To have four >5, we need to add one number >5. But we are adding three numbers. So we can add 5, and say 1 and 10. Then sorted: 1,2,3,4,5,6,9,10,14. Median is 5. So median=5 is achievable. Similarly, median=6: we can add numbers such that the fifth is 6. Currently, numbers <6: 2,3,4 (three), numbers >6: 9,14 (two). So to have four <6 and four >6, we need to add one number <6 and two numbers >6. For example, add 5,10,15. Then sorted: 2,3,4,5,6,9,10,14,15. Median=6. Similarly, median=7: add numbers so fifth is 7. Currently, <7: 2,3,4,6 (four), >7: 9,14 (two). So to have four <7 and four >7, we need to add zero <7 (since already four) and two >7. But we are adding three numbers. So we can add 7,8,15. Then sorted: 2,3,4,6,7,8,9,14,15. Median=7. Similarly, median=8: add 8, and two numbers, say 10,11. Then sorted: 2,3,4,6,8,9,10,11,14. Median=8? The fifth is 8. But to have four <8 and four >8: currently, without unknowns, <8: 2,3,4,6 (four), >8:9,14 (two). So to have four >8, we need to add two numbers >8. But we are adding three numbers. So we can add 8,9,10? But 9 is already there. Add 8,10,11. Then sorted: 2,3,4,6,8,9,10,11,14. The fifth is 8. So yes. Similarly, median=9: add numbers so fifth is 9. Currently, <9: 2,3,4,6 (four), >9:14 (one). So to have four >9, we need to add three numbers >9. But we are adding three numbers. So add 10,11,12. Then sorted: 2,3,4,6,9,10,11,12,14. Median=9. So possible medians: 4,5,6,7,8,9. That’s six values.  
ANSWER 1: C  

Problem 2:  
We are given the list 2,6,7,7,28. A number $ N $ is added. The mean becomes twice the median. The original list has five numbers. After adding $ N $, there are 

### Problem 1:
We have a set of 9 distinct integers, so the median (middle value when sorted) is the 5th element. The 6 known numbers sorted are [2,3,4,6,9,14], with 3 additional distinct integers to add. To find possible median values:
- The minimum possible median is 3: add 3 numbers smaller than 2, sorted set becomes [a,b,c,2,3,4,6,9,14], 5th term=3.
- The maximum possible median is 9: add 3 numbers larger than 14, sorted set becomes [2,3,4,6,9,14,x,y,z], 5th term=9.
- All integers between 3 and 9 are valid medians: 3,4,5,6,7,8,9, for 7 total possible values.
Tempting wrong choices: any count less than 7, from missing intermediate medians that can be formed by adding integers between the known values.
ANSWER 1: D

---

### Problem 2:
Original list sum: 2+6+7+7+28=50. Adding N gives a list of 6 elements, sum=50+N, mean=(50+N)/6. The median of 6 elements is the average of the 3rd and 4th sorted terms. The problem states mean=2×median, so 50+N=12×median.
- If N>28, sorted list is [2,6,7,7,28,N], median=(7+7)/2=7. Then 50+N=12×7=84 → N=34, which fits.
Testing other options confirms no other N satisfies the condition.
ANSWER 2: E

---

### Problem 3:
A number is divisible by 4 if its last two digits form a number divisible by 4. For 4-digit numbers with all even digits:
- Total count: 4 (non-zero thousands digits:2,4,6,8) ×5×5×5 (all even digits for other places)=500.
- Valid last two-digit pairs (both even, divisible by 4): For any even tens digit, 10×even is divisible by 4, so only units digits 0,4,8 work, giving 3 valid units digits per tens digit: 5×3=15 valid pairs.
- Fraction: (4×5×15)/500=300/500=3/5.
ANSWER 3: D

---

### Problem 4:
The tax rate difference is 6.5%-6%=0.5%. 0.5% of $20 is 0.005×20=$0.10.
ANSWER 4: B

---

### Problem 5:
Both sums have the same number of terms: 1993-1901+1=93 terms, 193-101+1=93 terms. Subtract term-by-term: (1901-101)+(1902-102)+...+(1993-193)=1800×93=167400.
ANSWER 5: A

---

### Problem 6:
Group terms in sets of 4: (1-2-3+4)+(5-6-7+8)+... each group sums to 0. There are 1996/4=499 groups, total sum=0.
ANSWER 6: C

---

### Problem 7:
Number of people with both caps and sunglasses: 35 cap-wearers × 2/5=14. Probability a sunglass-wearer has a cap:14/50=7/25.
ANSWER 7: B

---

### Problem 8:
An even 3-digit number with digit sum 26 must have an even units digit. The only possible even units digit is 8 (any smaller even digit leaves a sum of the first two digits >18, impossible). The only number is 998, so 1 valid number.
ANSWER 8: A

---

### Problem 9:
Three-digit numbers with digit sum 25 are all permutations of (9,9,7) and (9,8,8). There are 3 permutations of (9,9,7) and 3 permutations of (9,8,8), for 6 total numbers.
ANSWER 9: C

---

### Problem 10:
Between exit 5 and 26, there are 21 gaps. To maximize one gap, minimize the other 20 gaps at 5 miles each: 20×5=100 miles. Maximum gap=118-100=18 miles.
ANSWER 10: C

---

### Problem 11:
Total sum of 5 scores=5×90=450. Median=91, so sorted scores [a,b,91,c,d]. Mode=94, so 94 must appear at least twice, so c=d=94, sum c+d=188. Sum of the two lowest scores=450-91-188=171.
ANSWER 11: B

---

### Problem 12:
Using +,-,* exactly once, one valid calculation: 5-4+6×3=1+18=19, which uses all three operations and matches option E.
ANSWER 12: E

### Problem 1:
We need to find the number of possible median values for a set of 9 distinct integers, 6 of which are 2, 3, 4, 6, 9, 14. The median of 9 numbers is the 5th element when sorted in ascending order. We add 3 distinct integers (not in the given set) and analyze possible medians:
1.  **Case 1: All 3 new numbers < 2**: Sorted list is [a,b,c,2,3,4,6,9,14], median = 3.
2.  **Case 2: 2 new numbers < 2, 1 new number ≥4**: Sorted list is [a,b,2,3,4,...], median =4.
3.  **Case 3: 1 new number <2**:
    - If we add 5 (the only integer between 4 and 6): Sorted list is [a,2,3,4,5,6,...], median=5.
    - If we do not add 5: Sorted list is [a,2,3,4,6,...], median=6.
4.  **Case 4: 0 new numbers <2 (all new numbers ≥5)**:
    - Add 5: Sorted list [2,3,4,5,6,...], median=6.
    - Add a number between 6 and9 (7 or 8): Sorted list [2,3,4,6,7,...] or [2,3,4,6,8,...], median=7 or 8.
    - Add all new numbers ≥9: Sorted list [2,3,4,6,9,...], median=9.
The only impossible medians are 2 (needs 5 numbers ≤2, only 4 max) and values >9 (existing numbers already have 5 elements ≤9). The possible medians are 3,4,5,6,7,8,9: 7 values.
ANSWER 1: D
---
### Problem 2:
Original list: [2,6,7,7,28], sum = 2+6+7+7+28=50. Insert N to make 6 elements. The median of 6 elements is the average of the 3rd and 4th sorted elements. The mean is $\frac{50+N}{6}$, which equals twice the median:
$$\frac{50+N}{6} = 2 \times \text{median} \implies 50 + N = 12 \times \text{median} \implies N = 12 \times \text{median} -50.$$
Test possible median values from the sorted list with N:
- If N ≤7: Sorted list has 3rd=7, 4th=7, median=7. Then $N=12*7 -50=84-50=34$, which is >7, valid.
- Other median values (e.g., 6.5) give N=28, which does not fit the required sorted position.
Verify N=34: Sorted list [2,6,7,7,28,34], median=(7+7)/2=7, mean=(50+34)/6=14, which is twice 7.
ANSWER 2: E
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### Problem 3:
Positive 4-digit numbers with all even digits:
- Thousands place: 2,4,6,8 (4 choices, no 0).
- Hundreds, tens, units: 0,2,4,6,8 (5 choices each).
Total numbers: $4 \times 5 \times 5 \times5=500$.
A number is divisible by 4 iff its last two digits form a number divisible by 4. For even tens digits (0,2,4,6,8), a two-digit number $10a + b$ is divisible by 4 iff $b$ (units digit) is 0,4,8 (since $10a \equiv 0 \mod4$ for even a). So valid endings: 5 (tens choices) ×3 (units choices)=15.
Total valid numbers: $4 \times5 \times5 \times3=300$. Fraction: $\frac{300}{500}=\frac{3}{5}$.
ANSWER 3: D
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### Problem 4:
Difference between 6.5% and 6% tax on $20:
$$0.065*20 - 0.06*20 = (0.065-0.06)*20=0.005*20=0.10.$$
ANSWER 4: B
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### Problem 5:
Both sums have 93 terms (1993-1901+1=93, 193-101+1=93). Each term in the first sum is 1800 more than the corresponding term in the second sum:
Total difference = $93 \times 1800 = 167400$.
ANSWER 5: A
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### Problem 6:
The sign pattern repeats every 4 terms: $+ - - +$, and each block of 4 sums to 0:
$$(1-2-3+4)+(5-6-7+8)+\dots+(1993-1994-1995+1996)=0.$$
1996 is divisible by 4, so there are 499 such blocks, total sum 0.
ANSWER 6: C
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### Problem 7:
Let $S$ = sunglass wearers ($|S|=50$), $C$ = cap wearers ($|C|=35$). $P(S|C)=\frac{|S∩C|}{|C|}=\frac{2}{5}$, so $|S∩C|=\frac{2}{5}*35=14$.
The probability a sunglass wearer wears a cap is $P(C|S)=\frac{|S∩C|}{|S|}=\frac{14}{50}=\frac{7}{25}$.
ANSWER 7: B
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### Problem 8:
3-digit numbers with digit sum 26: maximum digit sum is 27 (999), so 26 is 1 less than maximum, meaning digits are permutations of (9,9,8). The only even permutation is 998 (units digit 8). So 1 number.
ANSWER 8: A
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### Problem 9:
3-digit numbers with digit sum 25: 25 is 2 less than 27, so digits are permutations of (9,9,7) and (9,8,8):
- Permutations of (9,9,7): 997,979,799 (3 numbers).
- Permutations of (9,8,8): 988,898,889 (3 numbers).
Total 6 numbers.
ANSWER 9: C
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### Problem 10:
Between exit 5 and 26, there are $26-5=21$ gaps, total distance 118 miles. Each gap ≥5 miles. To maximize one gap, minimize the other 20: $20*5=100$, so maximum gap = $118-100=18$.
ANSWER 10: C
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### Problem 11:
Five sorted scores: $a≤b≤c≤d≤e$. Median $c=91$, mean 90, so total sum $5*90=450$, so $a+b+d+e=450-91=359$. Mode is 94, so 94 appears at least twice. Since $c=91$, 94 can only be in $d$ and $e$, so $d=e=94$. Then $a+b=359-94-94=171$. This sum is fixed (valid examples exist, e.g., 81+90=171).
ANSWER 11: B
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### Problem 12:
Permute +, -, × in 5 __4 __6 __3, using standard order of operations (multiply first):
- $5 -4 +6×3 = 5-4+18=19$, which is an option.
All other permutations give values not in the choices.
ANSWER 12: E