🔬 Model Lab

Problem 1:
We are asked to estimate
[(487,000)(12,027,300) + (9,621,001)(487,000)] / [(19,367)(.05)]
to the nearest power of 10.

Factor the numerator:
487,000 × (12,027,300 + 9,621,001)
= 487,000 × 21,648,301.

Approximate for order of magnitude:
487,000 ≈ 5 × 10^5
21,648,301 ≈ 2.2 × 10^7
So numerator ≈ 5×10^5 × 2.2×10^7 ≈ 11×10^12 = 1.1×10^13.

Denominator:
19,367 × .05 ≈ 20,000 × .05 = 1,000 = 10^3.

Hence ratio ≈ 1.1×10^13 / 10^3 = 1.1×10^10 ≈ 11,000,000,000.

This is closest to 10,000,000,000, not 1,000,000,000 which is 10 times too small, nor 100,000,000,000 which is 10 times too large.
ANSWER 1: D

Problem 2:
We need whole numbers ◇, △ with product 36 to maximize the sum.

Whole numbers with product 36 must be positive factor pairs of 36. List them:
1 × 36, sum = 37
2 × 18, sum = 20
3 × 12, sum = 15
4 × 9, sum = 13
6 × 6, sum = 12

The sum gets smaller as the factors get closer together. The largest is obtained from the most extreme pair 1 and 36.

So 37 is largest, not 20 or 12.
ANSWER 2: E

Problem 3:
We are asked for
(1901+1902+...+1993) − (101+102+...+193).

Count terms: 1993−1901+1 = 93, and 193−101+1 = 93. So both sums have 93 terms.

Pair them in order:
1901−101 = 1800
1902−102 = 1800
...
1993−193 = 1800

There are 93 such differences, each 1800. So the total difference is
93 × 1800 = 93 × 18 × 100 = 1674 × 100 = 167,400.

This rules out e.g. adding instead of subtracting, or miscounting terms.
ANSWER 3: A

Problem 4:
We must place +, −, × each exactly once in 5 __ 4 __ 6 __ 3 and evaluate with normal precedence, × before + −.

List all 6 orderings:

1) 5+4−6×3 = 9−18 = −9
2) 5+4×6−3 = 5+24−3 = 26
3) 5−4+6×3 = 1+18 = 19
4) 5−4×6+3 = 5−24+3 = −16
5) 5×4+6−3 = 20+6−3 = 23
6) 5×4−6+3 = 20−6+3 = 17

The attainable values are −9, 26, 19, −16, 23, 17.

Of the choices 9,10,15,16,19, only 19 occurs, from 5−4+6×3.
ANSWER 4: E

Problem 5:
There are 1000 families. Mean = sum / 1000.

Only one income is wrong: true largest 98,000 entered as 980,000.
Increase in sum = 980,000 − 98,000 = 882,000.

Increase in mean = increase in sum divided by 1000:
882,000 / 1000 = 882.

So the incorrect mean is 882 dollars higher. This is not 980, which would ignore subtracting the correct value, nor hundreds of thousands, which forget to divide by 1000.
ANSWER 5: A

Problem 6:
Original prices total 80+40 = 120 dollars.

Saved on jacket: 40% of 80 = 0.40×80 = 32 dollars.
Saved on shirt: 55% of 40 = 0.55×40 = 22 dollars.
Total saved = 32+22 = 54 dollars.

As percent of original total:
54 / 120 = 9 / 20 = 45%.

Note 47½% is the simple average of 40% and 55%, tempting but wrong because the bases 80 and 40 are different.
ANSWER 6: A

Problem 7:
Original $180. Sale is 50% below original, so sale price is 50% of original:
0.50 × 180 = $90.

Saturday is an additional 20% off the sale price, so pay 100%−20% = 80% of sale price:
0.80 × 90 = $72.

$54 would be 70% off at once, which is wrong because successive discounts multiply, not add: 0.5×0.8 = 0.4 of original, not 0.3.
ANSWER 7: B

Problem 8:
Factor 2020 = 202×10 = 2×101×2×5 = 2^2 × 5^1 × 101^1.
Number of divisors = (2+1)(1+1)(1+1) = 12. They are:
1, 2, 4, 5, 10, 20, 101, 202, 404, 505, 1010, 2020.

A number has >3 factors unless it is 1 (1 factor), prime (2 factors), or square of a prime (3 factors).

Check:
1: 1 factor - no
2: prime - no
4 = 2^2: 3 factors - no, exactly 3 not more
5: prime - no
10 = 2×5: 4 factors - yes
20: 6 factors - yes
101: prime - no
202 = 2×101: 4 factors - yes
404 = 2^2×101: 6 factors - yes
505 = 5×101: 4 factors - yes
1010 = 2×5×101: 8 factors - yes
2020: 12 factors - yes

Yes count: 10,20,202,404,505,1010,2020 = 7.

Including 4 would give 8, but 4 has exactly 3, not more than 3.
ANSWER 8: B

Problem 9:
There are 4×4×4 = 64 small cubes. We want those that touch a vertical side or the bottom. The top does not count unless the cube also touches a side or bottom.

Count the complement: cubes touching neither a side nor the bottom.
To avoid the four sides, the cube must be interior in x and y: 2 by 2 = 4 per layer.
To avoid the bottom, it must not be in the bottom layer: so in the top 3 layers.

Complement = 2×2×3 = 12 cubes. These are the only ones that fail.

Hence desired = 64 − 12 = 52.

If all 6 faces counted, answer would be 64−8 = 56, which is not listed, confirming the top is excluded.
ANSWER 9: B

Problem 10:
6 hops, start and end at ground, one step up (U) or down (D) each time. To return, need 3 U and 3 D. We cannot go below ground, so height must stay ≥0.

In particular first hop must be U. This is counting Dyck paths of length 6.

Enumerate:
U U U D D D
U U D U D D
U U D D U D
U D U U D D
U D U D U D

That is 5 ways. This is the Catalan number C3 = 5.

Unrestricted 3U-3D would be C(6,3)=20, not listed, so the stay-above-ground condition is essential.
ANSWER 10: B

Here are the step-by-step solutions for each problem:

---

### Problem 1:

**Restatement of the problem:**  
We want to find the value closest to:
$$\frac{(487{,}000)(12{,}027{,}300) + (9{,}621{,}001)(487{,}000)}{(19{,}367)(0.05)}$$

**Step-by-step solution:**
1. Factor out the common term $487{,}000$ from the numerator:
   $$\text{Numerator} = 487{,}000 \times (12{,}027{,}300 + 9{,}621{,}001)$$
2. Add the terms inside the parentheses:
   $$12{,}027{,}300 + 9{,}621{,}001 = 21{,}648{,}301$$
3. Approximate the terms using scientific notation:
   - $487{,}000 \approx 5 \times 10^5$ (or more precisely, $4.87 \times 10^5$)
   - $21{,}648{,}301 \approx 2.16 \times 10^7$
   - Numerator $\approx (4.87 \times 10^5) \times (2.165 \times 10^7) \approx 1.05 \times 10^{13}$
4. Now calculate the denominator:
   $$\text{Denominator} = 19{,}367 \times 0.05 \approx 20{,}000 \times 0.05 = 1{,}000 = 10^3$$
5. Divide the numerator by the denominator:
   $$\frac{1.05 \times 10^{13}}{10^3} \approx 1.05 \times 10^{10} = 10{,}500{,}000{,}000$$
   This is closest to $10{,}000{,}000{,}000$ ($10^{10}$).

ANSWER 1: D

---

### Problem 2:

**Restatement of the problem:**  
Given that $\diamond$ and $\triangle$ are whole numbers and $\diamond \times \triangle = 36$, find the largest possible value of $\diamond + \triangle$.

**Step-by-step solution:**
1. The pairs of whole numbers (positive factors) that multiply to $36$ are:
   - $1 \times 36 = 36 \implies 1 + 36 = 37$
   - $2 \times 18 = 36 \implies 2 + 18 = 20$
   - $3 \times 12 = 36 \implies 3 + 12 = 15$
   - $4 \times 9 = 36 \implies 4 + 9 = 13$
   - $6 \times 6 = 36 \implies 6 + 6 = 12$
2. For two numbers with a fixed product, their sum is maximized when the two numbers are as far apart as possible.
3. The maximum sum occurs with the pair $1$ and $36$, which gives:
   $$1 + 36 = 37$$

ANSWER 2: E

---

### Problem 3:

**Restatement of the problem:**  
Calculate the difference:
$$(1901 + 1902 + 1903 + \dots + 1993) - (101 + 102 + 103 + \dots + 193)$$

**Step-by-step solution:**
1. Notice that both sums have the same number of terms: from $1$ to $93$, which is $93$ terms.
2. Group the terms into pairs by matching the $k$-th term of the first sum with the $k$-th term of the second sum:
   $$(1901 - 101) + (1902 - 102) + (1903 - 103) + \dots + (1993 - 193)$$
3. The difference between each corresponding pair is constant:
   $$1900 + k - (100 + k) = 1800$$
4. Since there are $93$ pairs, the total value is:
   $$93 \times 1800 = 93 \times 18 \times 100$$
5. Multiply $93 \times 18$:
   $$93 \times 18 = 93 \times (20 - 2) = 1860 - 186 = 1674$$
   $$1674 \times 100 = 167{,}400$$

ANSWER 3: A

---

### Problem 4:

**Restatement of the problem:**  
Each of the operations $+$, $-$, and $\times$ is used exactly once in the blanks of $5 \text{ \underline{ } } 4 \text{ \underline{ } } 6 \text{ \underline{ } } 3$. Determine which of the given choices could be the result.

**Step-by-step solution:**
1. Remember standard order of operations: multiplication ($\times$) is performed before addition and subtraction ($+$ and $-$).
2. Test all $6$ possible arrangements of the three signs:
   - $5 + 4 - 6 \times 3 = 5 + 4 - 18 = -9$
   - $5 + 4 \times 6 - 3 = 5 + 24 - 3 = 26$
   - $5 - 4 + 6 \times 3 = 5 - 4 + 18 = 19$
   - $5 - 4 \times 6 + 3 = 5 - 24 + 3 = -16$
   - $5 \times 4 + 6 - 3 = 20 + 6 - 3 = 23$
   - $5 \times 4 - 6 + 3 = 20 - 6 + 3 = 17$
3. Among the choices ($9, 10, 15, 16, 19$), the only value that appears in our list of possible outcomes is $19$.

ANSWER 4: E

---

### Problem 5:

**Restatement of the problem:**  
A set of $1{,}000$ incomes has one error: the largest income was entered as $\$980{,}000$ instead of $\$98{,}000$. Find the difference between the mean of the incorrect data and the mean of the actual data.

**Step-by-step solution:**
1. Let the correct sum of the $1{,}000$ incomes be $S_{\text{actual}}$.
2. The incorrect sum, $S_{\text{incorrect}}$, differs from the actual sum only by the error in the largest income:
   $$S_{\text{incorrect}} - S_{\text{actual}} = 980{,}000 - 98{,}000 = 882{,}000$$
3. The difference between the means is:
   $$\text{Mean}_{\text{incorrect}} - \text{Mean}_{\text{actual}} = \frac{S_{\text{incorrect}} - S_{\text{actual}}}{1{,}000} = \frac{882{,}000}{1{,}000} = 882\text{ dollars}$$

ANSWER 5: A

---

### Problem 6:

**Restatement of the problem:**  
A jacket originally cost $\$80$ and was discounted by $40\%$. A shirt originally cost $\$40$ and was discounted by $55\%$. Find the total amount saved as a percentage of the total original price.

**Step-by-step solution:**
1. Find the amount saved on the jacket:
   $$\text{Savings on jacket} = 40\% \text{ of } \$80 = 0.40 \times 80 = \$32$$
2. Find the amount saved on the shirt:
   $$\text{Savings on shirt} = 55\% \text{ of } \$40 = 0.55 \times 40 = \$22$$
3. Calculate the total savings:
   $$\text{Total savings} = \$32 + \$22 = \$54$$
4. Calculate the total original cost:
   $$\text{Total original price} = \$80 + \$40 = \$120$$
5. Find the percentage saved:
   $$\frac{54}{120} = \frac{9}{20} = \frac{45}{100} = 45\%$$

ANSWER 6: A

---

### Problem 7:

**Restatement of the problem:**  
A coat has an original price of $\$180$. The outlet sale price is $50\%$ off the original price, and on Saturday an additional $20\%$ is discounted from the sale price. Find the Saturday price.

**Step-by-step solution:**
1. Calculate the standard sale price:
   $$\text{Sale price} = \$180 - (0.50 \times \$180) = \$90$$
2. Calculate the Saturday price by taking an additional $20\%$ off the $\$90$ sale price:
   $$\text{Saturday discount} = 0.20 \times \$90 = \$18$$
   $$\text{Saturday price} = \$90 - \$18 = \$72$$
   *(Alternatively: $180 \times 0.50 \times 0.80 = 180 \times 0.40 = \$72$.)*

ANSWER 7: B

---

### Problem 8:

**Restatement of the problem:**  
Find how many factors of $2020$ have more than $3$ factors.

**Step-by-step solution:**
1. Find the prime factorization of $2020$:
   $$2020 = 20 \t

# Problem 1:

The expression is

$$\frac{(487{,}000)(12{,}027{,}300)+(9{,}621{,}001)(487{,}000)}{(19{,}367)(0.05)}$$

**Factor the numerator:** both terms share the factor 487,000.

$$487{,}000\,(12{,}027{,}300+9{,}621{,}001)=487{,}000\times 21{,}648{,}301$$

Estimate: $\approx (4.87\times10^5)(2.16\times10^7)\approx 1.05\times10^{13}$.

**Denominator:** $19{,}367\times0.05\approx 968$, i.e. about $9.7\times10^2$.

**Divide:**
$$\frac{1.05\times10^{13}}{9.7\times10^{2}}\approx 1.09\times10^{10}$$

That is about $10{,}900{,}000{,}000$, closest to $10{,}000{,}000{,}000$. (The next choice, $100{,}000{,}000{,}000$, is nearly ten times too big, and $1{,}000{,}000{,}000$ is nearly ten times too small.)

ANSWER 1: D

# Problem 2:

We need whole numbers with $\Diamond\times\triangle=36$, maximizing $\Diamond+\triangle$.

List factor pairs and sums:
- $1\cdot36 \Rightarrow 37$
- $2\cdot18 \Rightarrow 20$
- $3\cdot12 \Rightarrow 15$
- $4\cdot9 \Rightarrow 13$
- $6\cdot6 \Rightarrow 12$

The sum is largest when the factors are as far apart as possible: $1+36=37$. (Choices 12, 13, 15, 20 are the *other* pairs' sums — 12 is actually the smallest.)

ANSWER 2: E

# Problem 3:

Both sums have the same number of terms: from 1901 to 1993 there are $1993-1901+1=93$ terms, and from 101 to 193 there are also 93 terms.

Pair them term by term: $1901-101=1800$, $1902-102=1800$, …, $1993-193=1800$.

So the difference is
$$93\times1800=167{,}400.$$

(No need to compute either huge sum separately.)

ANSWER 3: A

# Problem 4:

We fill $5\ \_\ 4\ \_\ 6\ \_\ 3$ using $+,-,\times$ each exactly once. Try all 6 orders (remember $\times$ comes first):

- $+,-,\times$: $5+4-6\times3=5+4-18=-9$
- $+,\times,-$: $5+4\times6-3=5+24-3=26$
- $-,+,\times$: $5-4+6\times3=1+18=\mathbf{19}$
- $-,\times,+$: $5-4\times6+3=5-24+3=-16$
- $\times,+,-$: $5\times4+6-3=20+3=23$
- $\times,-,+$: $5\times4-6+3=20-6+3=17$

The only achievable value among the choices is $19$. (9, 10, 15, 16 never appear.)

ANSWER 4: E

# Problem 5:

Only one entry changed. The total was too large by
$$980{,}000-98{,}000=882{,}000\text{ dollars.}$$

The mean is the total divided by 1000 families, so the means differ by
$$\frac{882{,}000}{1000}=882\text{ dollars.}$$

(The $\$882{,}000$ choice is the error in the *total*, not the mean; $\$8200$ and the range are irrelevant.)

ANSWER 5: A

# Problem 6:

Savings:
- Jacket: $40\%$ of $80 = 0.40\times80=\$32$
- Shirt: $55\%$ of $40 = 0.55\times40=\$22$

Total saved $=32+22=\$54$. Total original price $=80+40=\$120$.

$$\frac{54}{120}=0.45=45\%$$

(Note it is **not** the average $\tfrac{40+55}{2}=47\tfrac12\%$, because the jacket costs more, pulling the percentage toward $40\%$.)

ANSWER 6: A

# Problem 7:

Sale price: $50\%$ below original means half price:
$$180\times0.50=\$90.$$

Saturday: an extra $20\%$ off that sale price, so you pay $80\%$ of \$90:
$$90\times0.80=\$72.$$

(\$54 would come from wrongly taking $70\%$ off the original; \$90 forgets the extra discount.)

ANSWER 7: B

# Problem 8:

Factor: $2020=2^2\cdot5\cdot101$, so it has $(2+1)(1+1)(1+1)=12$ divisors:
$$1,2,4,5,10,20,101,202,404,505,1010,2020.$$

A number has **more than 3** factors unless it is $1$ (one factor), a prime (two factors), or the square of a prime (three factors).

Among the list:
- $1$ → 1 factor (exclude)
- $2,5,101$ → primes, 2 factors (exclude)
- $4=2^2$ → 3 factors: $1,2,4$ (exclude)

All others ($10,20,202,404,505,1010,2020$) have more than 3 factors: that's $12-5=7$ numbers.

ANSWER 8: B

# Problem 9:

Count the cubes that touch **no** side and **not** the bottom, then subtract.

Set coordinates $1..4$ in each direction. A cube avoids all four vertical sides only if its horizontal position is $2$ or $3$ in both horizontal directions: $2\times2=4$ choices. It avoids the bottom if its level is $2,3,$ or $4$: 3 choices (the top is not mentioned, so touching the top is allowed).

Interior-and-not-bottom cubes: $2\times2\times3=12$.

Cubes touching a side or the bottom: $64-12=52$.

(48 would be the count of cubes on the outer shell of a $4\times4\times4$ block including the top face — but the top-face cubes not on a side don't touch a side or bottom, so 48 is wrong here.)

ANSWER 9: B

# Problem 10:

Each hop is up ($+1$) or down ($-1$). We need 6 hops summing to 0 (so 3 ups and 3 downs) and the height must never go below ground (he can't hop down from the ground).

These are exactly the "balanced sequences" counted by the Catalan number $C_3=\frac{1}{4}\binom{6}{3}=\frac{20}{4}=5$.

Listing them (U = up, D = down):
1. UUUDDD
2. UUDUDD
3. UUDDUD
4. UDUUDD
5. UDUDUD

Total: **5**. (The 20 sequences of three U's and three D's include 15 that dip below ground, like DUUUDD, which are impossible.)

ANSWER 10: B

Problem 1:

We are finding the value closest to the expression, which is intended to be

\[
\frac{(487{,}000)(12{,}027{,}300)+(9{,}621{,}001)(487{,}000)}
{(19{,}367)(0.05)}.
\]

Factor out \(487{,}000\) from the numerator:

\[
\frac{487{,}000(12{,}027{,}300+9{,}621{,}001)}
{19{,}367(0.05)}.
\]

Add inside the parentheses:

\[
12{,}027{,}300+9{,}621{,}001=21{,}648{,}301.
\]

Estimate the numerator:

\[
487{,}000\cdot 21{,}648{,}301\approx 4.87\times 10^5\cdot 2.165\times 10^7
\approx 1.05\times 10^{13}.
\]

Now estimate the denominator:

\[
19{,}367(0.05)=968.35\approx 10^3.
\]

Thus the whole value is approximately

\[
\frac{1.05\times 10^{13}}{10^3}\approx 1.05\times 10^{10}.
\]

So it is closest to \(10{,}000{,}000{,}000\).

\[
\boxed{\text{D}}
\]

ANSWER 1: D

Problem 2:

We need whole numbers \(\diamond\) and \(\triangle\) such that

\[
\diamond\cdot\triangle=36.
\]

To make their sum as large as possible, use the most unequal positive factor pair of 36:

\[
1\cdot 36=36.
\]

Then

\[
1+36=37.
\]

Other factor pairs give smaller sums:

\[
2+18=20,\qquad 3+12=15,\qquad 4+9=13,\qquad 6+6=12.
\]

The largest is \(37\).

\[
\boxed{\text{E}}
\]

ANSWER 2: E

Problem 3:

There are 93 terms in each sum:

\[
1901,1902,\dots,1993
\]

because

\[
1993-1901+1=93.
\]

Likewise,

\[
193-101+1=93.
\]

Match corresponding terms:

\[
1901-101=1800,
\]
\[
1902-102=1800,
\]

and so on. Every paired difference is \(1800\).

Therefore the total difference is

\[
93\cdot 1800=167{,}400.
\]

\[
\boxed{\text{A}}
\]

ANSWER 3: A

Problem 4:

We must use \(+\), \(-\), and \(\times\) exactly once in

\[
5\ \_\ 4\ \_\ 6\ \_\ 3.
\]

Remember that multiplication is done before addition and subtraction.

Try the possible placements:

\[
5+4-6\times3=9-18=-9
\]

\[
5+4\times6-3=5+24-3=26
\]

\[
5-4+6\times3=5-4+18=19
\]

This equals one of the choices.

For completeness, the remaining orders give:

\[
5-4\times6+3=5-24+3=-16,
\]

\[
5\times4+6-3=23,
\]

\[
5\times4-6+3=17.
\]

Only \(19\) appears among the answer choices.

\[
\boxed{\text{E}}
\]

ANSWER 4: E

Problem 5:

The actual largest income was

\[
98{,}000.
\]

It was incorrectly entered as

\[
980{,}000.
\]

So the total income in the computer data is too large by

\[
980{,}000-98{,}000=882{,}000.
\]

There are \(1000\) families, so the mean is too large by

\[
\frac{882{,}000}{1000}=882.
\]

Thus the difference between the two means is \(882\) dollars.

\[
\boxed{\text{A}}
\]

ANSWER 5: A

Problem 6:

The jacket originally cost \(80\) dollars and had a \(40\%\) discount.

Amount saved on jacket:

\[
0.40(80)=32.
\]

The shirt originally cost \(40\) dollars and had a \(55\%\) discount.

Amount saved on shirt:

\[
0.55(40)=22.
\]

Total saved:

\[
32+22=54.
\]

Total original price:

\[
80+40=120.
\]

The percent saved is

\[
\frac{54}{120}=0.45=45\%.
\]

\[
\boxed{\text{A}}
\]

ANSWER 6: A

Problem 7:

The original price is \(180\) dollars.

First, the store reduces the price by \(50\%\):

\[
180(0.50)=90.
\]

Then the Saturday discount takes another \(20\%\) off the sale price of \(90\):

\[
90(0.80)=72.
\]

The Saturday price is

\[
\$72.
\]

It is not \(\$54\), because the discounts are applied one after the other, not by adding \(50\%+20\%=70\%\) of the original price.

\[
\boxed{\text{B}}
\]

ANSWER 7: B

Problem 8:

First factor \(2020\):

\[
2020=2^2\cdot 5\cdot 101.
\]

So the number of factors of \(2020\) is

\[
(2+1)(1+1)(1+1)=12.
\]

We need to exclude factors that have at most 3 factors themselves.

- \(1\) has only 1 factor.
- The primes \(2,5,101\) each have 2 factors.
- \(4=2^2\) has 3 factors: \(1,2,4\).

Thus there are

\[
1+3+1=5
\]

factors of \(2020\) with at most 3 factors.

Therefore, the number with more than 3 factors is

\[
12-5=7.
\]

\[
\boxed{\text{B}}
\]

ANSWER 8: B

Problem 9:

There are

\[
4\cdot4\cdot4=64
\]

small cubes total.

It is easier to count the cubes that do **not** touch a side or the bottom.

To avoid touching a side, a cube must be in one of the middle \(2\times2\) positions horizontally:

\[
2\cdot2=4
\]

possible positions per layer.

To avoid touching the bottom, it can be in any of the upper 3 layers.

So the number that touch neither a side nor the bottom is

\[
2\cdot2\cdot3=12.
\]

Therefore the number that touch a side or the bottom is

\[
64-12=52.
\]

\[
\boxed{\text{B}}
\]

ANSWER 9: B

Problem 10:

Buzz makes 6 hops and ends where he started, on the ground. Therefore, he must make exactly 3 up-hops and 3 down-hops.

However, he cannot hop below the ground.

The valid sequences are:

\[
\text{up-up-up-down-down-down}
\]

\[
\text{up-up-down-up-down-down}
\]

\[
\text{up-up-down-down-up-down}
\]

\[
\text{up-down-up-up-down-down}
\]

\[
\text{up-down-up-down-up-down}
\]

There are \(5\) valid sequences.

For example, a sequence beginning with “down” is impossible, because Buzz would go below ground immediately.

\[
\boxed{\text{B}}
\]

ANSWER 10: B

Problem 1:
We are asked to estimate
[(487,000)(12,027,300) + (9,621,001)(487,000)] / [(19,367)(.05)]
to the nearest power of 10.

Factor the numerator:
487,000 × (12,027,300 + 9,621,001)
= 487,000 × 21,648,301.

Approximate for order of magnitude:
487,000 ≈ 5 × 10^5
21,648,301 ≈ 2.2 × 10^7
So numerator ≈ 5×10^5 × 2.2×10^7 ≈ 11×10^12 = 1.1×10^13.

Denominator:
19,367 × .05 ≈ 20,000 × .05 = 1,000 = 10^3.

Hence ratio ≈ 1.1×10^13 / 10^3 = 1.1×10^10 ≈ 11,000,000,000.

This is closest to 10,000,000,000, not 1,000,000,000 which is 10 times too small, nor 100,000,000,000 which is 10 times too large.
ANSWER 1: D

Problem 2:
We need whole numbers ◇, △ with product 36 to maximize the sum.

Whole numbers with product 36 must be positive factor pairs of 36. List them:
1 × 36, sum = 37
2 × 18, sum = 20
3 × 12, sum = 15
4 × 9, sum = 13
6 × 6, sum = 12

The sum gets smaller as the factors get closer together. The largest is obtained from the most extreme pair 1 and 36.

So 37 is largest, not 20 or 12.
ANSWER 2: E

Problem 3:
We are asked for
(1901+1902+...+1993) − (101+102+...+193).

Count terms: 1993−1901+1 = 93, and 193−101+1 = 93. So both sums have 93 terms.

Pair them in order:
1901−101 = 1800
1902−102 = 1800
...
1993−193 = 1800

There are 93 such differences, each 1800. So the total difference is
93 × 1800 = 93 × 18 × 100 = 1674 × 100 = 167,400.

This rules out e.g. adding instead of subtracting, or miscounting terms.
ANSWER 3: A

Problem 4:
We must place +, −, × each exactly once in 5 __ 4 __ 6 __ 3 and evaluate with normal precedence, × before + −.

List all 6 orderings:

1) 5+4−6×3 = 9−18 = −9
2) 5+4×6−3 = 5+24−3 = 26
3) 5−4+6×3 = 1+18 = 19
4) 5−4×6+3 = 5−24+3 = −16
5) 5×4+6−3 = 20+6−3 = 23
6) 5×4−6+3 = 20−6+3 = 17

The attainable values are −9, 26, 19, −16, 23, 17.

Of the choices 9,10,15,16,19, only 19 occurs, from 5−4+6×3.
ANSWER 4: E

Problem 5:
There are 1000 families. Mean = sum / 1000.

Only one income is wrong: true largest 98,000 entered as 980,000.
Increase in sum = 980,000 − 98,000 = 882,000.

Increase in mean = increase in sum divided by 1000:
882,000 / 1000 = 882.

So the incorrect mean is 882 dollars higher. This is not 980, which would ignore subtracting the correct value, nor hundreds of thousands, which forget to divide by 1000.
ANSWER 5: A

Problem 6:
Original prices total 80+40 = 120 dollars.

Saved on jacket: 40% of 80 = 0.40×80 = 32 dollars.
Saved on shirt: 55% of 40 = 0.55×40 = 22 dollars.
Total saved = 32+22 = 54 dollars.

As percent of original total:
54 / 120 = 9 / 20 = 45%.

Note 47½% is the simple average of 40% and 55%, tempting but wrong because the bases 80 and 40 are different.
ANSWER 6: A

Problem 7:
Original $180. Sale is 50% below original, so sale price is 50% of original:
0.50 × 180 = $90.

Saturday is an additional 20% off the sale price, so pay 100%−20% = 80% of sale price:
0.80 × 90 = $72.

$54 would be 70% off at once, which is wrong because successive discounts multiply, not add: 0.5×0.8 = 0.4 of original, not 0.3.
ANSWER 7: B

Problem 8:
Factor 2020 = 202×10 = 2×101×2×5 = 2^2 × 5^1 × 101^1.
Number of divisors = (2+1)(1+1)(1+1) = 12. They are:
1, 2, 4, 5, 10, 20, 101, 202, 404, 505, 1010, 2020.

A number has >3 factors unless it is 1 (1 factor), prime (2 factors), or square of a prime (3 factors).

Check:
1: 1 factor - no
2: prime - no
4 = 2^2: 3 factors - no, exactly 3 not more
5: prime - no
10 = 2×5: 4 factors - yes
20: 6 factors - yes
101: prime - no
202 = 2×101: 4 factors - yes
404 = 2^2×101: 6 factors - yes
505 = 5×101: 4 factors - yes
1010 = 2×5×101: 8 factors - yes
2020: 12 factors - yes

Yes count: 10,20,202,404,505,1010,2020 = 7.

Including 4 would give 8, but 4 has exactly 3, not more than 3.
ANSWER 8: B

Problem 9:
There are 4×4×4 = 64 small cubes. We want those that touch a vertical side or the bottom. The top does not count unless the cube also touches a side or bottom.

Count the complement: cubes touching neither a side nor the bottom.
To avoid the four sides, the cube must be interior in x and y: 2 by 2 = 4 per layer.
To avoid the bottom, it must not be in the bottom layer: so in the top 3 layers.

Complement = 2×2×3 = 12 cubes. These are the only ones that fail.

Hence desired = 64 − 12 = 52.

If all 6 faces counted, answer would be 64−8 = 56, which is not listed, confirming the top is excluded.
ANSWER 9: B

Problem 10:
6 hops, start and end at ground, one step up (U) or down (D) each time. To return, need 3 U and 3 D. We cannot go below ground, so height must stay ≥0.

In particular first hop must be U. This is counting Dyck paths of length 6.

Enumerate:
U U U D D D
U U D U D D
U U D D U D
U D U U D D
U D U D U D

That is 5 ways. This is the Catalan number C3 = 5.

Unrestricted 3U-3D would be C(6,3)=20, not listed, so the stay-above-ground condition is essential.
ANSWER 10: B

Here are the step-by-step solutions for each problem:

---

### Problem 1:

**Restatement of the problem:**  
We want to find the value closest to:
$$\frac{(487{,}000)(12{,}027{,}300) + (9{,}621{,}001)(487{,}000)}{(19{,}367)(0.05)}$$

**Step-by-step solution:**
1. Factor out the common term $487{,}000$ from the numerator:
   $$\text{Numerator} = 487{,}000 \times (12{,}027{,}300 + 9{,}621{,}001)$$
2. Add the terms inside the parentheses:
   $$12{,}027{,}300 + 9{,}621{,}001 = 21{,}648{,}301$$
3. Approximate the terms using scientific notation:
   - $487{,}000 \approx 5 \times 10^5$ (or more precisely, $4.87 \times 10^5$)
   - $21{,}648{,}301 \approx 2.16 \times 10^7$
   - Numerator $\approx (4.87 \times 10^5) \times (2.165 \times 10^7) \approx 1.05 \times 10^{13}$
4. Now calculate the denominator:
   $$\text{Denominator} = 19{,}367 \times 0.05 \approx 20{,}000 \times 0.05 = 1{,}000 = 10^3$$
5. Divide the numerator by the denominator:
   $$\frac{1.05 \times 10^{13}}{10^3} \approx 1.05 \times 10^{10} = 10{,}500{,}000{,}000$$
   This is closest to $10{,}000{,}000{,}000$ ($10^{10}$).

ANSWER 1: D

---

### Problem 2:

**Restatement of the problem:**  
Given that $\diamond$ and $\triangle$ are whole numbers and $\diamond \times \triangle = 36$, find the largest possible value of $\diamond + \triangle$.

**Step-by-step solution:**
1. The pairs of whole numbers (positive factors) that multiply to $36$ are:
   - $1 \times 36 = 36 \implies 1 + 36 = 37$
   - $2 \times 18 = 36 \implies 2 + 18 = 20$
   - $3 \times 12 = 36 \implies 3 + 12 = 15$
   - $4 \times 9 = 36 \implies 4 + 9 = 13$
   - $6 \times 6 = 36 \implies 6 + 6 = 12$
2. For two numbers with a fixed product, their sum is maximized when the two numbers are as far apart as possible.
3. The maximum sum occurs with the pair $1$ and $36$, which gives:
   $$1 + 36 = 37$$

ANSWER 2: E

---

### Problem 3:

**Restatement of the problem:**  
Calculate the difference:
$$(1901 + 1902 + 1903 + \dots + 1993) - (101 + 102 + 103 + \dots + 193)$$

**Step-by-step solution:**
1. Notice that both sums have the same number of terms: from $1$ to $93$, which is $93$ terms.
2. Group the terms into pairs by matching the $k$-th term of the first sum with the $k$-th term of the second sum:
   $$(1901 - 101) + (1902 - 102) + (1903 - 103) + \dots + (1993 - 193)$$
3. The difference between each corresponding pair is constant:
   $$1900 + k - (100 + k) = 1800$$
4. Since there are $93$ pairs, the total value is:
   $$93 \times 1800 = 93 \times 18 \times 100$$
5. Multiply $93 \times 18$:
   $$93 \times 18 = 93 \times (20 - 2) = 1860 - 186 = 1674$$
   $$1674 \times 100 = 167{,}400$$

ANSWER 3: A

---

### Problem 4:

**Restatement of the problem:**  
Each of the operations $+$, $-$, and $\times$ is used exactly once in the blanks of $5 \text{ \underline{ } } 4 \text{ \underline{ } } 6 \text{ \underline{ } } 3$. Determine which of the given choices could be the result.

**Step-by-step solution:**
1. Remember standard order of operations: multiplication ($\times$) is performed before addition and subtraction ($+$ and $-$).
2. Test all $6$ possible arrangements of the three signs:
   - $5 + 4 - 6 \times 3 = 5 + 4 - 18 = -9$
   - $5 + 4 \times 6 - 3 = 5 + 24 - 3 = 26$
   - $5 - 4 + 6 \times 3 = 5 - 4 + 18 = 19$
   - $5 - 4 \times 6 + 3 = 5 - 24 + 3 = -16$
   - $5 \times 4 + 6 - 3 = 20 + 6 - 3 = 23$
   - $5 \times 4 - 6 + 3 = 20 - 6 + 3 = 17$
3. Among the choices ($9, 10, 15, 16, 19$), the only value that appears in our list of possible outcomes is $19$.

ANSWER 4: E

---

### Problem 5:

**Restatement of the problem:**  
A set of $1{,}000$ incomes has one error: the largest income was entered as $\$980{,}000$ instead of $\$98{,}000$. Find the difference between the mean of the incorrect data and the mean of the actual data.

**Step-by-step solution:**
1. Let the correct sum of the $1{,}000$ incomes be $S_{\text{actual}}$.
2. The incorrect sum, $S_{\text{incorrect}}$, differs from the actual sum only by the error in the largest income:
   $$S_{\text{incorrect}} - S_{\text{actual}} = 980{,}000 - 98{,}000 = 882{,}000$$
3. The difference between the means is:
   $$\text{Mean}_{\text{incorrect}} - \text{Mean}_{\text{actual}} = \frac{S_{\text{incorrect}} - S_{\text{actual}}}{1{,}000} = \frac{882{,}000}{1{,}000} = 882\text{ dollars}$$

ANSWER 5: A

---

### Problem 6:

**Restatement of the problem:**  
A jacket originally cost $\$80$ and was discounted by $40\%$. A shirt originally cost $\$40$ and was discounted by $55\%$. Find the total amount saved as a percentage of the total original price.

**Step-by-step solution:**
1. Find the amount saved on the jacket:
   $$\text{Savings on jacket} = 40\% \text{ of } \$80 = 0.40 \times 80 = \$32$$
2. Find the amount saved on the shirt:
   $$\text{Savings on shirt} = 55\% \text{ of } \$40 = 0.55 \times 40 = \$22$$
3. Calculate the total savings:
   $$\text{Total savings} = \$32 + \$22 = \$54$$
4. Calculate the total original cost:
   $$\text{Total original price} = \$80 + \$40 = \$120$$
5. Find the percentage saved:
   $$\frac{54}{120} = \frac{9}{20} = \frac{45}{100} = 45\%$$

ANSWER 6: A

---

### Problem 7:

**Restatement of the problem:**  
A coat has an original price of $\$180$. The outlet sale price is $50\%$ off the original price, and on Saturday an additional $20\%$ is discounted from the sale price. Find the Saturday price.

**Step-by-step solution:**
1. Calculate the standard sale price:
   $$\text{Sale price} = \$180 - (0.50 \times \$180) = \$90$$
2. Calculate the Saturday price by taking an additional $20\%$ off the $\$90$ sale price:
   $$\text{Saturday discount} = 0.20 \times \$90 = \$18$$
   $$\text{Saturday price} = \$90 - \$18 = \$72$$
   *(Alternatively: $180 \times 0.50 \times 0.80 = 180 \times 0.40 = \$72$.)*

ANSWER 7: B

---

### Problem 8:

**Restatement of the problem:**  
Find how many factors of $2020$ have more than $3$ factors.

**Step-by-step solution:**
1. Find the prime factorization of $2020$:
   $$2020 = 20 \t

# Problem 1:

The expression is

$$\frac{(487{,}000)(12{,}027{,}300)+(9{,}621{,}001)(487{,}000)}{(19{,}367)(0.05)}$$

**Factor the numerator:** both terms share the factor 487,000.

$$487{,}000\,(12{,}027{,}300+9{,}621{,}001)=487{,}000\times 21{,}648{,}301$$

Estimate: $\approx (4.87\times10^5)(2.16\times10^7)\approx 1.05\times10^{13}$.

**Denominator:** $19{,}367\times0.05\approx 968$, i.e. about $9.7\times10^2$.

**Divide:**
$$\frac{1.05\times10^{13}}{9.7\times10^{2}}\approx 1.09\times10^{10}$$

That is about $10{,}900{,}000{,}000$, closest to $10{,}000{,}000{,}000$. (The next choice, $100{,}000{,}000{,}000$, is nearly ten times too big, and $1{,}000{,}000{,}000$ is nearly ten times too small.)

ANSWER 1: D

# Problem 2:

We need whole numbers with $\Diamond\times\triangle=36$, maximizing $\Diamond+\triangle$.

List factor pairs and sums:
- $1\cdot36 \Rightarrow 37$
- $2\cdot18 \Rightarrow 20$
- $3\cdot12 \Rightarrow 15$
- $4\cdot9 \Rightarrow 13$
- $6\cdot6 \Rightarrow 12$

The sum is largest when the factors are as far apart as possible: $1+36=37$. (Choices 12, 13, 15, 20 are the *other* pairs' sums — 12 is actually the smallest.)

ANSWER 2: E

# Problem 3:

Both sums have the same number of terms: from 1901 to 1993 there are $1993-1901+1=93$ terms, and from 101 to 193 there are also 93 terms.

Pair them term by term: $1901-101=1800$, $1902-102=1800$, …, $1993-193=1800$.

So the difference is
$$93\times1800=167{,}400.$$

(No need to compute either huge sum separately.)

ANSWER 3: A

# Problem 4:

We fill $5\ \_\ 4\ \_\ 6\ \_\ 3$ using $+,-,\times$ each exactly once. Try all 6 orders (remember $\times$ comes first):

- $+,-,\times$: $5+4-6\times3=5+4-18=-9$
- $+,\times,-$: $5+4\times6-3=5+24-3=26$
- $-,+,\times$: $5-4+6\times3=1+18=\mathbf{19}$
- $-,\times,+$: $5-4\times6+3=5-24+3=-16$
- $\times,+,-$: $5\times4+6-3=20+3=23$
- $\times,-,+$: $5\times4-6+3=20-6+3=17$

The only achievable value among the choices is $19$. (9, 10, 15, 16 never appear.)

ANSWER 4: E

# Problem 5:

Only one entry changed. The total was too large by
$$980{,}000-98{,}000=882{,}000\text{ dollars.}$$

The mean is the total divided by 1000 families, so the means differ by
$$\frac{882{,}000}{1000}=882\text{ dollars.}$$

(The $\$882{,}000$ choice is the error in the *total*, not the mean; $\$8200$ and the range are irrelevant.)

ANSWER 5: A

# Problem 6:

Savings:
- Jacket: $40\%$ of $80 = 0.40\times80=\$32$
- Shirt: $55\%$ of $40 = 0.55\times40=\$22$

Total saved $=32+22=\$54$. Total original price $=80+40=\$120$.

$$\frac{54}{120}=0.45=45\%$$

(Note it is **not** the average $\tfrac{40+55}{2}=47\tfrac12\%$, because the jacket costs more, pulling the percentage toward $40\%$.)

ANSWER 6: A

# Problem 7:

Sale price: $50\%$ below original means half price:
$$180\times0.50=\$90.$$

Saturday: an extra $20\%$ off that sale price, so you pay $80\%$ of \$90:
$$90\times0.80=\$72.$$

(\$54 would come from wrongly taking $70\%$ off the original; \$90 forgets the extra discount.)

ANSWER 7: B

# Problem 8:

Factor: $2020=2^2\cdot5\cdot101$, so it has $(2+1)(1+1)(1+1)=12$ divisors:
$$1,2,4,5,10,20,101,202,404,505,1010,2020.$$

A number has **more than 3** factors unless it is $1$ (one factor), a prime (two factors), or the square of a prime (three factors).

Among the list:
- $1$ → 1 factor (exclude)
- $2,5,101$ → primes, 2 factors (exclude)
- $4=2^2$ → 3 factors: $1,2,4$ (exclude)

All others ($10,20,202,404,505,1010,2020$) have more than 3 factors: that's $12-5=7$ numbers.

ANSWER 8: B

# Problem 9:

Count the cubes that touch **no** side and **not** the bottom, then subtract.

Set coordinates $1..4$ in each direction. A cube avoids all four vertical sides only if its horizontal position is $2$ or $3$ in both horizontal directions: $2\times2=4$ choices. It avoids the bottom if its level is $2,3,$ or $4$: 3 choices (the top is not mentioned, so touching the top is allowed).

Interior-and-not-bottom cubes: $2\times2\times3=12$.

Cubes touching a side or the bottom: $64-12=52$.

(48 would be the count of cubes on the outer shell of a $4\times4\times4$ block including the top face — but the top-face cubes not on a side don't touch a side or bottom, so 48 is wrong here.)

ANSWER 9: B

# Problem 10:

Each hop is up ($+1$) or down ($-1$). We need 6 hops summing to 0 (so 3 ups and 3 downs) and the height must never go below ground (he can't hop down from the ground).

These are exactly the "balanced sequences" counted by the Catalan number $C_3=\frac{1}{4}\binom{6}{3}=\frac{20}{4}=5$.

Listing them (U = up, D = down):
1. UUUDDD
2. UUDUDD
3. UUDDUD
4. UDUUDD
5. UDUDUD

Total: **5**. (The 20 sequences of three U's and three D's include 15 that dip below ground, like DUUUDD, which are impossible.)

ANSWER 10: B

Problem 1:

We are finding the value closest to the expression, which is intended to be

\[
\frac{(487{,}000)(12{,}027{,}300)+(9{,}621{,}001)(487{,}000)}
{(19{,}367)(0.05)}.
\]

Factor out \(487{,}000\) from the numerator:

\[
\frac{487{,}000(12{,}027{,}300+9{,}621{,}001)}
{19{,}367(0.05)}.
\]

Add inside the parentheses:

\[
12{,}027{,}300+9{,}621{,}001=21{,}648{,}301.
\]

Estimate the numerator:

\[
487{,}000\cdot 21{,}648{,}301\approx 4.87\times 10^5\cdot 2.165\times 10^7
\approx 1.05\times 10^{13}.
\]

Now estimate the denominator:

\[
19{,}367(0.05)=968.35\approx 10^3.
\]

Thus the whole value is approximately

\[
\frac{1.05\times 10^{13}}{10^3}\approx 1.05\times 10^{10}.
\]

So it is closest to \(10{,}000{,}000{,}000\).

\[
\boxed{\text{D}}
\]

ANSWER 1: D

Problem 2:

We need whole numbers \(\diamond\) and \(\triangle\) such that

\[
\diamond\cdot\triangle=36.
\]

To make their sum as large as possible, use the most unequal positive factor pair of 36:

\[
1\cdot 36=36.
\]

Then

\[
1+36=37.
\]

Other factor pairs give smaller sums:

\[
2+18=20,\qquad 3+12=15,\qquad 4+9=13,\qquad 6+6=12.
\]

The largest is \(37\).

\[
\boxed{\text{E}}
\]

ANSWER 2: E

Problem 3:

There are 93 terms in each sum:

\[
1901,1902,\dots,1993
\]

because

\[
1993-1901+1=93.
\]

Likewise,

\[
193-101+1=93.
\]

Match corresponding terms:

\[
1901-101=1800,
\]
\[
1902-102=1800,
\]

and so on. Every paired difference is \(1800\).

Therefore the total difference is

\[
93\cdot 1800=167{,}400.
\]

\[
\boxed{\text{A}}
\]

ANSWER 3: A

Problem 4:

We must use \(+\), \(-\), and \(\times\) exactly once in

\[
5\ \_\ 4\ \_\ 6\ \_\ 3.
\]

Remember that multiplication is done before addition and subtraction.

Try the possible placements:

\[
5+4-6\times3=9-18=-9
\]

\[
5+4\times6-3=5+24-3=26
\]

\[
5-4+6\times3=5-4+18=19
\]

This equals one of the choices.

For completeness, the remaining orders give:

\[
5-4\times6+3=5-24+3=-16,
\]

\[
5\times4+6-3=23,
\]

\[
5\times4-6+3=17.
\]

Only \(19\) appears among the answer choices.

\[
\boxed{\text{E}}
\]

ANSWER 4: E

Problem 5:

The actual largest income was

\[
98{,}000.
\]

It was incorrectly entered as

\[
980{,}000.
\]

So the total income in the computer data is too large by

\[
980{,}000-98{,}000=882{,}000.
\]

There are \(1000\) families, so the mean is too large by

\[
\frac{882{,}000}{1000}=882.
\]

Thus the difference between the two means is \(882\) dollars.

\[
\boxed{\text{A}}
\]

ANSWER 5: A

Problem 6:

The jacket originally cost \(80\) dollars and had a \(40\%\) discount.

Amount saved on jacket:

\[
0.40(80)=32.
\]

The shirt originally cost \(40\) dollars and had a \(55\%\) discount.

Amount saved on shirt:

\[
0.55(40)=22.
\]

Total saved:

\[
32+22=54.
\]

Total original price:

\[
80+40=120.
\]

The percent saved is

\[
\frac{54}{120}=0.45=45\%.
\]

\[
\boxed{\text{A}}
\]

ANSWER 6: A

Problem 7:

The original price is \(180\) dollars.

First, the store reduces the price by \(50\%\):

\[
180(0.50)=90.
\]

Then the Saturday discount takes another \(20\%\) off the sale price of \(90\):

\[
90(0.80)=72.
\]

The Saturday price is

\[
\$72.
\]

It is not \(\$54\), because the discounts are applied one after the other, not by adding \(50\%+20\%=70\%\) of the original price.

\[
\boxed{\text{B}}
\]

ANSWER 7: B

Problem 8:

First factor \(2020\):

\[
2020=2^2\cdot 5\cdot 101.
\]

So the number of factors of \(2020\) is

\[
(2+1)(1+1)(1+1)=12.
\]

We need to exclude factors that have at most 3 factors themselves.

- \(1\) has only 1 factor.
- The primes \(2,5,101\) each have 2 factors.
- \(4=2^2\) has 3 factors: \(1,2,4\).

Thus there are

\[
1+3+1=5
\]

factors of \(2020\) with at most 3 factors.

Therefore, the number with more than 3 factors is

\[
12-5=7.
\]

\[
\boxed{\text{B}}
\]

ANSWER 8: B

Problem 9:

There are

\[
4\cdot4\cdot4=64
\]

small cubes total.

It is easier to count the cubes that do **not** touch a side or the bottom.

To avoid touching a side, a cube must be in one of the middle \(2\times2\) positions horizontally:

\[
2\cdot2=4
\]

possible positions per layer.

To avoid touching the bottom, it can be in any of the upper 3 layers.

So the number that touch neither a side nor the bottom is

\[
2\cdot2\cdot3=12.
\]

Therefore the number that touch a side or the bottom is

\[
64-12=52.
\]

\[
\boxed{\text{B}}
\]

ANSWER 9: B

Problem 10:

Buzz makes 6 hops and ends where he started, on the ground. Therefore, he must make exactly 3 up-hops and 3 down-hops.

However, he cannot hop below the ground.

The valid sequences are:

\[
\text{up-up-up-down-down-down}
\]

\[
\text{up-up-down-up-down-down}
\]

\[
\text{up-up-down-down-up-down}
\]

\[
\text{up-down-up-up-down-down}
\]

\[
\text{up-down-up-down-up-down}
\]

There are \(5\) valid sequences.

For example, a sequence beginning with “down” is impossible, because Buzz would go below ground immediately.

\[
\boxed{\text{B}}
\]

ANSWER 10: B

Problem 1:
We are asked to estimate
[(487,000)(12,027,300) + (9,621,001)(487,000)] / [(19,367)(.05)]
to the nearest power of 10.

Factor the numerator:
487,000 × (12,027,300 + 9,621,001)
= 487,000 × 21,648,301.

Approximate for order of magnitude:
487,000 ≈ 5 × 10^5
21,648,301 ≈ 2.2 × 10^7
So numerator ≈ 5×10^5 × 2.2×10^7 ≈ 11×10^12 = 1.1×10^13.

Denominator:
19,367 × .05 ≈ 20,000 × .05 = 1,000 = 10^3.

Hence ratio ≈ 1.1×10^13 / 10^3 = 1.1×10^10 ≈ 11,000,000,000.

This is closest to 10,000,000,000, not 1,000,000,000 which is 10 times too small, nor 100,000,000,000 which is 10 times too large.
ANSWER 1: D

Problem 2:
We need whole numbers ◇, △ with product 36 to maximize the sum.

Whole numbers with product 36 must be positive factor pairs of 36. List them:
1 × 36, sum = 37
2 × 18, sum = 20
3 × 12, sum = 15
4 × 9, sum = 13
6 × 6, sum = 12

The sum gets smaller as the factors get closer together. The largest is obtained from the most extreme pair 1 and 36.

So 37 is largest, not 20 or 12.
ANSWER 2: E

Problem 3:
We are asked for
(1901+1902+...+1993) − (101+102+...+193).

Count terms: 1993−1901+1 = 93, and 193−101+1 = 93. So both sums have 93 terms.

Pair them in order:
1901−101 = 1800
1902−102 = 1800
...
1993−193 = 1800

There are 93 such differences, each 1800. So the total difference is
93 × 1800 = 93 × 18 × 100 = 1674 × 100 = 167,400.

This rules out e.g. adding instead of subtracting, or miscounting terms.
ANSWER 3: A

Problem 4:
We must place +, −, × each exactly once in 5 __ 4 __ 6 __ 3 and evaluate with normal precedence, × before + −.

List all 6 orderings:

1) 5+4−6×3 = 9−18 = −9
2) 5+4×6−3 = 5+24−3 = 26
3) 5−4+6×3 = 1+18 = 19
4) 5−4×6+3 = 5−24+3 = −16
5) 5×4+6−3 = 20+6−3 = 23
6) 5×4−6+3 = 20−6+3 = 17

The attainable values are −9, 26, 19, −16, 23, 17.

Of the choices 9,10,15,16,19, only 19 occurs, from 5−4+6×3.
ANSWER 4: E

Problem 5:
There are 1000 families. Mean = sum / 1000.

Only one income is wrong: true largest 98,000 entered as 980,000.
Increase in sum = 980,000 − 98,000 = 882,000.

Increase in mean = increase in sum divided by 1000:
882,000 / 1000 = 882.

So the incorrect mean is 882 dollars higher. This is not 980, which would ignore subtracting the correct value, nor hundreds of thousands, which forget to divide by 1000.
ANSWER 5: A

Problem 6:
Original prices total 80+40 = 120 dollars.

Saved on jacket: 40% of 80 = 0.40×80 = 32 dollars.
Saved on shirt: 55% of 40 = 0.55×40 = 22 dollars.
Total saved = 32+22 = 54 dollars.

As percent of original total:
54 / 120 = 9 / 20 = 45%.

Note 47½% is the simple average of 40% and 55%, tempting but wrong because the bases 80 and 40 are different.
ANSWER 6: A

Problem 7:
Original $180. Sale is 50% below original, so sale price is 50% of original:
0.50 × 180 = $90.

Saturday is an additional 20% off the sale price, so pay 100%−20% = 80% of sale price:
0.80 × 90 = $72.

$54 would be 70% off at once, which is wrong because successive discounts multiply, not add: 0.5×0.8 = 0.4 of original, not 0.3.
ANSWER 7: B

Problem 8:
Factor 2020 = 202×10 = 2×101×2×5 = 2^2 × 5^1 × 101^1.
Number of divisors = (2+1)(1+1)(1+1) = 12. They are:
1, 2, 4, 5, 10, 20, 101, 202, 404, 505, 1010, 2020.

A number has >3 factors unless it is 1 (1 factor), prime (2 factors), or square of a prime (3 factors).

Check:
1: 1 factor - no
2: prime - no
4 = 2^2: 3 factors - no, exactly 3 not more
5: prime - no
10 = 2×5: 4 factors - yes
20: 6 factors - yes
101: prime - no
202 = 2×101: 4 factors - yes
404 = 2^2×101: 6 factors - yes
505 = 5×101: 4 factors - yes
1010 = 2×5×101: 8 factors - yes
2020: 12 factors - yes

Yes count: 10,20,202,404,505,1010,2020 = 7.

Including 4 would give 8, but 4 has exactly 3, not more than 3.
ANSWER 8: B

Problem 9:
There are 4×4×4 = 64 small cubes. We want those that touch a vertical side or the bottom. The top does not count unless the cube also touches a side or bottom.

Count the complement: cubes touching neither a side nor the bottom.
To avoid the four sides, the cube must be interior in x and y: 2 by 2 = 4 per layer.
To avoid the bottom, it must not be in the bottom layer: so in the top 3 layers.

Complement = 2×2×3 = 12 cubes. These are the only ones that fail.

Hence desired = 64 − 12 = 52.

If all 6 faces counted, answer would be 64−8 = 56, which is not listed, confirming the top is excluded.
ANSWER 9: B

Problem 10:
6 hops, start and end at ground, one step up (U) or down (D) each time. To return, need 3 U and 3 D. We cannot go below ground, so height must stay ≥0.

In particular first hop must be U. This is counting Dyck paths of length 6.

Enumerate:
U U U D D D
U U D U D D
U U D D U D
U D U U D D
U D U D U D

That is 5 ways. This is the Catalan number C3 = 5.

Unrestricted 3U-3D would be C(6,3)=20, not listed, so the stay-above-ground condition is essential.
ANSWER 10: B

Here are the step-by-step solutions for each problem:

---

### Problem 1:

**Restatement of the problem:**  
We want to find the value closest to:
$$\frac{(487{,}000)(12{,}027{,}300) + (9{,}621{,}001)(487{,}000)}{(19{,}367)(0.05)}$$

**Step-by-step solution:**
1. Factor out the common term $487{,}000$ from the numerator:
   $$\text{Numerator} = 487{,}000 \times (12{,}027{,}300 + 9{,}621{,}001)$$
2. Add the terms inside the parentheses:
   $$12{,}027{,}300 + 9{,}621{,}001 = 21{,}648{,}301$$
3. Approximate the terms using scientific notation:
   - $487{,}000 \approx 5 \times 10^5$ (or more precisely, $4.87 \times 10^5$)
   - $21{,}648{,}301 \approx 2.16 \times 10^7$
   - Numerator $\approx (4.87 \times 10^5) \times (2.165 \times 10^7) \approx 1.05 \times 10^{13}$
4. Now calculate the denominator:
   $$\text{Denominator} = 19{,}367 \times 0.05 \approx 20{,}000 \times 0.05 = 1{,}000 = 10^3$$
5. Divide the numerator by the denominator:
   $$\frac{1.05 \times 10^{13}}{10^3} \approx 1.05 \times 10^{10} = 10{,}500{,}000{,}000$$
   This is closest to $10{,}000{,}000{,}000$ ($10^{10}$).

ANSWER 1: D

---

### Problem 2:

**Restatement of the problem:**  
Given that $\diamond$ and $\triangle$ are whole numbers and $\diamond \times \triangle = 36$, find the largest possible value of $\diamond + \triangle$.

**Step-by-step solution:**
1. The pairs of whole numbers (positive factors) that multiply to $36$ are:
   - $1 \times 36 = 36 \implies 1 + 36 = 37$
   - $2 \times 18 = 36 \implies 2 + 18 = 20$
   - $3 \times 12 = 36 \implies 3 + 12 = 15$
   - $4 \times 9 = 36 \implies 4 + 9 = 13$
   - $6 \times 6 = 36 \implies 6 + 6 = 12$
2. For two numbers with a fixed product, their sum is maximized when the two numbers are as far apart as possible.
3. The maximum sum occurs with the pair $1$ and $36$, which gives:
   $$1 + 36 = 37$$

ANSWER 2: E

---

### Problem 3:

**Restatement of the problem:**  
Calculate the difference:
$$(1901 + 1902 + 1903 + \dots + 1993) - (101 + 102 + 103 + \dots + 193)$$

**Step-by-step solution:**
1. Notice that both sums have the same number of terms: from $1$ to $93$, which is $93$ terms.
2. Group the terms into pairs by matching the $k$-th term of the first sum with the $k$-th term of the second sum:
   $$(1901 - 101) + (1902 - 102) + (1903 - 103) + \dots + (1993 - 193)$$
3. The difference between each corresponding pair is constant:
   $$1900 + k - (100 + k) = 1800$$
4. Since there are $93$ pairs, the total value is:
   $$93 \times 1800 = 93 \times 18 \times 100$$
5. Multiply $93 \times 18$:
   $$93 \times 18 = 93 \times (20 - 2) = 1860 - 186 = 1674$$
   $$1674 \times 100 = 167{,}400$$

ANSWER 3: A

---

### Problem 4:

**Restatement of the problem:**  
Each of the operations $+$, $-$, and $\times$ is used exactly once in the blanks of $5 \text{ \underline{ } } 4 \text{ \underline{ } } 6 \text{ \underline{ } } 3$. Determine which of the given choices could be the result.

**Step-by-step solution:**
1. Remember standard order of operations: multiplication ($\times$) is performed before addition and subtraction ($+$ and $-$).
2. Test all $6$ possible arrangements of the three signs:
   - $5 + 4 - 6 \times 3 = 5 + 4 - 18 = -9$
   - $5 + 4 \times 6 - 3 = 5 + 24 - 3 = 26$
   - $5 - 4 + 6 \times 3 = 5 - 4 + 18 = 19$
   - $5 - 4 \times 6 + 3 = 5 - 24 + 3 = -16$
   - $5 \times 4 + 6 - 3 = 20 + 6 - 3 = 23$
   - $5 \times 4 - 6 + 3 = 20 - 6 + 3 = 17$
3. Among the choices ($9, 10, 15, 16, 19$), the only value that appears in our list of possible outcomes is $19$.

ANSWER 4: E

---

### Problem 5:

**Restatement of the problem:**  
A set of $1{,}000$ incomes has one error: the largest income was entered as $\$980{,}000$ instead of $\$98{,}000$. Find the difference between the mean of the incorrect data and the mean of the actual data.

**Step-by-step solution:**
1. Let the correct sum of the $1{,}000$ incomes be $S_{\text{actual}}$.
2. The incorrect sum, $S_{\text{incorrect}}$, differs from the actual sum only by the error in the largest income:
   $$S_{\text{incorrect}} - S_{\text{actual}} = 980{,}000 - 98{,}000 = 882{,}000$$
3. The difference between the means is:
   $$\text{Mean}_{\text{incorrect}} - \text{Mean}_{\text{actual}} = \frac{S_{\text{incorrect}} - S_{\text{actual}}}{1{,}000} = \frac{882{,}000}{1{,}000} = 882\text{ dollars}$$

ANSWER 5: A

---

### Problem 6:

**Restatement of the problem:**  
A jacket originally cost $\$80$ and was discounted by $40\%$. A shirt originally cost $\$40$ and was discounted by $55\%$. Find the total amount saved as a percentage of the total original price.

**Step-by-step solution:**
1. Find the amount saved on the jacket:
   $$\text{Savings on jacket} = 40\% \text{ of } \$80 = 0.40 \times 80 = \$32$$
2. Find the amount saved on the shirt:
   $$\text{Savings on shirt} = 55\% \text{ of } \$40 = 0.55 \times 40 = \$22$$
3. Calculate the total savings:
   $$\text{Total savings} = \$32 + \$22 = \$54$$
4. Calculate the total original cost:
   $$\text{Total original price} = \$80 + \$40 = \$120$$
5. Find the percentage saved:
   $$\frac{54}{120} = \frac{9}{20} = \frac{45}{100} = 45\%$$

ANSWER 6: A

---

### Problem 7:

**Restatement of the problem:**  
A coat has an original price of $\$180$. The outlet sale price is $50\%$ off the original price, and on Saturday an additional $20\%$ is discounted from the sale price. Find the Saturday price.

**Step-by-step solution:**
1. Calculate the standard sale price:
   $$\text{Sale price} = \$180 - (0.50 \times \$180) = \$90$$
2. Calculate the Saturday price by taking an additional $20\%$ off the $\$90$ sale price:
   $$\text{Saturday discount} = 0.20 \times \$90 = \$18$$
   $$\text{Saturday price} = \$90 - \$18 = \$72$$
   *(Alternatively: $180 \times 0.50 \times 0.80 = 180 \times 0.40 = \$72$.)*

ANSWER 7: B

---

### Problem 8:

**Restatement of the problem:**  
Find how many factors of $2020$ have more than $3$ factors.

**Step-by-step solution:**
1. Find the prime factorization of $2020$:
   $$2020 = 20 \t

# Problem 1:

The expression is

$$\frac{(487{,}000)(12{,}027{,}300)+(9{,}621{,}001)(487{,}000)}{(19{,}367)(0.05)}$$

**Factor the numerator:** both terms share the factor 487,000.

$$487{,}000\,(12{,}027{,}300+9{,}621{,}001)=487{,}000\times 21{,}648{,}301$$

Estimate: $\approx (4.87\times10^5)(2.16\times10^7)\approx 1.05\times10^{13}$.

**Denominator:** $19{,}367\times0.05\approx 968$, i.e. about $9.7\times10^2$.

**Divide:**
$$\frac{1.05\times10^{13}}{9.7\times10^{2}}\approx 1.09\times10^{10}$$

That is about $10{,}900{,}000{,}000$, closest to $10{,}000{,}000{,}000$. (The next choice, $100{,}000{,}000{,}000$, is nearly ten times too big, and $1{,}000{,}000{,}000$ is nearly ten times too small.)

ANSWER 1: D

# Problem 2:

We need whole numbers with $\Diamond\times\triangle=36$, maximizing $\Diamond+\triangle$.

List factor pairs and sums:
- $1\cdot36 \Rightarrow 37$
- $2\cdot18 \Rightarrow 20$
- $3\cdot12 \Rightarrow 15$
- $4\cdot9 \Rightarrow 13$
- $6\cdot6 \Rightarrow 12$

The sum is largest when the factors are as far apart as possible: $1+36=37$. (Choices 12, 13, 15, 20 are the *other* pairs' sums — 12 is actually the smallest.)

ANSWER 2: E

# Problem 3:

Both sums have the same number of terms: from 1901 to 1993 there are $1993-1901+1=93$ terms, and from 101 to 193 there are also 93 terms.

Pair them term by term: $1901-101=1800$, $1902-102=1800$, …, $1993-193=1800$.

So the difference is
$$93\times1800=167{,}400.$$

(No need to compute either huge sum separately.)

ANSWER 3: A

# Problem 4:

We fill $5\ \_\ 4\ \_\ 6\ \_\ 3$ using $+,-,\times$ each exactly once. Try all 6 orders (remember $\times$ comes first):

- $+,-,\times$: $5+4-6\times3=5+4-18=-9$
- $+,\times,-$: $5+4\times6-3=5+24-3=26$
- $-,+,\times$: $5-4+6\times3=1+18=\mathbf{19}$
- $-,\times,+$: $5-4\times6+3=5-24+3=-16$
- $\times,+,-$: $5\times4+6-3=20+3=23$
- $\times,-,+$: $5\times4-6+3=20-6+3=17$

The only achievable value among the choices is $19$. (9, 10, 15, 16 never appear.)

ANSWER 4: E

# Problem 5:

Only one entry changed. The total was too large by
$$980{,}000-98{,}000=882{,}000\text{ dollars.}$$

The mean is the total divided by 1000 families, so the means differ by
$$\frac{882{,}000}{1000}=882\text{ dollars.}$$

(The $\$882{,}000$ choice is the error in the *total*, not the mean; $\$8200$ and the range are irrelevant.)

ANSWER 5: A

# Problem 6:

Savings:
- Jacket: $40\%$ of $80 = 0.40\times80=\$32$
- Shirt: $55\%$ of $40 = 0.55\times40=\$22$

Total saved $=32+22=\$54$. Total original price $=80+40=\$120$.

$$\frac{54}{120}=0.45=45\%$$

(Note it is **not** the average $\tfrac{40+55}{2}=47\tfrac12\%$, because the jacket costs more, pulling the percentage toward $40\%$.)

ANSWER 6: A

# Problem 7:

Sale price: $50\%$ below original means half price:
$$180\times0.50=\$90.$$

Saturday: an extra $20\%$ off that sale price, so you pay $80\%$ of \$90:
$$90\times0.80=\$72.$$

(\$54 would come from wrongly taking $70\%$ off the original; \$90 forgets the extra discount.)

ANSWER 7: B

# Problem 8:

Factor: $2020=2^2\cdot5\cdot101$, so it has $(2+1)(1+1)(1+1)=12$ divisors:
$$1,2,4,5,10,20,101,202,404,505,1010,2020.$$

A number has **more than 3** factors unless it is $1$ (one factor), a prime (two factors), or the square of a prime (three factors).

Among the list:
- $1$ → 1 factor (exclude)
- $2,5,101$ → primes, 2 factors (exclude)
- $4=2^2$ → 3 factors: $1,2,4$ (exclude)

All others ($10,20,202,404,505,1010,2020$) have more than 3 factors: that's $12-5=7$ numbers.

ANSWER 8: B

# Problem 9:

Count the cubes that touch **no** side and **not** the bottom, then subtract.

Set coordinates $1..4$ in each direction. A cube avoids all four vertical sides only if its horizontal position is $2$ or $3$ in both horizontal directions: $2\times2=4$ choices. It avoids the bottom if its level is $2,3,$ or $4$: 3 choices (the top is not mentioned, so touching the top is allowed).

Interior-and-not-bottom cubes: $2\times2\times3=12$.

Cubes touching a side or the bottom: $64-12=52$.

(48 would be the count of cubes on the outer shell of a $4\times4\times4$ block including the top face — but the top-face cubes not on a side don't touch a side or bottom, so 48 is wrong here.)

ANSWER 9: B

# Problem 10:

Each hop is up ($+1$) or down ($-1$). We need 6 hops summing to 0 (so 3 ups and 3 downs) and the height must never go below ground (he can't hop down from the ground).

These are exactly the "balanced sequences" counted by the Catalan number $C_3=\frac{1}{4}\binom{6}{3}=\frac{20}{4}=5$.

Listing them (U = up, D = down):
1. UUUDDD
2. UUDUDD
3. UUDDUD
4. UDUUDD
5. UDUDUD

Total: **5**. (The 20 sequences of three U's and three D's include 15 that dip below ground, like DUUUDD, which are impossible.)

ANSWER 10: B

Problem 1:

We are finding the value closest to the expression, which is intended to be

\[
\frac{(487{,}000)(12{,}027{,}300)+(9{,}621{,}001)(487{,}000)}
{(19{,}367)(0.05)}.
\]

Factor out \(487{,}000\) from the numerator:

\[
\frac{487{,}000(12{,}027{,}300+9{,}621{,}001)}
{19{,}367(0.05)}.
\]

Add inside the parentheses:

\[
12{,}027{,}300+9{,}621{,}001=21{,}648{,}301.
\]

Estimate the numerator:

\[
487{,}000\cdot 21{,}648{,}301\approx 4.87\times 10^5\cdot 2.165\times 10^7
\approx 1.05\times 10^{13}.
\]

Now estimate the denominator:

\[
19{,}367(0.05)=968.35\approx 10^3.
\]

Thus the whole value is approximately

\[
\frac{1.05\times 10^{13}}{10^3}\approx 1.05\times 10^{10}.
\]

So it is closest to \(10{,}000{,}000{,}000\).

\[
\boxed{\text{D}}
\]

ANSWER 1: D

Problem 2:

We need whole numbers \(\diamond\) and \(\triangle\) such that

\[
\diamond\cdot\triangle=36.
\]

To make their sum as large as possible, use the most unequal positive factor pair of 36:

\[
1\cdot 36=36.
\]

Then

\[
1+36=37.
\]

Other factor pairs give smaller sums:

\[
2+18=20,\qquad 3+12=15,\qquad 4+9=13,\qquad 6+6=12.
\]

The largest is \(37\).

\[
\boxed{\text{E}}
\]

ANSWER 2: E

Problem 3:

There are 93 terms in each sum:

\[
1901,1902,\dots,1993
\]

because

\[
1993-1901+1=93.
\]

Likewise,

\[
193-101+1=93.
\]

Match corresponding terms:

\[
1901-101=1800,
\]
\[
1902-102=1800,
\]

and so on. Every paired difference is \(1800\).

Therefore the total difference is

\[
93\cdot 1800=167{,}400.
\]

\[
\boxed{\text{A}}
\]

ANSWER 3: A

Problem 4:

We must use \(+\), \(-\), and \(\times\) exactly once in

\[
5\ \_\ 4\ \_\ 6\ \_\ 3.
\]

Remember that multiplication is done before addition and subtraction.

Try the possible placements:

\[
5+4-6\times3=9-18=-9
\]

\[
5+4\times6-3=5+24-3=26
\]

\[
5-4+6\times3=5-4+18=19
\]

This equals one of the choices.

For completeness, the remaining orders give:

\[
5-4\times6+3=5-24+3=-16,
\]

\[
5\times4+6-3=23,
\]

\[
5\times4-6+3=17.
\]

Only \(19\) appears among the answer choices.

\[
\boxed{\text{E}}
\]

ANSWER 4: E

Problem 5:

The actual largest income was

\[
98{,}000.
\]

It was incorrectly entered as

\[
980{,}000.
\]

So the total income in the computer data is too large by

\[
980{,}000-98{,}000=882{,}000.
\]

There are \(1000\) families, so the mean is too large by

\[
\frac{882{,}000}{1000}=882.
\]

Thus the difference between the two means is \(882\) dollars.

\[
\boxed{\text{A}}
\]

ANSWER 5: A

Problem 6:

The jacket originally cost \(80\) dollars and had a \(40\%\) discount.

Amount saved on jacket:

\[
0.40(80)=32.
\]

The shirt originally cost \(40\) dollars and had a \(55\%\) discount.

Amount saved on shirt:

\[
0.55(40)=22.
\]

Total saved:

\[
32+22=54.
\]

Total original price:

\[
80+40=120.
\]

The percent saved is

\[
\frac{54}{120}=0.45=45\%.
\]

\[
\boxed{\text{A}}
\]

ANSWER 6: A

Problem 7:

The original price is \(180\) dollars.

First, the store reduces the price by \(50\%\):

\[
180(0.50)=90.
\]

Then the Saturday discount takes another \(20\%\) off the sale price of \(90\):

\[
90(0.80)=72.
\]

The Saturday price is

\[
\$72.
\]

It is not \(\$54\), because the discounts are applied one after the other, not by adding \(50\%+20\%=70\%\) of the original price.

\[
\boxed{\text{B}}
\]

ANSWER 7: B

Problem 8:

First factor \(2020\):

\[
2020=2^2\cdot 5\cdot 101.
\]

So the number of factors of \(2020\) is

\[
(2+1)(1+1)(1+1)=12.
\]

We need to exclude factors that have at most 3 factors themselves.

- \(1\) has only 1 factor.
- The primes \(2,5,101\) each have 2 factors.
- \(4=2^2\) has 3 factors: \(1,2,4\).

Thus there are

\[
1+3+1=5
\]

factors of \(2020\) with at most 3 factors.

Therefore, the number with more than 3 factors is

\[
12-5=7.
\]

\[
\boxed{\text{B}}
\]

ANSWER 8: B

Problem 9:

There are

\[
4\cdot4\cdot4=64
\]

small cubes total.

It is easier to count the cubes that do **not** touch a side or the bottom.

To avoid touching a side, a cube must be in one of the middle \(2\times2\) positions horizontally:

\[
2\cdot2=4
\]

possible positions per layer.

To avoid touching the bottom, it can be in any of the upper 3 layers.

So the number that touch neither a side nor the bottom is

\[
2\cdot2\cdot3=12.
\]

Therefore the number that touch a side or the bottom is

\[
64-12=52.
\]

\[
\boxed{\text{B}}
\]

ANSWER 9: B

Problem 10:

Buzz makes 6 hops and ends where he started, on the ground. Therefore, he must make exactly 3 up-hops and 3 down-hops.

However, he cannot hop below the ground.

The valid sequences are:

\[
\text{up-up-up-down-down-down}
\]

\[
\text{up-up-down-up-down-down}
\]

\[
\text{up-up-down-down-up-down}
\]

\[
\text{up-down-up-up-down-down}
\]

\[
\text{up-down-up-down-up-down}
\]

There are \(5\) valid sequences.

For example, a sequence beginning with “down” is impossible, because Buzz would go below ground immediately.

\[
\boxed{\text{B}}
\]

ANSWER 10: B

Problem 1:
We are asked to estimate
[(487,000)(12,027,300) + (9,621,001)(487,000)] / [(19,367)(.05)]
to the nearest power of 10.

Factor the numerator:
487,000 × (12,027,300 + 9,621,001)
= 487,000 × 21,648,301.

Approximate for order of magnitude:
487,000 ≈ 5 × 10^5
21,648,301 ≈ 2.2 × 10^7
So numerator ≈ 5×10^5 × 2.2×10^7 ≈ 11×10^12 = 1.1×10^13.

Denominator:
19,367 × .05 ≈ 20,000 × .05 = 1,000 = 10^3.

Hence ratio ≈ 1.1×10^13 / 10^3 = 1.1×10^10 ≈ 11,000,000,000.

This is closest to 10,000,000,000, not 1,000,000,000 which is 10 times too small, nor 100,000,000,000 which is 10 times too large.
ANSWER 1: D

Problem 2:
We need whole numbers ◇, △ with product 36 to maximize the sum.

Whole numbers with product 36 must be positive factor pairs of 36. List them:
1 × 36, sum = 37
2 × 18, sum = 20
3 × 12, sum = 15
4 × 9, sum = 13
6 × 6, sum = 12

The sum gets smaller as the factors get closer together. The largest is obtained from the most extreme pair 1 and 36.

So 37 is largest, not 20 or 12.
ANSWER 2: E

Problem 3:
We are asked for
(1901+1902+...+1993) − (101+102+...+193).

Count terms: 1993−1901+1 = 93, and 193−101+1 = 93. So both sums have 93 terms.

Pair them in order:
1901−101 = 1800
1902−102 = 1800
...
1993−193 = 1800

There are 93 such differences, each 1800. So the total difference is
93 × 1800 = 93 × 18 × 100 = 1674 × 100 = 167,400.

This rules out e.g. adding instead of subtracting, or miscounting terms.
ANSWER 3: A

Problem 4:
We must place +, −, × each exactly once in 5 __ 4 __ 6 __ 3 and evaluate with normal precedence, × before + −.

List all 6 orderings:

1) 5+4−6×3 = 9−18 = −9
2) 5+4×6−3 = 5+24−3 = 26
3) 5−4+6×3 = 1+18 = 19
4) 5−4×6+3 = 5−24+3 = −16
5) 5×4+6−3 = 20+6−3 = 23
6) 5×4−6+3 = 20−6+3 = 17

The attainable values are −9, 26, 19, −16, 23, 17.

Of the choices 9,10,15,16,19, only 19 occurs, from 5−4+6×3.
ANSWER 4: E

Problem 5:
There are 1000 families. Mean = sum / 1000.

Only one income is wrong: true largest 98,000 entered as 980,000.
Increase in sum = 980,000 − 98,000 = 882,000.

Increase in mean = increase in sum divided by 1000:
882,000 / 1000 = 882.

So the incorrect mean is 882 dollars higher. This is not 980, which would ignore subtracting the correct value, nor hundreds of thousands, which forget to divide by 1000.
ANSWER 5: A

Problem 6:
Original prices total 80+40 = 120 dollars.

Saved on jacket: 40% of 80 = 0.40×80 = 32 dollars.
Saved on shirt: 55% of 40 = 0.55×40 = 22 dollars.
Total saved = 32+22 = 54 dollars.

As percent of original total:
54 / 120 = 9 / 20 = 45%.

Note 47½% is the simple average of 40% and 55%, tempting but wrong because the bases 80 and 40 are different.
ANSWER 6: A

Problem 7:
Original $180. Sale is 50% below original, so sale price is 50% of original:
0.50 × 180 = $90.

Saturday is an additional 20% off the sale price, so pay 100%−20% = 80% of sale price:
0.80 × 90 = $72.

$54 would be 70% off at once, which is wrong because successive discounts multiply, not add: 0.5×0.8 = 0.4 of original, not 0.3.
ANSWER 7: B

Problem 8:
Factor 2020 = 202×10 = 2×101×2×5 = 2^2 × 5^1 × 101^1.
Number of divisors = (2+1)(1+1)(1+1) = 12. They are:
1, 2, 4, 5, 10, 20, 101, 202, 404, 505, 1010, 2020.

A number has >3 factors unless it is 1 (1 factor), prime (2 factors), or square of a prime (3 factors).

Check:
1: 1 factor - no
2: prime - no
4 = 2^2: 3 factors - no, exactly 3 not more
5: prime - no
10 = 2×5: 4 factors - yes
20: 6 factors - yes
101: prime - no
202 = 2×101: 4 factors - yes
404 = 2^2×101: 6 factors - yes
505 = 5×101: 4 factors - yes
1010 = 2×5×101: 8 factors - yes
2020: 12 factors - yes

Yes count: 10,20,202,404,505,1010,2020 = 7.

Including 4 would give 8, but 4 has exactly 3, not more than 3.
ANSWER 8: B

Problem 9:
There are 4×4×4 = 64 small cubes. We want those that touch a vertical side or the bottom. The top does not count unless the cube also touches a side or bottom.

Count the complement: cubes touching neither a side nor the bottom.
To avoid the four sides, the cube must be interior in x and y: 2 by 2 = 4 per layer.
To avoid the bottom, it must not be in the bottom layer: so in the top 3 layers.

Complement = 2×2×3 = 12 cubes. These are the only ones that fail.

Hence desired = 64 − 12 = 52.

If all 6 faces counted, answer would be 64−8 = 56, which is not listed, confirming the top is excluded.
ANSWER 9: B

Problem 10:
6 hops, start and end at ground, one step up (U) or down (D) each time. To return, need 3 U and 3 D. We cannot go below ground, so height must stay ≥0.

In particular first hop must be U. This is counting Dyck paths of length 6.

Enumerate:
U U U D D D
U U D U D D
U U D D U D
U D U U D D
U D U D U D

That is 5 ways. This is the Catalan number C3 = 5.

Unrestricted 3U-3D would be C(6,3)=20, not listed, so the stay-above-ground condition is essential.
ANSWER 10: B

Here are the step-by-step solutions for each problem:

---

### Problem 1:

**Restatement of the problem:**  
We want to find the value closest to:
$$\frac{(487{,}000)(12{,}027{,}300) + (9{,}621{,}001)(487{,}000)}{(19{,}367)(0.05)}$$

**Step-by-step solution:**
1. Factor out the common term $487{,}000$ from the numerator:
   $$\text{Numerator} = 487{,}000 \times (12{,}027{,}300 + 9{,}621{,}001)$$
2. Add the terms inside the parentheses:
   $$12{,}027{,}300 + 9{,}621{,}001 = 21{,}648{,}301$$
3. Approximate the terms using scientific notation:
   - $487{,}000 \approx 5 \times 10^5$ (or more precisely, $4.87 \times 10^5$)
   - $21{,}648{,}301 \approx 2.16 \times 10^7$
   - Numerator $\approx (4.87 \times 10^5) \times (2.165 \times 10^7) \approx 1.05 \times 10^{13}$
4. Now calculate the denominator:
   $$\text{Denominator} = 19{,}367 \times 0.05 \approx 20{,}000 \times 0.05 = 1{,}000 = 10^3$$
5. Divide the numerator by the denominator:
   $$\frac{1.05 \times 10^{13}}{10^3} \approx 1.05 \times 10^{10} = 10{,}500{,}000{,}000$$
   This is closest to $10{,}000{,}000{,}000$ ($10^{10}$).

ANSWER 1: D

---

### Problem 2:

**Restatement of the problem:**  
Given that $\diamond$ and $\triangle$ are whole numbers and $\diamond \times \triangle = 36$, find the largest possible value of $\diamond + \triangle$.

**Step-by-step solution:**
1. The pairs of whole numbers (positive factors) that multiply to $36$ are:
   - $1 \times 36 = 36 \implies 1 + 36 = 37$
   - $2 \times 18 = 36 \implies 2 + 18 = 20$
   - $3 \times 12 = 36 \implies 3 + 12 = 15$
   - $4 \times 9 = 36 \implies 4 + 9 = 13$
   - $6 \times 6 = 36 \implies 6 + 6 = 12$
2. For two numbers with a fixed product, their sum is maximized when the two numbers are as far apart as possible.
3. The maximum sum occurs with the pair $1$ and $36$, which gives:
   $$1 + 36 = 37$$

ANSWER 2: E

---

### Problem 3:

**Restatement of the problem:**  
Calculate the difference:
$$(1901 + 1902 + 1903 + \dots + 1993) - (101 + 102 + 103 + \dots + 193)$$

**Step-by-step solution:**
1. Notice that both sums have the same number of terms: from $1$ to $93$, which is $93$ terms.
2. Group the terms into pairs by matching the $k$-th term of the first sum with the $k$-th term of the second sum:
   $$(1901 - 101) + (1902 - 102) + (1903 - 103) + \dots + (1993 - 193)$$
3. The difference between each corresponding pair is constant:
   $$1900 + k - (100 + k) = 1800$$
4. Since there are $93$ pairs, the total value is:
   $$93 \times 1800 = 93 \times 18 \times 100$$
5. Multiply $93 \times 18$:
   $$93 \times 18 = 93 \times (20 - 2) = 1860 - 186 = 1674$$
   $$1674 \times 100 = 167{,}400$$

ANSWER 3: A

---

### Problem 4:

**Restatement of the problem:**  
Each of the operations $+$, $-$, and $\times$ is used exactly once in the blanks of $5 \text{ \underline{ } } 4 \text{ \underline{ } } 6 \text{ \underline{ } } 3$. Determine which of the given choices could be the result.

**Step-by-step solution:**
1. Remember standard order of operations: multiplication ($\times$) is performed before addition and subtraction ($+$ and $-$).
2. Test all $6$ possible arrangements of the three signs:
   - $5 + 4 - 6 \times 3 = 5 + 4 - 18 = -9$
   - $5 + 4 \times 6 - 3 = 5 + 24 - 3 = 26$
   - $5 - 4 + 6 \times 3 = 5 - 4 + 18 = 19$
   - $5 - 4 \times 6 + 3 = 5 - 24 + 3 = -16$
   - $5 \times 4 + 6 - 3 = 20 + 6 - 3 = 23$
   - $5 \times 4 - 6 + 3 = 20 - 6 + 3 = 17$
3. Among the choices ($9, 10, 15, 16, 19$), the only value that appears in our list of possible outcomes is $19$.

ANSWER 4: E

---

### Problem 5:

**Restatement of the problem:**  
A set of $1{,}000$ incomes has one error: the largest income was entered as $\$980{,}000$ instead of $\$98{,}000$. Find the difference between the mean of the incorrect data and the mean of the actual data.

**Step-by-step solution:**
1. Let the correct sum of the $1{,}000$ incomes be $S_{\text{actual}}$.
2. The incorrect sum, $S_{\text{incorrect}}$, differs from the actual sum only by the error in the largest income:
   $$S_{\text{incorrect}} - S_{\text{actual}} = 980{,}000 - 98{,}000 = 882{,}000$$
3. The difference between the means is:
   $$\text{Mean}_{\text{incorrect}} - \text{Mean}_{\text{actual}} = \frac{S_{\text{incorrect}} - S_{\text{actual}}}{1{,}000} = \frac{882{,}000}{1{,}000} = 882\text{ dollars}$$

ANSWER 5: A

---

### Problem 6:

**Restatement of the problem:**  
A jacket originally cost $\$80$ and was discounted by $40\%$. A shirt originally cost $\$40$ and was discounted by $55\%$. Find the total amount saved as a percentage of the total original price.

**Step-by-step solution:**
1. Find the amount saved on the jacket:
   $$\text{Savings on jacket} = 40\% \text{ of } \$80 = 0.40 \times 80 = \$32$$
2. Find the amount saved on the shirt:
   $$\text{Savings on shirt} = 55\% \text{ of } \$40 = 0.55 \times 40 = \$22$$
3. Calculate the total savings:
   $$\text{Total savings} = \$32 + \$22 = \$54$$
4. Calculate the total original cost:
   $$\text{Total original price} = \$80 + \$40 = \$120$$
5. Find the percentage saved:
   $$\frac{54}{120} = \frac{9}{20} = \frac{45}{100} = 45\%$$

ANSWER 6: A

---

### Problem 7:

**Restatement of the problem:**  
A coat has an original price of $\$180$. The outlet sale price is $50\%$ off the original price, and on Saturday an additional $20\%$ is discounted from the sale price. Find the Saturday price.

**Step-by-step solution:**
1. Calculate the standard sale price:
   $$\text{Sale price} = \$180 - (0.50 \times \$180) = \$90$$
2. Calculate the Saturday price by taking an additional $20\%$ off the $\$90$ sale price:
   $$\text{Saturday discount} = 0.20 \times \$90 = \$18$$
   $$\text{Saturday price} = \$90 - \$18 = \$72$$
   *(Alternatively: $180 \times 0.50 \times 0.80 = 180 \times 0.40 = \$72$.)*

ANSWER 7: B

---

### Problem 8:

**Restatement of the problem:**  
Find how many factors of $2020$ have more than $3$ factors.

**Step-by-step solution:**
1. Find the prime factorization of $2020$:
   $$2020 = 20 \t

# Problem 1:

The expression is

$$\frac{(487{,}000)(12{,}027{,}300)+(9{,}621{,}001)(487{,}000)}{(19{,}367)(0.05)}$$

**Factor the numerator:** both terms share the factor 487,000.

$$487{,}000\,(12{,}027{,}300+9{,}621{,}001)=487{,}000\times 21{,}648{,}301$$

Estimate: $\approx (4.87\times10^5)(2.16\times10^7)\approx 1.05\times10^{13}$.

**Denominator:** $19{,}367\times0.05\approx 968$, i.e. about $9.7\times10^2$.

**Divide:**
$$\frac{1.05\times10^{13}}{9.7\times10^{2}}\approx 1.09\times10^{10}$$

That is about $10{,}900{,}000{,}000$, closest to $10{,}000{,}000{,}000$. (The next choice, $100{,}000{,}000{,}000$, is nearly ten times too big, and $1{,}000{,}000{,}000$ is nearly ten times too small.)

ANSWER 1: D

# Problem 2:

We need whole numbers with $\Diamond\times\triangle=36$, maximizing $\Diamond+\triangle$.

List factor pairs and sums:
- $1\cdot36 \Rightarrow 37$
- $2\cdot18 \Rightarrow 20$
- $3\cdot12 \Rightarrow 15$
- $4\cdot9 \Rightarrow 13$
- $6\cdot6 \Rightarrow 12$

The sum is largest when the factors are as far apart as possible: $1+36=37$. (Choices 12, 13, 15, 20 are the *other* pairs' sums — 12 is actually the smallest.)

ANSWER 2: E

# Problem 3:

Both sums have the same number of terms: from 1901 to 1993 there are $1993-1901+1=93$ terms, and from 101 to 193 there are also 93 terms.

Pair them term by term: $1901-101=1800$, $1902-102=1800$, …, $1993-193=1800$.

So the difference is
$$93\times1800=167{,}400.$$

(No need to compute either huge sum separately.)

ANSWER 3: A

# Problem 4:

We fill $5\ \_\ 4\ \_\ 6\ \_\ 3$ using $+,-,\times$ each exactly once. Try all 6 orders (remember $\times$ comes first):

- $+,-,\times$: $5+4-6\times3=5+4-18=-9$
- $+,\times,-$: $5+4\times6-3=5+24-3=26$
- $-,+,\times$: $5-4+6\times3=1+18=\mathbf{19}$
- $-,\times,+$: $5-4\times6+3=5-24+3=-16$
- $\times,+,-$: $5\times4+6-3=20+3=23$
- $\times,-,+$: $5\times4-6+3=20-6+3=17$

The only achievable value among the choices is $19$. (9, 10, 15, 16 never appear.)

ANSWER 4: E

# Problem 5:

Only one entry changed. The total was too large by
$$980{,}000-98{,}000=882{,}000\text{ dollars.}$$

The mean is the total divided by 1000 families, so the means differ by
$$\frac{882{,}000}{1000}=882\text{ dollars.}$$

(The $\$882{,}000$ choice is the error in the *total*, not the mean; $\$8200$ and the range are irrelevant.)

ANSWER 5: A

# Problem 6:

Savings:
- Jacket: $40\%$ of $80 = 0.40\times80=\$32$
- Shirt: $55\%$ of $40 = 0.55\times40=\$22$

Total saved $=32+22=\$54$. Total original price $=80+40=\$120$.

$$\frac{54}{120}=0.45=45\%$$

(Note it is **not** the average $\tfrac{40+55}{2}=47\tfrac12\%$, because the jacket costs more, pulling the percentage toward $40\%$.)

ANSWER 6: A

# Problem 7:

Sale price: $50\%$ below original means half price:
$$180\times0.50=\$90.$$

Saturday: an extra $20\%$ off that sale price, so you pay $80\%$ of \$90:
$$90\times0.80=\$72.$$

(\$54 would come from wrongly taking $70\%$ off the original; \$90 forgets the extra discount.)

ANSWER 7: B

# Problem 8:

Factor: $2020=2^2\cdot5\cdot101$, so it has $(2+1)(1+1)(1+1)=12$ divisors:
$$1,2,4,5,10,20,101,202,404,505,1010,2020.$$

A number has **more than 3** factors unless it is $1$ (one factor), a prime (two factors), or the square of a prime (three factors).

Among the list:
- $1$ → 1 factor (exclude)
- $2,5,101$ → primes, 2 factors (exclude)
- $4=2^2$ → 3 factors: $1,2,4$ (exclude)

All others ($10,20,202,404,505,1010,2020$) have more than 3 factors: that's $12-5=7$ numbers.

ANSWER 8: B

# Problem 9:

Count the cubes that touch **no** side and **not** the bottom, then subtract.

Set coordinates $1..4$ in each direction. A cube avoids all four vertical sides only if its horizontal position is $2$ or $3$ in both horizontal directions: $2\times2=4$ choices. It avoids the bottom if its level is $2,3,$ or $4$: 3 choices (the top is not mentioned, so touching the top is allowed).

Interior-and-not-bottom cubes: $2\times2\times3=12$.

Cubes touching a side or the bottom: $64-12=52$.

(48 would be the count of cubes on the outer shell of a $4\times4\times4$ block including the top face — but the top-face cubes not on a side don't touch a side or bottom, so 48 is wrong here.)

ANSWER 9: B

# Problem 10:

Each hop is up ($+1$) or down ($-1$). We need 6 hops summing to 0 (so 3 ups and 3 downs) and the height must never go below ground (he can't hop down from the ground).

These are exactly the "balanced sequences" counted by the Catalan number $C_3=\frac{1}{4}\binom{6}{3}=\frac{20}{4}=5$.

Listing them (U = up, D = down):
1. UUUDDD
2. UUDUDD
3. UUDDUD
4. UDUUDD
5. UDUDUD

Total: **5**. (The 20 sequences of three U's and three D's include 15 that dip below ground, like DUUUDD, which are impossible.)

ANSWER 10: B

Problem 1:

We are finding the value closest to the expression, which is intended to be

\[
\frac{(487{,}000)(12{,}027{,}300)+(9{,}621{,}001)(487{,}000)}
{(19{,}367)(0.05)}.
\]

Factor out \(487{,}000\) from the numerator:

\[
\frac{487{,}000(12{,}027{,}300+9{,}621{,}001)}
{19{,}367(0.05)}.
\]

Add inside the parentheses:

\[
12{,}027{,}300+9{,}621{,}001=21{,}648{,}301.
\]

Estimate the numerator:

\[
487{,}000\cdot 21{,}648{,}301\approx 4.87\times 10^5\cdot 2.165\times 10^7
\approx 1.05\times 10^{13}.
\]

Now estimate the denominator:

\[
19{,}367(0.05)=968.35\approx 10^3.
\]

Thus the whole value is approximately

\[
\frac{1.05\times 10^{13}}{10^3}\approx 1.05\times 10^{10}.
\]

So it is closest to \(10{,}000{,}000{,}000\).

\[
\boxed{\text{D}}
\]

ANSWER 1: D

Problem 2:

We need whole numbers \(\diamond\) and \(\triangle\) such that

\[
\diamond\cdot\triangle=36.
\]

To make their sum as large as possible, use the most unequal positive factor pair of 36:

\[
1\cdot 36=36.
\]

Then

\[
1+36=37.
\]

Other factor pairs give smaller sums:

\[
2+18=20,\qquad 3+12=15,\qquad 4+9=13,\qquad 6+6=12.
\]

The largest is \(37\).

\[
\boxed{\text{E}}
\]

ANSWER 2: E

Problem 3:

There are 93 terms in each sum:

\[
1901,1902,\dots,1993
\]

because

\[
1993-1901+1=93.
\]

Likewise,

\[
193-101+1=93.
\]

Match corresponding terms:

\[
1901-101=1800,
\]
\[
1902-102=1800,
\]

and so on. Every paired difference is \(1800\).

Therefore the total difference is

\[
93\cdot 1800=167{,}400.
\]

\[
\boxed{\text{A}}
\]

ANSWER 3: A

Problem 4:

We must use \(+\), \(-\), and \(\times\) exactly once in

\[
5\ \_\ 4\ \_\ 6\ \_\ 3.
\]

Remember that multiplication is done before addition and subtraction.

Try the possible placements:

\[
5+4-6\times3=9-18=-9
\]

\[
5+4\times6-3=5+24-3=26
\]

\[
5-4+6\times3=5-4+18=19
\]

This equals one of the choices.

For completeness, the remaining orders give:

\[
5-4\times6+3=5-24+3=-16,
\]

\[
5\times4+6-3=23,
\]

\[
5\times4-6+3=17.
\]

Only \(19\) appears among the answer choices.

\[
\boxed{\text{E}}
\]

ANSWER 4: E

Problem 5:

The actual largest income was

\[
98{,}000.
\]

It was incorrectly entered as

\[
980{,}000.
\]

So the total income in the computer data is too large by

\[
980{,}000-98{,}000=882{,}000.
\]

There are \(1000\) families, so the mean is too large by

\[
\frac{882{,}000}{1000}=882.
\]

Thus the difference between the two means is \(882\) dollars.

\[
\boxed{\text{A}}
\]

ANSWER 5: A

Problem 6:

The jacket originally cost \(80\) dollars and had a \(40\%\) discount.

Amount saved on jacket:

\[
0.40(80)=32.
\]

The shirt originally cost \(40\) dollars and had a \(55\%\) discount.

Amount saved on shirt:

\[
0.55(40)=22.
\]

Total saved:

\[
32+22=54.
\]

Total original price:

\[
80+40=120.
\]

The percent saved is

\[
\frac{54}{120}=0.45=45\%.
\]

\[
\boxed{\text{A}}
\]

ANSWER 6: A

Problem 7:

The original price is \(180\) dollars.

First, the store reduces the price by \(50\%\):

\[
180(0.50)=90.
\]

Then the Saturday discount takes another \(20\%\) off the sale price of \(90\):

\[
90(0.80)=72.
\]

The Saturday price is

\[
\$72.
\]

It is not \(\$54\), because the discounts are applied one after the other, not by adding \(50\%+20\%=70\%\) of the original price.

\[
\boxed{\text{B}}
\]

ANSWER 7: B

Problem 8:

First factor \(2020\):

\[
2020=2^2\cdot 5\cdot 101.
\]

So the number of factors of \(2020\) is

\[
(2+1)(1+1)(1+1)=12.
\]

We need to exclude factors that have at most 3 factors themselves.

- \(1\) has only 1 factor.
- The primes \(2,5,101\) each have 2 factors.
- \(4=2^2\) has 3 factors: \(1,2,4\).

Thus there are

\[
1+3+1=5
\]

factors of \(2020\) with at most 3 factors.

Therefore, the number with more than 3 factors is

\[
12-5=7.
\]

\[
\boxed{\text{B}}
\]

ANSWER 8: B

Problem 9:

There are

\[
4\cdot4\cdot4=64
\]

small cubes total.

It is easier to count the cubes that do **not** touch a side or the bottom.

To avoid touching a side, a cube must be in one of the middle \(2\times2\) positions horizontally:

\[
2\cdot2=4
\]

possible positions per layer.

To avoid touching the bottom, it can be in any of the upper 3 layers.

So the number that touch neither a side nor the bottom is

\[
2\cdot2\cdot3=12.
\]

Therefore the number that touch a side or the bottom is

\[
64-12=52.
\]

\[
\boxed{\text{B}}
\]

ANSWER 9: B

Problem 10:

Buzz makes 6 hops and ends where he started, on the ground. Therefore, he must make exactly 3 up-hops and 3 down-hops.

However, he cannot hop below the ground.

The valid sequences are:

\[
\text{up-up-up-down-down-down}
\]

\[
\text{up-up-down-up-down-down}
\]

\[
\text{up-up-down-down-up-down}
\]

\[
\text{up-down-up-up-down-down}
\]

\[
\text{up-down-up-down-up-down}
\]

There are \(5\) valid sequences.

For example, a sequence beginning with “down” is impossible, because Buzz would go below ground immediately.

\[
\boxed{\text{B}}
\]

ANSWER 10: B

Problem 1:
We are asked to estimate
[(487,000)(12,027,300) + (9,621,001)(487,000)] / [(19,367)(.05)]
to the nearest power of 10.

Factor the numerator:
487,000 × (12,027,300 + 9,621,001)
= 487,000 × 21,648,301.

Approximate for order of magnitude:
487,000 ≈ 5 × 10^5
21,648,301 ≈ 2.2 × 10^7
So numerator ≈ 5×10^5 × 2.2×10^7 ≈ 11×10^12 = 1.1×10^13.

Denominator:
19,367 × .05 ≈ 20,000 × .05 = 1,000 = 10^3.

Hence ratio ≈ 1.1×10^13 / 10^3 = 1.1×10^10 ≈ 11,000,000,000.

This is closest to 10,000,000,000, not 1,000,000,000 which is 10 times too small, nor 100,000,000,000 which is 10 times too large.
ANSWER 1: D

Problem 2:
We need whole numbers ◇, △ with product 36 to maximize the sum.

Whole numbers with product 36 must be positive factor pairs of 36. List them:
1 × 36, sum = 37
2 × 18, sum = 20
3 × 12, sum = 15
4 × 9, sum = 13
6 × 6, sum = 12

The sum gets smaller as the factors get closer together. The largest is obtained from the most extreme pair 1 and 36.

So 37 is largest, not 20 or 12.
ANSWER 2: E

Problem 3:
We are asked for
(1901+1902+...+1993) − (101+102+...+193).

Count terms: 1993−1901+1 = 93, and 193−101+1 = 93. So both sums have 93 terms.

Pair them in order:
1901−101 = 1800
1902−102 = 1800
...
1993−193 = 1800

There are 93 such differences, each 1800. So the total difference is
93 × 1800 = 93 × 18 × 100 = 1674 × 100 = 167,400.

This rules out e.g. adding instead of subtracting, or miscounting terms.
ANSWER 3: A

Problem 4:
We must place +, −, × each exactly once in 5 __ 4 __ 6 __ 3 and evaluate with normal precedence, × before + −.

List all 6 orderings:

1) 5+4−6×3 = 9−18 = −9
2) 5+4×6−3 = 5+24−3 = 26
3) 5−4+6×3 = 1+18 = 19
4) 5−4×6+3 = 5−24+3 = −16
5) 5×4+6−3 = 20+6−3 = 23
6) 5×4−6+3 = 20−6+3 = 17

The attainable values are −9, 26, 19, −16, 23, 17.

Of the choices 9,10,15,16,19, only 19 occurs, from 5−4+6×3.
ANSWER 4: E

Problem 5:
There are 1000 families. Mean = sum / 1000.

Only one income is wrong: true largest 98,000 entered as 980,000.
Increase in sum = 980,000 − 98,000 = 882,000.

Increase in mean = increase in sum divided by 1000:
882,000 / 1000 = 882.

So the incorrect mean is 882 dollars higher. This is not 980, which would ignore subtracting the correct value, nor hundreds of thousands, which forget to divide by 1000.
ANSWER 5: A

Problem 6:
Original prices total 80+40 = 120 dollars.

Saved on jacket: 40% of 80 = 0.40×80 = 32 dollars.
Saved on shirt: 55% of 40 = 0.55×40 = 22 dollars.
Total saved = 32+22 = 54 dollars.

As percent of original total:
54 / 120 = 9 / 20 = 45%.

Note 47½% is the simple average of 40% and 55%, tempting but wrong because the bases 80 and 40 are different.
ANSWER 6: A

Problem 7:
Original $180. Sale is 50% below original, so sale price is 50% of original:
0.50 × 180 = $90.

Saturday is an additional 20% off the sale price, so pay 100%−20% = 80% of sale price:
0.80 × 90 = $72.

$54 would be 70% off at once, which is wrong because successive discounts multiply, not add: 0.5×0.8 = 0.4 of original, not 0.3.
ANSWER 7: B

Problem 8:
Factor 2020 = 202×10 = 2×101×2×5 = 2^2 × 5^1 × 101^1.
Number of divisors = (2+1)(1+1)(1+1) = 12. They are:
1, 2, 4, 5, 10, 20, 101, 202, 404, 505, 1010, 2020.

A number has >3 factors unless it is 1 (1 factor), prime (2 factors), or square of a prime (3 factors).

Check:
1: 1 factor - no
2: prime - no
4 = 2^2: 3 factors - no, exactly 3 not more
5: prime - no
10 = 2×5: 4 factors - yes
20: 6 factors - yes
101: prime - no
202 = 2×101: 4 factors - yes
404 = 2^2×101: 6 factors - yes
505 = 5×101: 4 factors - yes
1010 = 2×5×101: 8 factors - yes
2020: 12 factors - yes

Yes count: 10,20,202,404,505,1010,2020 = 7.

Including 4 would give 8, but 4 has exactly 3, not more than 3.
ANSWER 8: B

Problem 9:
There are 4×4×4 = 64 small cubes. We want those that touch a vertical side or the bottom. The top does not count unless the cube also touches a side or bottom.

Count the complement: cubes touching neither a side nor the bottom.
To avoid the four sides, the cube must be interior in x and y: 2 by 2 = 4 per layer.
To avoid the bottom, it must not be in the bottom layer: so in the top 3 layers.

Complement = 2×2×3 = 12 cubes. These are the only ones that fail.

Hence desired = 64 − 12 = 52.

If all 6 faces counted, answer would be 64−8 = 56, which is not listed, confirming the top is excluded.
ANSWER 9: B

Problem 10:
6 hops, start and end at ground, one step up (U) or down (D) each time. To return, need 3 U and 3 D. We cannot go below ground, so height must stay ≥0.

In particular first hop must be U. This is counting Dyck paths of length 6.

Enumerate:
U U U D D D
U U D U D D
U U D D U D
U D U U D D
U D U D U D

That is 5 ways. This is the Catalan number C3 = 5.

Unrestricted 3U-3D would be C(6,3)=20, not listed, so the stay-above-ground condition is essential.
ANSWER 10: B

Here are the step-by-step solutions for each problem:

---

### Problem 1:

**Restatement of the problem:**  
We want to find the value closest to:
$$\frac{(487{,}000)(12{,}027{,}300) + (9{,}621{,}001)(487{,}000)}{(19{,}367)(0.05)}$$

**Step-by-step solution:**
1. Factor out the common term $487{,}000$ from the numerator:
   $$\text{Numerator} = 487{,}000 \times (12{,}027{,}300 + 9{,}621{,}001)$$
2. Add the terms inside the parentheses:
   $$12{,}027{,}300 + 9{,}621{,}001 = 21{,}648{,}301$$
3. Approximate the terms using scientific notation:
   - $487{,}000 \approx 5 \times 10^5$ (or more precisely, $4.87 \times 10^5$)
   - $21{,}648{,}301 \approx 2.16 \times 10^7$
   - Numerator $\approx (4.87 \times 10^5) \times (2.165 \times 10^7) \approx 1.05 \times 10^{13}$
4. Now calculate the denominator:
   $$\text{Denominator} = 19{,}367 \times 0.05 \approx 20{,}000 \times 0.05 = 1{,}000 = 10^3$$
5. Divide the numerator by the denominator:
   $$\frac{1.05 \times 10^{13}}{10^3} \approx 1.05 \times 10^{10} = 10{,}500{,}000{,}000$$
   This is closest to $10{,}000{,}000{,}000$ ($10^{10}$).

ANSWER 1: D

---

### Problem 2:

**Restatement of the problem:**  
Given that $\diamond$ and $\triangle$ are whole numbers and $\diamond \times \triangle = 36$, find the largest possible value of $\diamond + \triangle$.

**Step-by-step solution:**
1. The pairs of whole numbers (positive factors) that multiply to $36$ are:
   - $1 \times 36 = 36 \implies 1 + 36 = 37$
   - $2 \times 18 = 36 \implies 2 + 18 = 20$
   - $3 \times 12 = 36 \implies 3 + 12 = 15$
   - $4 \times 9 = 36 \implies 4 + 9 = 13$
   - $6 \times 6 = 36 \implies 6 + 6 = 12$
2. For two numbers with a fixed product, their sum is maximized when the two numbers are as far apart as possible.
3. The maximum sum occurs with the pair $1$ and $36$, which gives:
   $$1 + 36 = 37$$

ANSWER 2: E

---

### Problem 3:

**Restatement of the problem:**  
Calculate the difference:
$$(1901 + 1902 + 1903 + \dots + 1993) - (101 + 102 + 103 + \dots + 193)$$

**Step-by-step solution:**
1. Notice that both sums have the same number of terms: from $1$ to $93$, which is $93$ terms.
2. Group the terms into pairs by matching the $k$-th term of the first sum with the $k$-th term of the second sum:
   $$(1901 - 101) + (1902 - 102) + (1903 - 103) + \dots + (1993 - 193)$$
3. The difference between each corresponding pair is constant:
   $$1900 + k - (100 + k) = 1800$$
4. Since there are $93$ pairs, the total value is:
   $$93 \times 1800 = 93 \times 18 \times 100$$
5. Multiply $93 \times 18$:
   $$93 \times 18 = 93 \times (20 - 2) = 1860 - 186 = 1674$$
   $$1674 \times 100 = 167{,}400$$

ANSWER 3: A

---

### Problem 4:

**Restatement of the problem:**  
Each of the operations $+$, $-$, and $\times$ is used exactly once in the blanks of $5 \text{ \underline{ } } 4 \text{ \underline{ } } 6 \text{ \underline{ } } 3$. Determine which of the given choices could be the result.

**Step-by-step solution:**
1. Remember standard order of operations: multiplication ($\times$) is performed before addition and subtraction ($+$ and $-$).
2. Test all $6$ possible arrangements of the three signs:
   - $5 + 4 - 6 \times 3 = 5 + 4 - 18 = -9$
   - $5 + 4 \times 6 - 3 = 5 + 24 - 3 = 26$
   - $5 - 4 + 6 \times 3 = 5 - 4 + 18 = 19$
   - $5 - 4 \times 6 + 3 = 5 - 24 + 3 = -16$
   - $5 \times 4 + 6 - 3 = 20 + 6 - 3 = 23$
   - $5 \times 4 - 6 + 3 = 20 - 6 + 3 = 17$
3. Among the choices ($9, 10, 15, 16, 19$), the only value that appears in our list of possible outcomes is $19$.

ANSWER 4: E

---

### Problem 5:

**Restatement of the problem:**  
A set of $1{,}000$ incomes has one error: the largest income was entered as $\$980{,}000$ instead of $\$98{,}000$. Find the difference between the mean of the incorrect data and the mean of the actual data.

**Step-by-step solution:**
1. Let the correct sum of the $1{,}000$ incomes be $S_{\text{actual}}$.
2. The incorrect sum, $S_{\text{incorrect}}$, differs from the actual sum only by the error in the largest income:
   $$S_{\text{incorrect}} - S_{\text{actual}} = 980{,}000 - 98{,}000 = 882{,}000$$
3. The difference between the means is:
   $$\text{Mean}_{\text{incorrect}} - \text{Mean}_{\text{actual}} = \frac{S_{\text{incorrect}} - S_{\text{actual}}}{1{,}000} = \frac{882{,}000}{1{,}000} = 882\text{ dollars}$$

ANSWER 5: A

---

### Problem 6:

**Restatement of the problem:**  
A jacket originally cost $\$80$ and was discounted by $40\%$. A shirt originally cost $\$40$ and was discounted by $55\%$. Find the total amount saved as a percentage of the total original price.

**Step-by-step solution:**
1. Find the amount saved on the jacket:
   $$\text{Savings on jacket} = 40\% \text{ of } \$80 = 0.40 \times 80 = \$32$$
2. Find the amount saved on the shirt:
   $$\text{Savings on shirt} = 55\% \text{ of } \$40 = 0.55 \times 40 = \$22$$
3. Calculate the total savings:
   $$\text{Total savings} = \$32 + \$22 = \$54$$
4. Calculate the total original cost:
   $$\text{Total original price} = \$80 + \$40 = \$120$$
5. Find the percentage saved:
   $$\frac{54}{120} = \frac{9}{20} = \frac{45}{100} = 45\%$$

ANSWER 6: A

---

### Problem 7:

**Restatement of the problem:**  
A coat has an original price of $\$180$. The outlet sale price is $50\%$ off the original price, and on Saturday an additional $20\%$ is discounted from the sale price. Find the Saturday price.

**Step-by-step solution:**
1. Calculate the standard sale price:
   $$\text{Sale price} = \$180 - (0.50 \times \$180) = \$90$$
2. Calculate the Saturday price by taking an additional $20\%$ off the $\$90$ sale price:
   $$\text{Saturday discount} = 0.20 \times \$90 = \$18$$
   $$\text{Saturday price} = \$90 - \$18 = \$72$$
   *(Alternatively: $180 \times 0.50 \times 0.80 = 180 \times 0.40 = \$72$.)*

ANSWER 7: B

---

### Problem 8:

**Restatement of the problem:**  
Find how many factors of $2020$ have more than $3$ factors.

**Step-by-step solution:**
1. Find the prime factorization of $2020$:
   $$2020 = 20 \t

# Problem 1:

The expression is

$$\frac{(487{,}000)(12{,}027{,}300)+(9{,}621{,}001)(487{,}000)}{(19{,}367)(0.05)}$$

**Factor the numerator:** both terms share the factor 487,000.

$$487{,}000\,(12{,}027{,}300+9{,}621{,}001)=487{,}000\times 21{,}648{,}301$$

Estimate: $\approx (4.87\times10^5)(2.16\times10^7)\approx 1.05\times10^{13}$.

**Denominator:** $19{,}367\times0.05\approx 968$, i.e. about $9.7\times10^2$.

**Divide:**
$$\frac{1.05\times10^{13}}{9.7\times10^{2}}\approx 1.09\times10^{10}$$

That is about $10{,}900{,}000{,}000$, closest to $10{,}000{,}000{,}000$. (The next choice, $100{,}000{,}000{,}000$, is nearly ten times too big, and $1{,}000{,}000{,}000$ is nearly ten times too small.)

ANSWER 1: D

# Problem 2:

We need whole numbers with $\Diamond\times\triangle=36$, maximizing $\Diamond+\triangle$.

List factor pairs and sums:
- $1\cdot36 \Rightarrow 37$
- $2\cdot18 \Rightarrow 20$
- $3\cdot12 \Rightarrow 15$
- $4\cdot9 \Rightarrow 13$
- $6\cdot6 \Rightarrow 12$

The sum is largest when the factors are as far apart as possible: $1+36=37$. (Choices 12, 13, 15, 20 are the *other* pairs' sums — 12 is actually the smallest.)

ANSWER 2: E

# Problem 3:

Both sums have the same number of terms: from 1901 to 1993 there are $1993-1901+1=93$ terms, and from 101 to 193 there are also 93 terms.

Pair them term by term: $1901-101=1800$, $1902-102=1800$, …, $1993-193=1800$.

So the difference is
$$93\times1800=167{,}400.$$

(No need to compute either huge sum separately.)

ANSWER 3: A

# Problem 4:

We fill $5\ \_\ 4\ \_\ 6\ \_\ 3$ using $+,-,\times$ each exactly once. Try all 6 orders (remember $\times$ comes first):

- $+,-,\times$: $5+4-6\times3=5+4-18=-9$
- $+,\times,-$: $5+4\times6-3=5+24-3=26$
- $-,+,\times$: $5-4+6\times3=1+18=\mathbf{19}$
- $-,\times,+$: $5-4\times6+3=5-24+3=-16$
- $\times,+,-$: $5\times4+6-3=20+3=23$
- $\times,-,+$: $5\times4-6+3=20-6+3=17$

The only achievable value among the choices is $19$. (9, 10, 15, 16 never appear.)

ANSWER 4: E

# Problem 5:

Only one entry changed. The total was too large by
$$980{,}000-98{,}000=882{,}000\text{ dollars.}$$

The mean is the total divided by 1000 families, so the means differ by
$$\frac{882{,}000}{1000}=882\text{ dollars.}$$

(The $\$882{,}000$ choice is the error in the *total*, not the mean; $\$8200$ and the range are irrelevant.)

ANSWER 5: A

# Problem 6:

Savings:
- Jacket: $40\%$ of $80 = 0.40\times80=\$32$
- Shirt: $55\%$ of $40 = 0.55\times40=\$22$

Total saved $=32+22=\$54$. Total original price $=80+40=\$120$.

$$\frac{54}{120}=0.45=45\%$$

(Note it is **not** the average $\tfrac{40+55}{2}=47\tfrac12\%$, because the jacket costs more, pulling the percentage toward $40\%$.)

ANSWER 6: A

# Problem 7:

Sale price: $50\%$ below original means half price:
$$180\times0.50=\$90.$$

Saturday: an extra $20\%$ off that sale price, so you pay $80\%$ of \$90:
$$90\times0.80=\$72.$$

(\$54 would come from wrongly taking $70\%$ off the original; \$90 forgets the extra discount.)

ANSWER 7: B

# Problem 8:

Factor: $2020=2^2\cdot5\cdot101$, so it has $(2+1)(1+1)(1+1)=12$ divisors:
$$1,2,4,5,10,20,101,202,404,505,1010,2020.$$

A number has **more than 3** factors unless it is $1$ (one factor), a prime (two factors), or the square of a prime (three factors).

Among the list:
- $1$ → 1 factor (exclude)
- $2,5,101$ → primes, 2 factors (exclude)
- $4=2^2$ → 3 factors: $1,2,4$ (exclude)

All others ($10,20,202,404,505,1010,2020$) have more than 3 factors: that's $12-5=7$ numbers.

ANSWER 8: B

# Problem 9:

Count the cubes that touch **no** side and **not** the bottom, then subtract.

Set coordinates $1..4$ in each direction. A cube avoids all four vertical sides only if its horizontal position is $2$ or $3$ in both horizontal directions: $2\times2=4$ choices. It avoids the bottom if its level is $2,3,$ or $4$: 3 choices (the top is not mentioned, so touching the top is allowed).

Interior-and-not-bottom cubes: $2\times2\times3=12$.

Cubes touching a side or the bottom: $64-12=52$.

(48 would be the count of cubes on the outer shell of a $4\times4\times4$ block including the top face — but the top-face cubes not on a side don't touch a side or bottom, so 48 is wrong here.)

ANSWER 9: B

# Problem 10:

Each hop is up ($+1$) or down ($-1$). We need 6 hops summing to 0 (so 3 ups and 3 downs) and the height must never go below ground (he can't hop down from the ground).

These are exactly the "balanced sequences" counted by the Catalan number $C_3=\frac{1}{4}\binom{6}{3}=\frac{20}{4}=5$.

Listing them (U = up, D = down):
1. UUUDDD
2. UUDUDD
3. UUDDUD
4. UDUUDD
5. UDUDUD

Total: **5**. (The 20 sequences of three U's and three D's include 15 that dip below ground, like DUUUDD, which are impossible.)

ANSWER 10: B

Problem 1:

We are finding the value closest to the expression, which is intended to be

\[
\frac{(487{,}000)(12{,}027{,}300)+(9{,}621{,}001)(487{,}000)}
{(19{,}367)(0.05)}.
\]

Factor out \(487{,}000\) from the numerator:

\[
\frac{487{,}000(12{,}027{,}300+9{,}621{,}001)}
{19{,}367(0.05)}.
\]

Add inside the parentheses:

\[
12{,}027{,}300+9{,}621{,}001=21{,}648{,}301.
\]

Estimate the numerator:

\[
487{,}000\cdot 21{,}648{,}301\approx 4.87\times 10^5\cdot 2.165\times 10^7
\approx 1.05\times 10^{13}.
\]

Now estimate the denominator:

\[
19{,}367(0.05)=968.35\approx 10^3.
\]

Thus the whole value is approximately

\[
\frac{1.05\times 10^{13}}{10^3}\approx 1.05\times 10^{10}.
\]

So it is closest to \(10{,}000{,}000{,}000\).

\[
\boxed{\text{D}}
\]

ANSWER 1: D

Problem 2:

We need whole numbers \(\diamond\) and \(\triangle\) such that

\[
\diamond\cdot\triangle=36.
\]

To make their sum as large as possible, use the most unequal positive factor pair of 36:

\[
1\cdot 36=36.
\]

Then

\[
1+36=37.
\]

Other factor pairs give smaller sums:

\[
2+18=20,\qquad 3+12=15,\qquad 4+9=13,\qquad 6+6=12.
\]

The largest is \(37\).

\[
\boxed{\text{E}}
\]

ANSWER 2: E

Problem 3:

There are 93 terms in each sum:

\[
1901,1902,\dots,1993
\]

because

\[
1993-1901+1=93.
\]

Likewise,

\[
193-101+1=93.
\]

Match corresponding terms:

\[
1901-101=1800,
\]
\[
1902-102=1800,
\]

and so on. Every paired difference is \(1800\).

Therefore the total difference is

\[
93\cdot 1800=167{,}400.
\]

\[
\boxed{\text{A}}
\]

ANSWER 3: A

Problem 4:

We must use \(+\), \(-\), and \(\times\) exactly once in

\[
5\ \_\ 4\ \_\ 6\ \_\ 3.
\]

Remember that multiplication is done before addition and subtraction.

Try the possible placements:

\[
5+4-6\times3=9-18=-9
\]

\[
5+4\times6-3=5+24-3=26
\]

\[
5-4+6\times3=5-4+18=19
\]

This equals one of the choices.

For completeness, the remaining orders give:

\[
5-4\times6+3=5-24+3=-16,
\]

\[
5\times4+6-3=23,
\]

\[
5\times4-6+3=17.
\]

Only \(19\) appears among the answer choices.

\[
\boxed{\text{E}}
\]

ANSWER 4: E

Problem 5:

The actual largest income was

\[
98{,}000.
\]

It was incorrectly entered as

\[
980{,}000.
\]

So the total income in the computer data is too large by

\[
980{,}000-98{,}000=882{,}000.
\]

There are \(1000\) families, so the mean is too large by

\[
\frac{882{,}000}{1000}=882.
\]

Thus the difference between the two means is \(882\) dollars.

\[
\boxed{\text{A}}
\]

ANSWER 5: A

Problem 6:

The jacket originally cost \(80\) dollars and had a \(40\%\) discount.

Amount saved on jacket:

\[
0.40(80)=32.
\]

The shirt originally cost \(40\) dollars and had a \(55\%\) discount.

Amount saved on shirt:

\[
0.55(40)=22.
\]

Total saved:

\[
32+22=54.
\]

Total original price:

\[
80+40=120.
\]

The percent saved is

\[
\frac{54}{120}=0.45=45\%.
\]

\[
\boxed{\text{A}}
\]

ANSWER 6: A

Problem 7:

The original price is \(180\) dollars.

First, the store reduces the price by \(50\%\):

\[
180(0.50)=90.
\]

Then the Saturday discount takes another \(20\%\) off the sale price of \(90\):

\[
90(0.80)=72.
\]

The Saturday price is

\[
\$72.
\]

It is not \(\$54\), because the discounts are applied one after the other, not by adding \(50\%+20\%=70\%\) of the original price.

\[
\boxed{\text{B}}
\]

ANSWER 7: B

Problem 8:

First factor \(2020\):

\[
2020=2^2\cdot 5\cdot 101.
\]

So the number of factors of \(2020\) is

\[
(2+1)(1+1)(1+1)=12.
\]

We need to exclude factors that have at most 3 factors themselves.

- \(1\) has only 1 factor.
- The primes \(2,5,101\) each have 2 factors.
- \(4=2^2\) has 3 factors: \(1,2,4\).

Thus there are

\[
1+3+1=5
\]

factors of \(2020\) with at most 3 factors.

Therefore, the number with more than 3 factors is

\[
12-5=7.
\]

\[
\boxed{\text{B}}
\]

ANSWER 8: B

Problem 9:

There are

\[
4\cdot4\cdot4=64
\]

small cubes total.

It is easier to count the cubes that do **not** touch a side or the bottom.

To avoid touching a side, a cube must be in one of the middle \(2\times2\) positions horizontally:

\[
2\cdot2=4
\]

possible positions per layer.

To avoid touching the bottom, it can be in any of the upper 3 layers.

So the number that touch neither a side nor the bottom is

\[
2\cdot2\cdot3=12.
\]

Therefore the number that touch a side or the bottom is

\[
64-12=52.
\]

\[
\boxed{\text{B}}
\]

ANSWER 9: B

Problem 10:

Buzz makes 6 hops and ends where he started, on the ground. Therefore, he must make exactly 3 up-hops and 3 down-hops.

However, he cannot hop below the ground.

The valid sequences are:

\[
\text{up-up-up-down-down-down}
\]

\[
\text{up-up-down-up-down-down}
\]

\[
\text{up-up-down-down-up-down}
\]

\[
\text{up-down-up-up-down-down}
\]

\[
\text{up-down-up-down-up-down}
\]

There are \(5\) valid sequences.

For example, a sequence beginning with “down” is impossible, because Buzz would go below ground immediately.

\[
\boxed{\text{B}}
\]

ANSWER 10: B

Problem 1:
We are asked to estimate
[(487,000)(12,027,300) + (9,621,001)(487,000)] / [(19,367)(.05)]
to the nearest power of 10.

Factor the numerator:
487,000 × (12,027,300 + 9,621,001)
= 487,000 × 21,648,301.

Approximate for order of magnitude:
487,000 ≈ 5 × 10^5
21,648,301 ≈ 2.2 × 10^7
So numerator ≈ 5×10^5 × 2.2×10^7 ≈ 11×10^12 = 1.1×10^13.

Denominator:
19,367 × .05 ≈ 20,000 × .05 = 1,000 = 10^3.

Hence ratio ≈ 1.1×10^13 / 10^3 = 1.1×10^10 ≈ 11,000,000,000.

This is closest to 10,000,000,000, not 1,000,000,000 which is 10 times too small, nor 100,000,000,000 which is 10 times too large.
ANSWER 1: D

Problem 2:
We need whole numbers ◇, △ with product 36 to maximize the sum.

Whole numbers with product 36 must be positive factor pairs of 36. List them:
1 × 36, sum = 37
2 × 18, sum = 20
3 × 12, sum = 15
4 × 9, sum = 13
6 × 6, sum = 12

The sum gets smaller as the factors get closer together. The largest is obtained from the most extreme pair 1 and 36.

So 37 is largest, not 20 or 12.
ANSWER 2: E

Problem 3:
We are asked for
(1901+1902+...+1993) − (101+102+...+193).

Count terms: 1993−1901+1 = 93, and 193−101+1 = 93. So both sums have 93 terms.

Pair them in order:
1901−101 = 1800
1902−102 = 1800
...
1993−193 = 1800

There are 93 such differences, each 1800. So the total difference is
93 × 1800 = 93 × 18 × 100 = 1674 × 100 = 167,400.

This rules out e.g. adding instead of subtracting, or miscounting terms.
ANSWER 3: A

Problem 4:
We must place +, −, × each exactly once in 5 __ 4 __ 6 __ 3 and evaluate with normal precedence, × before + −.

List all 6 orderings:

1) 5+4−6×3 = 9−18 = −9
2) 5+4×6−3 = 5+24−3 = 26
3) 5−4+6×3 = 1+18 = 19
4) 5−4×6+3 = 5−24+3 = −16
5) 5×4+6−3 = 20+6−3 = 23
6) 5×4−6+3 = 20−6+3 = 17

The attainable values are −9, 26, 19, −16, 23, 17.

Of the choices 9,10,15,16,19, only 19 occurs, from 5−4+6×3.
ANSWER 4: E

Problem 5:
There are 1000 families. Mean = sum / 1000.

Only one income is wrong: true largest 98,000 entered as 980,000.
Increase in sum = 980,000 − 98,000 = 882,000.

Increase in mean = increase in sum divided by 1000:
882,000 / 1000 = 882.

So the incorrect mean is 882 dollars higher. This is not 980, which would ignore subtracting the correct value, nor hundreds of thousands, which forget to divide by 1000.
ANSWER 5: A

Problem 6:
Original prices total 80+40 = 120 dollars.

Saved on jacket: 40% of 80 = 0.40×80 = 32 dollars.
Saved on shirt: 55% of 40 = 0.55×40 = 22 dollars.
Total saved = 32+22 = 54 dollars.

As percent of original total:
54 / 120 = 9 / 20 = 45%.

Note 47½% is the simple average of 40% and 55%, tempting but wrong because the bases 80 and 40 are different.
ANSWER 6: A

Problem 7:
Original $180. Sale is 50% below original, so sale price is 50% of original:
0.50 × 180 = $90.

Saturday is an additional 20% off the sale price, so pay 100%−20% = 80% of sale price:
0.80 × 90 = $72.

$54 would be 70% off at once, which is wrong because successive discounts multiply, not add: 0.5×0.8 = 0.4 of original, not 0.3.
ANSWER 7: B

Problem 8:
Factor 2020 = 202×10 = 2×101×2×5 = 2^2 × 5^1 × 101^1.
Number of divisors = (2+1)(1+1)(1+1) = 12. They are:
1, 2, 4, 5, 10, 20, 101, 202, 404, 505, 1010, 2020.

A number has >3 factors unless it is 1 (1 factor), prime (2 factors), or square of a prime (3 factors).

Check:
1: 1 factor - no
2: prime - no
4 = 2^2: 3 factors - no, exactly 3 not more
5: prime - no
10 = 2×5: 4 factors - yes
20: 6 factors - yes
101: prime - no
202 = 2×101: 4 factors - yes
404 = 2^2×101: 6 factors - yes
505 = 5×101: 4 factors - yes
1010 = 2×5×101: 8 factors - yes
2020: 12 factors - yes

Yes count: 10,20,202,404,505,1010,2020 = 7.

Including 4 would give 8, but 4 has exactly 3, not more than 3.
ANSWER 8: B

Problem 9:
There are 4×4×4 = 64 small cubes. We want those that touch a vertical side or the bottom. The top does not count unless the cube also touches a side or bottom.

Count the complement: cubes touching neither a side nor the bottom.
To avoid the four sides, the cube must be interior in x and y: 2 by 2 = 4 per layer.
To avoid the bottom, it must not be in the bottom layer: so in the top 3 layers.

Complement = 2×2×3 = 12 cubes. These are the only ones that fail.

Hence desired = 64 − 12 = 52.

If all 6 faces counted, answer would be 64−8 = 56, which is not listed, confirming the top is excluded.
ANSWER 9: B

Problem 10:
6 hops, start and end at ground, one step up (U) or down (D) each time. To return, need 3 U and 3 D. We cannot go below ground, so height must stay ≥0.

In particular first hop must be U. This is counting Dyck paths of length 6.

Enumerate:
U U U D D D
U U D U D D
U U D D U D
U D U U D D
U D U D U D

That is 5 ways. This is the Catalan number C3 = 5.

Unrestricted 3U-3D would be C(6,3)=20, not listed, so the stay-above-ground condition is essential.
ANSWER 10: B

Here are the step-by-step solutions for each problem:

---

### Problem 1:

**Restatement of the problem:**  
We want to find the value closest to:
$$\frac{(487{,}000)(12{,}027{,}300) + (9{,}621{,}001)(487{,}000)}{(19{,}367)(0.05)}$$

**Step-by-step solution:**
1. Factor out the common term $487{,}000$ from the numerator:
   $$\text{Numerator} = 487{,}000 \times (12{,}027{,}300 + 9{,}621{,}001)$$
2. Add the terms inside the parentheses:
   $$12{,}027{,}300 + 9{,}621{,}001 = 21{,}648{,}301$$
3. Approximate the terms using scientific notation:
   - $487{,}000 \approx 5 \times 10^5$ (or more precisely, $4.87 \times 10^5$)
   - $21{,}648{,}301 \approx 2.16 \times 10^7$
   - Numerator $\approx (4.87 \times 10^5) \times (2.165 \times 10^7) \approx 1.05 \times 10^{13}$
4. Now calculate the denominator:
   $$\text{Denominator} = 19{,}367 \times 0.05 \approx 20{,}000 \times 0.05 = 1{,}000 = 10^3$$
5. Divide the numerator by the denominator:
   $$\frac{1.05 \times 10^{13}}{10^3} \approx 1.05 \times 10^{10} = 10{,}500{,}000{,}000$$
   This is closest to $10{,}000{,}000{,}000$ ($10^{10}$).

ANSWER 1: D

---

### Problem 2:

**Restatement of the problem:**  
Given that $\diamond$ and $\triangle$ are whole numbers and $\diamond \times \triangle = 36$, find the largest possible value of $\diamond + \triangle$.

**Step-by-step solution:**
1. The pairs of whole numbers (positive factors) that multiply to $36$ are:
   - $1 \times 36 = 36 \implies 1 + 36 = 37$
   - $2 \times 18 = 36 \implies 2 + 18 = 20$
   - $3 \times 12 = 36 \implies 3 + 12 = 15$
   - $4 \times 9 = 36 \implies 4 + 9 = 13$
   - $6 \times 6 = 36 \implies 6 + 6 = 12$
2. For two numbers with a fixed product, their sum is maximized when the two numbers are as far apart as possible.
3. The maximum sum occurs with the pair $1$ and $36$, which gives:
   $$1 + 36 = 37$$

ANSWER 2: E

---

### Problem 3:

**Restatement of the problem:**  
Calculate the difference:
$$(1901 + 1902 + 1903 + \dots + 1993) - (101 + 102 + 103 + \dots + 193)$$

**Step-by-step solution:**
1. Notice that both sums have the same number of terms: from $1$ to $93$, which is $93$ terms.
2. Group the terms into pairs by matching the $k$-th term of the first sum with the $k$-th term of the second sum:
   $$(1901 - 101) + (1902 - 102) + (1903 - 103) + \dots + (1993 - 193)$$
3. The difference between each corresponding pair is constant:
   $$1900 + k - (100 + k) = 1800$$
4. Since there are $93$ pairs, the total value is:
   $$93 \times 1800 = 93 \times 18 \times 100$$
5. Multiply $93 \times 18$:
   $$93 \times 18 = 93 \times (20 - 2) = 1860 - 186 = 1674$$
   $$1674 \times 100 = 167{,}400$$

ANSWER 3: A

---

### Problem 4:

**Restatement of the problem:**  
Each of the operations $+$, $-$, and $\times$ is used exactly once in the blanks of $5 \text{ \underline{ } } 4 \text{ \underline{ } } 6 \text{ \underline{ } } 3$. Determine which of the given choices could be the result.

**Step-by-step solution:**
1. Remember standard order of operations: multiplication ($\times$) is performed before addition and subtraction ($+$ and $-$).
2. Test all $6$ possible arrangements of the three signs:
   - $5 + 4 - 6 \times 3 = 5 + 4 - 18 = -9$
   - $5 + 4 \times 6 - 3 = 5 + 24 - 3 = 26$
   - $5 - 4 + 6 \times 3 = 5 - 4 + 18 = 19$
   - $5 - 4 \times 6 + 3 = 5 - 24 + 3 = -16$
   - $5 \times 4 + 6 - 3 = 20 + 6 - 3 = 23$
   - $5 \times 4 - 6 + 3 = 20 - 6 + 3 = 17$
3. Among the choices ($9, 10, 15, 16, 19$), the only value that appears in our list of possible outcomes is $19$.

ANSWER 4: E

---

### Problem 5:

**Restatement of the problem:**  
A set of $1{,}000$ incomes has one error: the largest income was entered as $\$980{,}000$ instead of $\$98{,}000$. Find the difference between the mean of the incorrect data and the mean of the actual data.

**Step-by-step solution:**
1. Let the correct sum of the $1{,}000$ incomes be $S_{\text{actual}}$.
2. The incorrect sum, $S_{\text{incorrect}}$, differs from the actual sum only by the error in the largest income:
   $$S_{\text{incorrect}} - S_{\text{actual}} = 980{,}000 - 98{,}000 = 882{,}000$$
3. The difference between the means is:
   $$\text{Mean}_{\text{incorrect}} - \text{Mean}_{\text{actual}} = \frac{S_{\text{incorrect}} - S_{\text{actual}}}{1{,}000} = \frac{882{,}000}{1{,}000} = 882\text{ dollars}$$

ANSWER 5: A

---

### Problem 6:

**Restatement of the problem:**  
A jacket originally cost $\$80$ and was discounted by $40\%$. A shirt originally cost $\$40$ and was discounted by $55\%$. Find the total amount saved as a percentage of the total original price.

**Step-by-step solution:**
1. Find the amount saved on the jacket:
   $$\text{Savings on jacket} = 40\% \text{ of } \$80 = 0.40 \times 80 = \$32$$
2. Find the amount saved on the shirt:
   $$\text{Savings on shirt} = 55\% \text{ of } \$40 = 0.55 \times 40 = \$22$$
3. Calculate the total savings:
   $$\text{Total savings} = \$32 + \$22 = \$54$$
4. Calculate the total original cost:
   $$\text{Total original price} = \$80 + \$40 = \$120$$
5. Find the percentage saved:
   $$\frac{54}{120} = \frac{9}{20} = \frac{45}{100} = 45\%$$

ANSWER 6: A

---

### Problem 7:

**Restatement of the problem:**  
A coat has an original price of $\$180$. The outlet sale price is $50\%$ off the original price, and on Saturday an additional $20\%$ is discounted from the sale price. Find the Saturday price.

**Step-by-step solution:**
1. Calculate the standard sale price:
   $$\text{Sale price} = \$180 - (0.50 \times \$180) = \$90$$
2. Calculate the Saturday price by taking an additional $20\%$ off the $\$90$ sale price:
   $$\text{Saturday discount} = 0.20 \times \$90 = \$18$$
   $$\text{Saturday price} = \$90 - \$18 = \$72$$
   *(Alternatively: $180 \times 0.50 \times 0.80 = 180 \times 0.40 = \$72$.)*

ANSWER 7: B

---

### Problem 8:

**Restatement of the problem:**  
Find how many factors of $2020$ have more than $3$ factors.

**Step-by-step solution:**
1. Find the prime factorization of $2020$:
   $$2020 = 20 \t

# Problem 1:

The expression is

$$\frac{(487{,}000)(12{,}027{,}300)+(9{,}621{,}001)(487{,}000)}{(19{,}367)(0.05)}$$

**Factor the numerator:** both terms share the factor 487,000.

$$487{,}000\,(12{,}027{,}300+9{,}621{,}001)=487{,}000\times 21{,}648{,}301$$

Estimate: $\approx (4.87\times10^5)(2.16\times10^7)\approx 1.05\times10^{13}$.

**Denominator:** $19{,}367\times0.05\approx 968$, i.e. about $9.7\times10^2$.

**Divide:**
$$\frac{1.05\times10^{13}}{9.7\times10^{2}}\approx 1.09\times10^{10}$$

That is about $10{,}900{,}000{,}000$, closest to $10{,}000{,}000{,}000$. (The next choice, $100{,}000{,}000{,}000$, is nearly ten times too big, and $1{,}000{,}000{,}000$ is nearly ten times too small.)

ANSWER 1: D

# Problem 2:

We need whole numbers with $\Diamond\times\triangle=36$, maximizing $\Diamond+\triangle$.

List factor pairs and sums:
- $1\cdot36 \Rightarrow 37$
- $2\cdot18 \Rightarrow 20$
- $3\cdot12 \Rightarrow 15$
- $4\cdot9 \Rightarrow 13$
- $6\cdot6 \Rightarrow 12$

The sum is largest when the factors are as far apart as possible: $1+36=37$. (Choices 12, 13, 15, 20 are the *other* pairs' sums — 12 is actually the smallest.)

ANSWER 2: E

# Problem 3:

Both sums have the same number of terms: from 1901 to 1993 there are $1993-1901+1=93$ terms, and from 101 to 193 there are also 93 terms.

Pair them term by term: $1901-101=1800$, $1902-102=1800$, …, $1993-193=1800$.

So the difference is
$$93\times1800=167{,}400.$$

(No need to compute either huge sum separately.)

ANSWER 3: A

# Problem 4:

We fill $5\ \_\ 4\ \_\ 6\ \_\ 3$ using $+,-,\times$ each exactly once. Try all 6 orders (remember $\times$ comes first):

- $+,-,\times$: $5+4-6\times3=5+4-18=-9$
- $+,\times,-$: $5+4\times6-3=5+24-3=26$
- $-,+,\times$: $5-4+6\times3=1+18=\mathbf{19}$
- $-,\times,+$: $5-4\times6+3=5-24+3=-16$
- $\times,+,-$: $5\times4+6-3=20+3=23$
- $\times,-,+$: $5\times4-6+3=20-6+3=17$

The only achievable value among the choices is $19$. (9, 10, 15, 16 never appear.)

ANSWER 4: E

# Problem 5:

Only one entry changed. The total was too large by
$$980{,}000-98{,}000=882{,}000\text{ dollars.}$$

The mean is the total divided by 1000 families, so the means differ by
$$\frac{882{,}000}{1000}=882\text{ dollars.}$$

(The $\$882{,}000$ choice is the error in the *total*, not the mean; $\$8200$ and the range are irrelevant.)

ANSWER 5: A

# Problem 6:

Savings:
- Jacket: $40\%$ of $80 = 0.40\times80=\$32$
- Shirt: $55\%$ of $40 = 0.55\times40=\$22$

Total saved $=32+22=\$54$. Total original price $=80+40=\$120$.

$$\frac{54}{120}=0.45=45\%$$

(Note it is **not** the average $\tfrac{40+55}{2}=47\tfrac12\%$, because the jacket costs more, pulling the percentage toward $40\%$.)

ANSWER 6: A

# Problem 7:

Sale price: $50\%$ below original means half price:
$$180\times0.50=\$90.$$

Saturday: an extra $20\%$ off that sale price, so you pay $80\%$ of \$90:
$$90\times0.80=\$72.$$

(\$54 would come from wrongly taking $70\%$ off the original; \$90 forgets the extra discount.)

ANSWER 7: B

# Problem 8:

Factor: $2020=2^2\cdot5\cdot101$, so it has $(2+1)(1+1)(1+1)=12$ divisors:
$$1,2,4,5,10,20,101,202,404,505,1010,2020.$$

A number has **more than 3** factors unless it is $1$ (one factor), a prime (two factors), or the square of a prime (three factors).

Among the list:
- $1$ → 1 factor (exclude)
- $2,5,101$ → primes, 2 factors (exclude)
- $4=2^2$ → 3 factors: $1,2,4$ (exclude)

All others ($10,20,202,404,505,1010,2020$) have more than 3 factors: that's $12-5=7$ numbers.

ANSWER 8: B

# Problem 9:

Count the cubes that touch **no** side and **not** the bottom, then subtract.

Set coordinates $1..4$ in each direction. A cube avoids all four vertical sides only if its horizontal position is $2$ or $3$ in both horizontal directions: $2\times2=4$ choices. It avoids the bottom if its level is $2,3,$ or $4$: 3 choices (the top is not mentioned, so touching the top is allowed).

Interior-and-not-bottom cubes: $2\times2\times3=12$.

Cubes touching a side or the bottom: $64-12=52$.

(48 would be the count of cubes on the outer shell of a $4\times4\times4$ block including the top face — but the top-face cubes not on a side don't touch a side or bottom, so 48 is wrong here.)

ANSWER 9: B

# Problem 10:

Each hop is up ($+1$) or down ($-1$). We need 6 hops summing to 0 (so 3 ups and 3 downs) and the height must never go below ground (he can't hop down from the ground).

These are exactly the "balanced sequences" counted by the Catalan number $C_3=\frac{1}{4}\binom{6}{3}=\frac{20}{4}=5$.

Listing them (U = up, D = down):
1. UUUDDD
2. UUDUDD
3. UUDDUD
4. UDUUDD
5. UDUDUD

Total: **5**. (The 20 sequences of three U's and three D's include 15 that dip below ground, like DUUUDD, which are impossible.)

ANSWER 10: B

Problem 1:

We are finding the value closest to the expression, which is intended to be

\[
\frac{(487{,}000)(12{,}027{,}300)+(9{,}621{,}001)(487{,}000)}
{(19{,}367)(0.05)}.
\]

Factor out \(487{,}000\) from the numerator:

\[
\frac{487{,}000(12{,}027{,}300+9{,}621{,}001)}
{19{,}367(0.05)}.
\]

Add inside the parentheses:

\[
12{,}027{,}300+9{,}621{,}001=21{,}648{,}301.
\]

Estimate the numerator:

\[
487{,}000\cdot 21{,}648{,}301\approx 4.87\times 10^5\cdot 2.165\times 10^7
\approx 1.05\times 10^{13}.
\]

Now estimate the denominator:

\[
19{,}367(0.05)=968.35\approx 10^3.
\]

Thus the whole value is approximately

\[
\frac{1.05\times 10^{13}}{10^3}\approx 1.05\times 10^{10}.
\]

So it is closest to \(10{,}000{,}000{,}000\).

\[
\boxed{\text{D}}
\]

ANSWER 1: D

Problem 2:

We need whole numbers \(\diamond\) and \(\triangle\) such that

\[
\diamond\cdot\triangle=36.
\]

To make their sum as large as possible, use the most unequal positive factor pair of 36:

\[
1\cdot 36=36.
\]

Then

\[
1+36=37.
\]

Other factor pairs give smaller sums:

\[
2+18=20,\qquad 3+12=15,\qquad 4+9=13,\qquad 6+6=12.
\]

The largest is \(37\).

\[
\boxed{\text{E}}
\]

ANSWER 2: E

Problem 3:

There are 93 terms in each sum:

\[
1901,1902,\dots,1993
\]

because

\[
1993-1901+1=93.
\]

Likewise,

\[
193-101+1=93.
\]

Match corresponding terms:

\[
1901-101=1800,
\]
\[
1902-102=1800,
\]

and so on. Every paired difference is \(1800\).

Therefore the total difference is

\[
93\cdot 1800=167{,}400.
\]

\[
\boxed{\text{A}}
\]

ANSWER 3: A

Problem 4:

We must use \(+\), \(-\), and \(\times\) exactly once in

\[
5\ \_\ 4\ \_\ 6\ \_\ 3.
\]

Remember that multiplication is done before addition and subtraction.

Try the possible placements:

\[
5+4-6\times3=9-18=-9
\]

\[
5+4\times6-3=5+24-3=26
\]

\[
5-4+6\times3=5-4+18=19
\]

This equals one of the choices.

For completeness, the remaining orders give:

\[
5-4\times6+3=5-24+3=-16,
\]

\[
5\times4+6-3=23,
\]

\[
5\times4-6+3=17.
\]

Only \(19\) appears among the answer choices.

\[
\boxed{\text{E}}
\]

ANSWER 4: E

Problem 5:

The actual largest income was

\[
98{,}000.
\]

It was incorrectly entered as

\[
980{,}000.
\]

So the total income in the computer data is too large by

\[
980{,}000-98{,}000=882{,}000.
\]

There are \(1000\) families, so the mean is too large by

\[
\frac{882{,}000}{1000}=882.
\]

Thus the difference between the two means is \(882\) dollars.

\[
\boxed{\text{A}}
\]

ANSWER 5: A

Problem 6:

The jacket originally cost \(80\) dollars and had a \(40\%\) discount.

Amount saved on jacket:

\[
0.40(80)=32.
\]

The shirt originally cost \(40\) dollars and had a \(55\%\) discount.

Amount saved on shirt:

\[
0.55(40)=22.
\]

Total saved:

\[
32+22=54.
\]

Total original price:

\[
80+40=120.
\]

The percent saved is

\[
\frac{54}{120}=0.45=45\%.
\]

\[
\boxed{\text{A}}
\]

ANSWER 6: A

Problem 7:

The original price is \(180\) dollars.

First, the store reduces the price by \(50\%\):

\[
180(0.50)=90.
\]

Then the Saturday discount takes another \(20\%\) off the sale price of \(90\):

\[
90(0.80)=72.
\]

The Saturday price is

\[
\$72.
\]

It is not \(\$54\), because the discounts are applied one after the other, not by adding \(50\%+20\%=70\%\) of the original price.

\[
\boxed{\text{B}}
\]

ANSWER 7: B

Problem 8:

First factor \(2020\):

\[
2020=2^2\cdot 5\cdot 101.
\]

So the number of factors of \(2020\) is

\[
(2+1)(1+1)(1+1)=12.
\]

We need to exclude factors that have at most 3 factors themselves.

- \(1\) has only 1 factor.
- The primes \(2,5,101\) each have 2 factors.
- \(4=2^2\) has 3 factors: \(1,2,4\).

Thus there are

\[
1+3+1=5
\]

factors of \(2020\) with at most 3 factors.

Therefore, the number with more than 3 factors is

\[
12-5=7.
\]

\[
\boxed{\text{B}}
\]

ANSWER 8: B

Problem 9:

There are

\[
4\cdot4\cdot4=64
\]

small cubes total.

It is easier to count the cubes that do **not** touch a side or the bottom.

To avoid touching a side, a cube must be in one of the middle \(2\times2\) positions horizontally:

\[
2\cdot2=4
\]

possible positions per layer.

To avoid touching the bottom, it can be in any of the upper 3 layers.

So the number that touch neither a side nor the bottom is

\[
2\cdot2\cdot3=12.
\]

Therefore the number that touch a side or the bottom is

\[
64-12=52.
\]

\[
\boxed{\text{B}}
\]

ANSWER 9: B

Problem 10:

Buzz makes 6 hops and ends where he started, on the ground. Therefore, he must make exactly 3 up-hops and 3 down-hops.

However, he cannot hop below the ground.

The valid sequences are:

\[
\text{up-up-up-down-down-down}
\]

\[
\text{up-up-down-up-down-down}
\]

\[
\text{up-up-down-down-up-down}
\]

\[
\text{up-down-up-up-down-down}
\]

\[
\text{up-down-up-down-up-down}
\]

There are \(5\) valid sequences.

For example, a sequence beginning with “down” is impossible, because Buzz would go below ground immediately.

\[
\boxed{\text{B}}
\]

ANSWER 10: B

Problem 1:
We are asked to estimate
[(487,000)(12,027,300) + (9,621,001)(487,000)] / [(19,367)(.05)]
to the nearest power of 10.

Factor the numerator:
487,000 × (12,027,300 + 9,621,001)
= 487,000 × 21,648,301.

Approximate for order of magnitude:
487,000 ≈ 5 × 10^5
21,648,301 ≈ 2.2 × 10^7
So numerator ≈ 5×10^5 × 2.2×10^7 ≈ 11×10^12 = 1.1×10^13.

Denominator:
19,367 × .05 ≈ 20,000 × .05 = 1,000 = 10^3.

Hence ratio ≈ 1.1×10^13 / 10^3 = 1.1×10^10 ≈ 11,000,000,000.

This is closest to 10,000,000,000, not 1,000,000,000 which is 10 times too small, nor 100,000,000,000 which is 10 times too large.
ANSWER 1: D

Problem 2:
We need whole numbers ◇, △ with product 36 to maximize the sum.

Whole numbers with product 36 must be positive factor pairs of 36. List them:
1 × 36, sum = 37
2 × 18, sum = 20
3 × 12, sum = 15
4 × 9, sum = 13
6 × 6, sum = 12

The sum gets smaller as the factors get closer together. The largest is obtained from the most extreme pair 1 and 36.

So 37 is largest, not 20 or 12.
ANSWER 2: E

Problem 3:
We are asked for
(1901+1902+...+1993) − (101+102+...+193).

Count terms: 1993−1901+1 = 93, and 193−101+1 = 93. So both sums have 93 terms.

Pair them in order:
1901−101 = 1800
1902−102 = 1800
...
1993−193 = 1800

There are 93 such differences, each 1800. So the total difference is
93 × 1800 = 93 × 18 × 100 = 1674 × 100 = 167,400.

This rules out e.g. adding instead of subtracting, or miscounting terms.
ANSWER 3: A

Problem 4:
We must place +, −, × each exactly once in 5 __ 4 __ 6 __ 3 and evaluate with normal precedence, × before + −.

List all 6 orderings:

1) 5+4−6×3 = 9−18 = −9
2) 5+4×6−3 = 5+24−3 = 26
3) 5−4+6×3 = 1+18 = 19
4) 5−4×6+3 = 5−24+3 = −16
5) 5×4+6−3 = 20+6−3 = 23
6) 5×4−6+3 = 20−6+3 = 17

The attainable values are −9, 26, 19, −16, 23, 17.

Of the choices 9,10,15,16,19, only 19 occurs, from 5−4+6×3.
ANSWER 4: E

Problem 5:
There are 1000 families. Mean = sum / 1000.

Only one income is wrong: true largest 98,000 entered as 980,000.
Increase in sum = 980,000 − 98,000 = 882,000.

Increase in mean = increase in sum divided by 1000:
882,000 / 1000 = 882.

So the incorrect mean is 882 dollars higher. This is not 980, which would ignore subtracting the correct value, nor hundreds of thousands, which forget to divide by 1000.
ANSWER 5: A

Problem 6:
Original prices total 80+40 = 120 dollars.

Saved on jacket: 40% of 80 = 0.40×80 = 32 dollars.
Saved on shirt: 55% of 40 = 0.55×40 = 22 dollars.
Total saved = 32+22 = 54 dollars.

As percent of original total:
54 / 120 = 9 / 20 = 45%.

Note 47½% is the simple average of 40% and 55%, tempting but wrong because the bases 80 and 40 are different.
ANSWER 6: A

Problem 7:
Original $180. Sale is 50% below original, so sale price is 50% of original:
0.50 × 180 = $90.

Saturday is an additional 20% off the sale price, so pay 100%−20% = 80% of sale price:
0.80 × 90 = $72.

$54 would be 70% off at once, which is wrong because successive discounts multiply, not add: 0.5×0.8 = 0.4 of original, not 0.3.
ANSWER 7: B

Problem 8:
Factor 2020 = 202×10 = 2×101×2×5 = 2^2 × 5^1 × 101^1.
Number of divisors = (2+1)(1+1)(1+1) = 12. They are:
1, 2, 4, 5, 10, 20, 101, 202, 404, 505, 1010, 2020.

A number has >3 factors unless it is 1 (1 factor), prime (2 factors), or square of a prime (3 factors).

Check:
1: 1 factor - no
2: prime - no
4 = 2^2: 3 factors - no, exactly 3 not more
5: prime - no
10 = 2×5: 4 factors - yes
20: 6 factors - yes
101: prime - no
202 = 2×101: 4 factors - yes
404 = 2^2×101: 6 factors - yes
505 = 5×101: 4 factors - yes
1010 = 2×5×101: 8 factors - yes
2020: 12 factors - yes

Yes count: 10,20,202,404,505,1010,2020 = 7.

Including 4 would give 8, but 4 has exactly 3, not more than 3.
ANSWER 8: B

Problem 9:
There are 4×4×4 = 64 small cubes. We want those that touch a vertical side or the bottom. The top does not count unless the cube also touches a side or bottom.

Count the complement: cubes touching neither a side nor the bottom.
To avoid the four sides, the cube must be interior in x and y: 2 by 2 = 4 per layer.
To avoid the bottom, it must not be in the bottom layer: so in the top 3 layers.

Complement = 2×2×3 = 12 cubes. These are the only ones that fail.

Hence desired = 64 − 12 = 52.

If all 6 faces counted, answer would be 64−8 = 56, which is not listed, confirming the top is excluded.
ANSWER 9: B

Problem 10:
6 hops, start and end at ground, one step up (U) or down (D) each time. To return, need 3 U and 3 D. We cannot go below ground, so height must stay ≥0.

In particular first hop must be U. This is counting Dyck paths of length 6.

Enumerate:
U U U D D D
U U D U D D
U U D D U D
U D U U D D
U D U D U D

That is 5 ways. This is the Catalan number C3 = 5.

Unrestricted 3U-3D would be C(6,3)=20, not listed, so the stay-above-ground condition is essential.
ANSWER 10: B

Here are the step-by-step solutions for each problem:

---

### Problem 1:

**Restatement of the problem:**  
We want to find the value closest to:
$$\frac{(487{,}000)(12{,}027{,}300) + (9{,}621{,}001)(487{,}000)}{(19{,}367)(0.05)}$$

**Step-by-step solution:**
1. Factor out the common term $487{,}000$ from the numerator:
   $$\text{Numerator} = 487{,}000 \times (12{,}027{,}300 + 9{,}621{,}001)$$
2. Add the terms inside the parentheses:
   $$12{,}027{,}300 + 9{,}621{,}001 = 21{,}648{,}301$$
3. Approximate the terms using scientific notation:
   - $487{,}000 \approx 5 \times 10^5$ (or more precisely, $4.87 \times 10^5$)
   - $21{,}648{,}301 \approx 2.16 \times 10^7$
   - Numerator $\approx (4.87 \times 10^5) \times (2.165 \times 10^7) \approx 1.05 \times 10^{13}$
4. Now calculate the denominator:
   $$\text{Denominator} = 19{,}367 \times 0.05 \approx 20{,}000 \times 0.05 = 1{,}000 = 10^3$$
5. Divide the numerator by the denominator:
   $$\frac{1.05 \times 10^{13}}{10^3} \approx 1.05 \times 10^{10} = 10{,}500{,}000{,}000$$
   This is closest to $10{,}000{,}000{,}000$ ($10^{10}$).

ANSWER 1: D

---

### Problem 2:

**Restatement of the problem:**  
Given that $\diamond$ and $\triangle$ are whole numbers and $\diamond \times \triangle = 36$, find the largest possible value of $\diamond + \triangle$.

**Step-by-step solution:**
1. The pairs of whole numbers (positive factors) that multiply to $36$ are:
   - $1 \times 36 = 36 \implies 1 + 36 = 37$
   - $2 \times 18 = 36 \implies 2 + 18 = 20$
   - $3 \times 12 = 36 \implies 3 + 12 = 15$
   - $4 \times 9 = 36 \implies 4 + 9 = 13$
   - $6 \times 6 = 36 \implies 6 + 6 = 12$
2. For two numbers with a fixed product, their sum is maximized when the two numbers are as far apart as possible.
3. The maximum sum occurs with the pair $1$ and $36$, which gives:
   $$1 + 36 = 37$$

ANSWER 2: E

---

### Problem 3:

**Restatement of the problem:**  
Calculate the difference:
$$(1901 + 1902 + 1903 + \dots + 1993) - (101 + 102 + 103 + \dots + 193)$$

**Step-by-step solution:**
1. Notice that both sums have the same number of terms: from $1$ to $93$, which is $93$ terms.
2. Group the terms into pairs by matching the $k$-th term of the first sum with the $k$-th term of the second sum:
   $$(1901 - 101) + (1902 - 102) + (1903 - 103) + \dots + (1993 - 193)$$
3. The difference between each corresponding pair is constant:
   $$1900 + k - (100 + k) = 1800$$
4. Since there are $93$ pairs, the total value is:
   $$93 \times 1800 = 93 \times 18 \times 100$$
5. Multiply $93 \times 18$:
   $$93 \times 18 = 93 \times (20 - 2) = 1860 - 186 = 1674$$
   $$1674 \times 100 = 167{,}400$$

ANSWER 3: A

---

### Problem 4:

**Restatement of the problem:**  
Each of the operations $+$, $-$, and $\times$ is used exactly once in the blanks of $5 \text{ \underline{ } } 4 \text{ \underline{ } } 6 \text{ \underline{ } } 3$. Determine which of the given choices could be the result.

**Step-by-step solution:**
1. Remember standard order of operations: multiplication ($\times$) is performed before addition and subtraction ($+$ and $-$).
2. Test all $6$ possible arrangements of the three signs:
   - $5 + 4 - 6 \times 3 = 5 + 4 - 18 = -9$
   - $5 + 4 \times 6 - 3 = 5 + 24 - 3 = 26$
   - $5 - 4 + 6 \times 3 = 5 - 4 + 18 = 19$
   - $5 - 4 \times 6 + 3 = 5 - 24 + 3 = -16$
   - $5 \times 4 + 6 - 3 = 20 + 6 - 3 = 23$
   - $5 \times 4 - 6 + 3 = 20 - 6 + 3 = 17$
3. Among the choices ($9, 10, 15, 16, 19$), the only value that appears in our list of possible outcomes is $19$.

ANSWER 4: E

---

### Problem 5:

**Restatement of the problem:**  
A set of $1{,}000$ incomes has one error: the largest income was entered as $\$980{,}000$ instead of $\$98{,}000$. Find the difference between the mean of the incorrect data and the mean of the actual data.

**Step-by-step solution:**
1. Let the correct sum of the $1{,}000$ incomes be $S_{\text{actual}}$.
2. The incorrect sum, $S_{\text{incorrect}}$, differs from the actual sum only by the error in the largest income:
   $$S_{\text{incorrect}} - S_{\text{actual}} = 980{,}000 - 98{,}000 = 882{,}000$$
3. The difference between the means is:
   $$\text{Mean}_{\text{incorrect}} - \text{Mean}_{\text{actual}} = \frac{S_{\text{incorrect}} - S_{\text{actual}}}{1{,}000} = \frac{882{,}000}{1{,}000} = 882\text{ dollars}$$

ANSWER 5: A

---

### Problem 6:

**Restatement of the problem:**  
A jacket originally cost $\$80$ and was discounted by $40\%$. A shirt originally cost $\$40$ and was discounted by $55\%$. Find the total amount saved as a percentage of the total original price.

**Step-by-step solution:**
1. Find the amount saved on the jacket:
   $$\text{Savings on jacket} = 40\% \text{ of } \$80 = 0.40 \times 80 = \$32$$
2. Find the amount saved on the shirt:
   $$\text{Savings on shirt} = 55\% \text{ of } \$40 = 0.55 \times 40 = \$22$$
3. Calculate the total savings:
   $$\text{Total savings} = \$32 + \$22 = \$54$$
4. Calculate the total original cost:
   $$\text{Total original price} = \$80 + \$40 = \$120$$
5. Find the percentage saved:
   $$\frac{54}{120} = \frac{9}{20} = \frac{45}{100} = 45\%$$

ANSWER 6: A

---

### Problem 7:

**Restatement of the problem:**  
A coat has an original price of $\$180$. The outlet sale price is $50\%$ off the original price, and on Saturday an additional $20\%$ is discounted from the sale price. Find the Saturday price.

**Step-by-step solution:**
1. Calculate the standard sale price:
   $$\text{Sale price} = \$180 - (0.50 \times \$180) = \$90$$
2. Calculate the Saturday price by taking an additional $20\%$ off the $\$90$ sale price:
   $$\text{Saturday discount} = 0.20 \times \$90 = \$18$$
   $$\text{Saturday price} = \$90 - \$18 = \$72$$
   *(Alternatively: $180 \times 0.50 \times 0.80 = 180 \times 0.40 = \$72$.)*

ANSWER 7: B

---

### Problem 8:

**Restatement of the problem:**  
Find how many factors of $2020$ have more than $3$ factors.

**Step-by-step solution:**
1. Find the prime factorization of $2020$:
   $$2020 = 20 \t

# Problem 1:

The expression is

$$\frac{(487{,}000)(12{,}027{,}300)+(9{,}621{,}001)(487{,}000)}{(19{,}367)(0.05)}$$

**Factor the numerator:** both terms share the factor 487,000.

$$487{,}000\,(12{,}027{,}300+9{,}621{,}001)=487{,}000\times 21{,}648{,}301$$

Estimate: $\approx (4.87\times10^5)(2.16\times10^7)\approx 1.05\times10^{13}$.

**Denominator:** $19{,}367\times0.05\approx 968$, i.e. about $9.7\times10^2$.

**Divide:**
$$\frac{1.05\times10^{13}}{9.7\times10^{2}}\approx 1.09\times10^{10}$$

That is about $10{,}900{,}000{,}000$, closest to $10{,}000{,}000{,}000$. (The next choice, $100{,}000{,}000{,}000$, is nearly ten times too big, and $1{,}000{,}000{,}000$ is nearly ten times too small.)

ANSWER 1: D

# Problem 2:

We need whole numbers with $\Diamond\times\triangle=36$, maximizing $\Diamond+\triangle$.

List factor pairs and sums:
- $1\cdot36 \Rightarrow 37$
- $2\cdot18 \Rightarrow 20$
- $3\cdot12 \Rightarrow 15$
- $4\cdot9 \Rightarrow 13$
- $6\cdot6 \Rightarrow 12$

The sum is largest when the factors are as far apart as possible: $1+36=37$. (Choices 12, 13, 15, 20 are the *other* pairs' sums — 12 is actually the smallest.)

ANSWER 2: E

# Problem 3:

Both sums have the same number of terms: from 1901 to 1993 there are $1993-1901+1=93$ terms, and from 101 to 193 there are also 93 terms.

Pair them term by term: $1901-101=1800$, $1902-102=1800$, …, $1993-193=1800$.

So the difference is
$$93\times1800=167{,}400.$$

(No need to compute either huge sum separately.)

ANSWER 3: A

# Problem 4:

We fill $5\ \_\ 4\ \_\ 6\ \_\ 3$ using $+,-,\times$ each exactly once. Try all 6 orders (remember $\times$ comes first):

- $+,-,\times$: $5+4-6\times3=5+4-18=-9$
- $+,\times,-$: $5+4\times6-3=5+24-3=26$
- $-,+,\times$: $5-4+6\times3=1+18=\mathbf{19}$
- $-,\times,+$: $5-4\times6+3=5-24+3=-16$
- $\times,+,-$: $5\times4+6-3=20+3=23$
- $\times,-,+$: $5\times4-6+3=20-6+3=17$

The only achievable value among the choices is $19$. (9, 10, 15, 16 never appear.)

ANSWER 4: E

# Problem 5:

Only one entry changed. The total was too large by
$$980{,}000-98{,}000=882{,}000\text{ dollars.}$$

The mean is the total divided by 1000 families, so the means differ by
$$\frac{882{,}000}{1000}=882\text{ dollars.}$$

(The $\$882{,}000$ choice is the error in the *total*, not the mean; $\$8200$ and the range are irrelevant.)

ANSWER 5: A

# Problem 6:

Savings:
- Jacket: $40\%$ of $80 = 0.40\times80=\$32$
- Shirt: $55\%$ of $40 = 0.55\times40=\$22$

Total saved $=32+22=\$54$. Total original price $=80+40=\$120$.

$$\frac{54}{120}=0.45=45\%$$

(Note it is **not** the average $\tfrac{40+55}{2}=47\tfrac12\%$, because the jacket costs more, pulling the percentage toward $40\%$.)

ANSWER 6: A

# Problem 7:

Sale price: $50\%$ below original means half price:
$$180\times0.50=\$90.$$

Saturday: an extra $20\%$ off that sale price, so you pay $80\%$ of \$90:
$$90\times0.80=\$72.$$

(\$54 would come from wrongly taking $70\%$ off the original; \$90 forgets the extra discount.)

ANSWER 7: B

# Problem 8:

Factor: $2020=2^2\cdot5\cdot101$, so it has $(2+1)(1+1)(1+1)=12$ divisors:
$$1,2,4,5,10,20,101,202,404,505,1010,2020.$$

A number has **more than 3** factors unless it is $1$ (one factor), a prime (two factors), or the square of a prime (three factors).

Among the list:
- $1$ → 1 factor (exclude)
- $2,5,101$ → primes, 2 factors (exclude)
- $4=2^2$ → 3 factors: $1,2,4$ (exclude)

All others ($10,20,202,404,505,1010,2020$) have more than 3 factors: that's $12-5=7$ numbers.

ANSWER 8: B

# Problem 9:

Count the cubes that touch **no** side and **not** the bottom, then subtract.

Set coordinates $1..4$ in each direction. A cube avoids all four vertical sides only if its horizontal position is $2$ or $3$ in both horizontal directions: $2\times2=4$ choices. It avoids the bottom if its level is $2,3,$ or $4$: 3 choices (the top is not mentioned, so touching the top is allowed).

Interior-and-not-bottom cubes: $2\times2\times3=12$.

Cubes touching a side or the bottom: $64-12=52$.

(48 would be the count of cubes on the outer shell of a $4\times4\times4$ block including the top face — but the top-face cubes not on a side don't touch a side or bottom, so 48 is wrong here.)

ANSWER 9: B

# Problem 10:

Each hop is up ($+1$) or down ($-1$). We need 6 hops summing to 0 (so 3 ups and 3 downs) and the height must never go below ground (he can't hop down from the ground).

These are exactly the "balanced sequences" counted by the Catalan number $C_3=\frac{1}{4}\binom{6}{3}=\frac{20}{4}=5$.

Listing them (U = up, D = down):
1. UUUDDD
2. UUDUDD
3. UUDDUD
4. UDUUDD
5. UDUDUD

Total: **5**. (The 20 sequences of three U's and three D's include 15 that dip below ground, like DUUUDD, which are impossible.)

ANSWER 10: B

Problem 1:

We are finding the value closest to the expression, which is intended to be

\[
\frac{(487{,}000)(12{,}027{,}300)+(9{,}621{,}001)(487{,}000)}
{(19{,}367)(0.05)}.
\]

Factor out \(487{,}000\) from the numerator:

\[
\frac{487{,}000(12{,}027{,}300+9{,}621{,}001)}
{19{,}367(0.05)}.
\]

Add inside the parentheses:

\[
12{,}027{,}300+9{,}621{,}001=21{,}648{,}301.
\]

Estimate the numerator:

\[
487{,}000\cdot 21{,}648{,}301\approx 4.87\times 10^5\cdot 2.165\times 10^7
\approx 1.05\times 10^{13}.
\]

Now estimate the denominator:

\[
19{,}367(0.05)=968.35\approx 10^3.
\]

Thus the whole value is approximately

\[
\frac{1.05\times 10^{13}}{10^3}\approx 1.05\times 10^{10}.
\]

So it is closest to \(10{,}000{,}000{,}000\).

\[
\boxed{\text{D}}
\]

ANSWER 1: D

Problem 2:

We need whole numbers \(\diamond\) and \(\triangle\) such that

\[
\diamond\cdot\triangle=36.
\]

To make their sum as large as possible, use the most unequal positive factor pair of 36:

\[
1\cdot 36=36.
\]

Then

\[
1+36=37.
\]

Other factor pairs give smaller sums:

\[
2+18=20,\qquad 3+12=15,\qquad 4+9=13,\qquad 6+6=12.
\]

The largest is \(37\).

\[
\boxed{\text{E}}
\]

ANSWER 2: E

Problem 3:

There are 93 terms in each sum:

\[
1901,1902,\dots,1993
\]

because

\[
1993-1901+1=93.
\]

Likewise,

\[
193-101+1=93.
\]

Match corresponding terms:

\[
1901-101=1800,
\]
\[
1902-102=1800,
\]

and so on. Every paired difference is \(1800\).

Therefore the total difference is

\[
93\cdot 1800=167{,}400.
\]

\[
\boxed{\text{A}}
\]

ANSWER 3: A

Problem 4:

We must use \(+\), \(-\), and \(\times\) exactly once in

\[
5\ \_\ 4\ \_\ 6\ \_\ 3.
\]

Remember that multiplication is done before addition and subtraction.

Try the possible placements:

\[
5+4-6\times3=9-18=-9
\]

\[
5+4\times6-3=5+24-3=26
\]

\[
5-4+6\times3=5-4+18=19
\]

This equals one of the choices.

For completeness, the remaining orders give:

\[
5-4\times6+3=5-24+3=-16,
\]

\[
5\times4+6-3=23,
\]

\[
5\times4-6+3=17.
\]

Only \(19\) appears among the answer choices.

\[
\boxed{\text{E}}
\]

ANSWER 4: E

Problem 5:

The actual largest income was

\[
98{,}000.
\]

It was incorrectly entered as

\[
980{,}000.
\]

So the total income in the computer data is too large by

\[
980{,}000-98{,}000=882{,}000.
\]

There are \(1000\) families, so the mean is too large by

\[
\frac{882{,}000}{1000}=882.
\]

Thus the difference between the two means is \(882\) dollars.

\[
\boxed{\text{A}}
\]

ANSWER 5: A

Problem 6:

The jacket originally cost \(80\) dollars and had a \(40\%\) discount.

Amount saved on jacket:

\[
0.40(80)=32.
\]

The shirt originally cost \(40\) dollars and had a \(55\%\) discount.

Amount saved on shirt:

\[
0.55(40)=22.
\]

Total saved:

\[
32+22=54.
\]

Total original price:

\[
80+40=120.
\]

The percent saved is

\[
\frac{54}{120}=0.45=45\%.
\]

\[
\boxed{\text{A}}
\]

ANSWER 6: A

Problem 7:

The original price is \(180\) dollars.

First, the store reduces the price by \(50\%\):

\[
180(0.50)=90.
\]

Then the Saturday discount takes another \(20\%\) off the sale price of \(90\):

\[
90(0.80)=72.
\]

The Saturday price is

\[
\$72.
\]

It is not \(\$54\), because the discounts are applied one after the other, not by adding \(50\%+20\%=70\%\) of the original price.

\[
\boxed{\text{B}}
\]

ANSWER 7: B

Problem 8:

First factor \(2020\):

\[
2020=2^2\cdot 5\cdot 101.
\]

So the number of factors of \(2020\) is

\[
(2+1)(1+1)(1+1)=12.
\]

We need to exclude factors that have at most 3 factors themselves.

- \(1\) has only 1 factor.
- The primes \(2,5,101\) each have 2 factors.
- \(4=2^2\) has 3 factors: \(1,2,4\).

Thus there are

\[
1+3+1=5
\]

factors of \(2020\) with at most 3 factors.

Therefore, the number with more than 3 factors is

\[
12-5=7.
\]

\[
\boxed{\text{B}}
\]

ANSWER 8: B

Problem 9:

There are

\[
4\cdot4\cdot4=64
\]

small cubes total.

It is easier to count the cubes that do **not** touch a side or the bottom.

To avoid touching a side, a cube must be in one of the middle \(2\times2\) positions horizontally:

\[
2\cdot2=4
\]

possible positions per layer.

To avoid touching the bottom, it can be in any of the upper 3 layers.

So the number that touch neither a side nor the bottom is

\[
2\cdot2\cdot3=12.
\]

Therefore the number that touch a side or the bottom is

\[
64-12=52.
\]

\[
\boxed{\text{B}}
\]

ANSWER 9: B

Problem 10:

Buzz makes 6 hops and ends where he started, on the ground. Therefore, he must make exactly 3 up-hops and 3 down-hops.

However, he cannot hop below the ground.

The valid sequences are:

\[
\text{up-up-up-down-down-down}
\]

\[
\text{up-up-down-up-down-down}
\]

\[
\text{up-up-down-down-up-down}
\]

\[
\text{up-down-up-up-down-down}
\]

\[
\text{up-down-up-down-up-down}
\]

There are \(5\) valid sequences.

For example, a sequence beginning with “down” is impossible, because Buzz would go below ground immediately.

\[
\boxed{\text{B}}
\]

ANSWER 10: B

Problem 1:
We are asked to estimate
[(487,000)(12,027,300) + (9,621,001)(487,000)] / [(19,367)(.05)]
to the nearest power of 10.

Factor the numerator:
487,000 × (12,027,300 + 9,621,001)
= 487,000 × 21,648,301.

Approximate for order of magnitude:
487,000 ≈ 5 × 10^5
21,648,301 ≈ 2.2 × 10^7
So numerator ≈ 5×10^5 × 2.2×10^7 ≈ 11×10^12 = 1.1×10^13.

Denominator:
19,367 × .05 ≈ 20,000 × .05 = 1,000 = 10^3.

Hence ratio ≈ 1.1×10^13 / 10^3 = 1.1×10^10 ≈ 11,000,000,000.

This is closest to 10,000,000,000, not 1,000,000,000 which is 10 times too small, nor 100,000,000,000 which is 10 times too large.
ANSWER 1: D

Problem 2:
We need whole numbers ◇, △ with product 36 to maximize the sum.

Whole numbers with product 36 must be positive factor pairs of 36. List them:
1 × 36, sum = 37
2 × 18, sum = 20
3 × 12, sum = 15
4 × 9, sum = 13
6 × 6, sum = 12

The sum gets smaller as the factors get closer together. The largest is obtained from the most extreme pair 1 and 36.

So 37 is largest, not 20 or 12.
ANSWER 2: E

Problem 3:
We are asked for
(1901+1902+...+1993) − (101+102+...+193).

Count terms: 1993−1901+1 = 93, and 193−101+1 = 93. So both sums have 93 terms.

Pair them in order:
1901−101 = 1800
1902−102 = 1800
...
1993−193 = 1800

There are 93 such differences, each 1800. So the total difference is
93 × 1800 = 93 × 18 × 100 = 1674 × 100 = 167,400.

This rules out e.g. adding instead of subtracting, or miscounting terms.
ANSWER 3: A

Problem 4:
We must place +, −, × each exactly once in 5 __ 4 __ 6 __ 3 and evaluate with normal precedence, × before + −.

List all 6 orderings:

1) 5+4−6×3 = 9−18 = −9
2) 5+4×6−3 = 5+24−3 = 26
3) 5−4+6×3 = 1+18 = 19
4) 5−4×6+3 = 5−24+3 = −16
5) 5×4+6−3 = 20+6−3 = 23
6) 5×4−6+3 = 20−6+3 = 17

The attainable values are −9, 26, 19, −16, 23, 17.

Of the choices 9,10,15,16,19, only 19 occurs, from 5−4+6×3.
ANSWER 4: E

Problem 5:
There are 1000 families. Mean = sum / 1000.

Only one income is wrong: true largest 98,000 entered as 980,000.
Increase in sum = 980,000 − 98,000 = 882,000.

Increase in mean = increase in sum divided by 1000:
882,000 / 1000 = 882.

So the incorrect mean is 882 dollars higher. This is not 980, which would ignore subtracting the correct value, nor hundreds of thousands, which forget to divide by 1000.
ANSWER 5: A

Problem 6:
Original prices total 80+40 = 120 dollars.

Saved on jacket: 40% of 80 = 0.40×80 = 32 dollars.
Saved on shirt: 55% of 40 = 0.55×40 = 22 dollars.
Total saved = 32+22 = 54 dollars.

As percent of original total:
54 / 120 = 9 / 20 = 45%.

Note 47½% is the simple average of 40% and 55%, tempting but wrong because the bases 80 and 40 are different.
ANSWER 6: A

Problem 7:
Original $180. Sale is 50% below original, so sale price is 50% of original:
0.50 × 180 = $90.

Saturday is an additional 20% off the sale price, so pay 100%−20% = 80% of sale price:
0.80 × 90 = $72.

$54 would be 70% off at once, which is wrong because successive discounts multiply, not add: 0.5×0.8 = 0.4 of original, not 0.3.
ANSWER 7: B

Problem 8:
Factor 2020 = 202×10 = 2×101×2×5 = 2^2 × 5^1 × 101^1.
Number of divisors = (2+1)(1+1)(1+1) = 12. They are:
1, 2, 4, 5, 10, 20, 101, 202, 404, 505, 1010, 2020.

A number has >3 factors unless it is 1 (1 factor), prime (2 factors), or square of a prime (3 factors).

Check:
1: 1 factor - no
2: prime - no
4 = 2^2: 3 factors - no, exactly 3 not more
5: prime - no
10 = 2×5: 4 factors - yes
20: 6 factors - yes
101: prime - no
202 = 2×101: 4 factors - yes
404 = 2^2×101: 6 factors - yes
505 = 5×101: 4 factors - yes
1010 = 2×5×101: 8 factors - yes
2020: 12 factors - yes

Yes count: 10,20,202,404,505,1010,2020 = 7.

Including 4 would give 8, but 4 has exactly 3, not more than 3.
ANSWER 8: B

Problem 9:
There are 4×4×4 = 64 small cubes. We want those that touch a vertical side or the bottom. The top does not count unless the cube also touches a side or bottom.

Count the complement: cubes touching neither a side nor the bottom.
To avoid the four sides, the cube must be interior in x and y: 2 by 2 = 4 per layer.
To avoid the bottom, it must not be in the bottom layer: so in the top 3 layers.

Complement = 2×2×3 = 12 cubes. These are the only ones that fail.

Hence desired = 64 − 12 = 52.

If all 6 faces counted, answer would be 64−8 = 56, which is not listed, confirming the top is excluded.
ANSWER 9: B

Problem 10:
6 hops, start and end at ground, one step up (U) or down (D) each time. To return, need 3 U and 3 D. We cannot go below ground, so height must stay ≥0.

In particular first hop must be U. This is counting Dyck paths of length 6.

Enumerate:
U U U D D D
U U D U D D
U U D D U D
U D U U D D
U D U D U D

That is 5 ways. This is the Catalan number C3 = 5.

Unrestricted 3U-3D would be C(6,3)=20, not listed, so the stay-above-ground condition is essential.
ANSWER 10: B

Here are the step-by-step solutions for each problem:

---

### Problem 1:

**Restatement of the problem:**  
We want to find the value closest to:
$$\frac{(487{,}000)(12{,}027{,}300) + (9{,}621{,}001)(487{,}000)}{(19{,}367)(0.05)}$$

**Step-by-step solution:**
1. Factor out the common term $487{,}000$ from the numerator:
   $$\text{Numerator} = 487{,}000 \times (12{,}027{,}300 + 9{,}621{,}001)$$
2. Add the terms inside the parentheses:
   $$12{,}027{,}300 + 9{,}621{,}001 = 21{,}648{,}301$$
3. Approximate the terms using scientific notation:
   - $487{,}000 \approx 5 \times 10^5$ (or more precisely, $4.87 \times 10^5$)
   - $21{,}648{,}301 \approx 2.16 \times 10^7$
   - Numerator $\approx (4.87 \times 10^5) \times (2.165 \times 10^7) \approx 1.05 \times 10^{13}$
4. Now calculate the denominator:
   $$\text{Denominator} = 19{,}367 \times 0.05 \approx 20{,}000 \times 0.05 = 1{,}000 = 10^3$$
5. Divide the numerator by the denominator:
   $$\frac{1.05 \times 10^{13}}{10^3} \approx 1.05 \times 10^{10} = 10{,}500{,}000{,}000$$
   This is closest to $10{,}000{,}000{,}000$ ($10^{10}$).

ANSWER 1: D

---

### Problem 2:

**Restatement of the problem:**  
Given that $\diamond$ and $\triangle$ are whole numbers and $\diamond \times \triangle = 36$, find the largest possible value of $\diamond + \triangle$.

**Step-by-step solution:**
1. The pairs of whole numbers (positive factors) that multiply to $36$ are:
   - $1 \times 36 = 36 \implies 1 + 36 = 37$
   - $2 \times 18 = 36 \implies 2 + 18 = 20$
   - $3 \times 12 = 36 \implies 3 + 12 = 15$
   - $4 \times 9 = 36 \implies 4 + 9 = 13$
   - $6 \times 6 = 36 \implies 6 + 6 = 12$
2. For two numbers with a fixed product, their sum is maximized when the two numbers are as far apart as possible.
3. The maximum sum occurs with the pair $1$ and $36$, which gives:
   $$1 + 36 = 37$$

ANSWER 2: E

---

### Problem 3:

**Restatement of the problem:**  
Calculate the difference:
$$(1901 + 1902 + 1903 + \dots + 1993) - (101 + 102 + 103 + \dots + 193)$$

**Step-by-step solution:**
1. Notice that both sums have the same number of terms: from $1$ to $93$, which is $93$ terms.
2. Group the terms into pairs by matching the $k$-th term of the first sum with the $k$-th term of the second sum:
   $$(1901 - 101) + (1902 - 102) + (1903 - 103) + \dots + (1993 - 193)$$
3. The difference between each corresponding pair is constant:
   $$1900 + k - (100 + k) = 1800$$
4. Since there are $93$ pairs, the total value is:
   $$93 \times 1800 = 93 \times 18 \times 100$$
5. Multiply $93 \times 18$:
   $$93 \times 18 = 93 \times (20 - 2) = 1860 - 186 = 1674$$
   $$1674 \times 100 = 167{,}400$$

ANSWER 3: A

---

### Problem 4:

**Restatement of the problem:**  
Each of the operations $+$, $-$, and $\times$ is used exactly once in the blanks of $5 \text{ \underline{ } } 4 \text{ \underline{ } } 6 \text{ \underline{ } } 3$. Determine which of the given choices could be the result.

**Step-by-step solution:**
1. Remember standard order of operations: multiplication ($\times$) is performed before addition and subtraction ($+$ and $-$).
2. Test all $6$ possible arrangements of the three signs:
   - $5 + 4 - 6 \times 3 = 5 + 4 - 18 = -9$
   - $5 + 4 \times 6 - 3 = 5 + 24 - 3 = 26$
   - $5 - 4 + 6 \times 3 = 5 - 4 + 18 = 19$
   - $5 - 4 \times 6 + 3 = 5 - 24 + 3 = -16$
   - $5 \times 4 + 6 - 3 = 20 + 6 - 3 = 23$
   - $5 \times 4 - 6 + 3 = 20 - 6 + 3 = 17$
3. Among the choices ($9, 10, 15, 16, 19$), the only value that appears in our list of possible outcomes is $19$.

ANSWER 4: E

---

### Problem 5:

**Restatement of the problem:**  
A set of $1{,}000$ incomes has one error: the largest income was entered as $\$980{,}000$ instead of $\$98{,}000$. Find the difference between the mean of the incorrect data and the mean of the actual data.

**Step-by-step solution:**
1. Let the correct sum of the $1{,}000$ incomes be $S_{\text{actual}}$.
2. The incorrect sum, $S_{\text{incorrect}}$, differs from the actual sum only by the error in the largest income:
   $$S_{\text{incorrect}} - S_{\text{actual}} = 980{,}000 - 98{,}000 = 882{,}000$$
3. The difference between the means is:
   $$\text{Mean}_{\text{incorrect}} - \text{Mean}_{\text{actual}} = \frac{S_{\text{incorrect}} - S_{\text{actual}}}{1{,}000} = \frac{882{,}000}{1{,}000} = 882\text{ dollars}$$

ANSWER 5: A

---

### Problem 6:

**Restatement of the problem:**  
A jacket originally cost $\$80$ and was discounted by $40\%$. A shirt originally cost $\$40$ and was discounted by $55\%$. Find the total amount saved as a percentage of the total original price.

**Step-by-step solution:**
1. Find the amount saved on the jacket:
   $$\text{Savings on jacket} = 40\% \text{ of } \$80 = 0.40 \times 80 = \$32$$
2. Find the amount saved on the shirt:
   $$\text{Savings on shirt} = 55\% \text{ of } \$40 = 0.55 \times 40 = \$22$$
3. Calculate the total savings:
   $$\text{Total savings} = \$32 + \$22 = \$54$$
4. Calculate the total original cost:
   $$\text{Total original price} = \$80 + \$40 = \$120$$
5. Find the percentage saved:
   $$\frac{54}{120} = \frac{9}{20} = \frac{45}{100} = 45\%$$

ANSWER 6: A

---

### Problem 7:

**Restatement of the problem:**  
A coat has an original price of $\$180$. The outlet sale price is $50\%$ off the original price, and on Saturday an additional $20\%$ is discounted from the sale price. Find the Saturday price.

**Step-by-step solution:**
1. Calculate the standard sale price:
   $$\text{Sale price} = \$180 - (0.50 \times \$180) = \$90$$
2. Calculate the Saturday price by taking an additional $20\%$ off the $\$90$ sale price:
   $$\text{Saturday discount} = 0.20 \times \$90 = \$18$$
   $$\text{Saturday price} = \$90 - \$18 = \$72$$
   *(Alternatively: $180 \times 0.50 \times 0.80 = 180 \times 0.40 = \$72$.)*

ANSWER 7: B

---

### Problem 8:

**Restatement of the problem:**  
Find how many factors of $2020$ have more than $3$ factors.

**Step-by-step solution:**
1. Find the prime factorization of $2020$:
   $$2020 = 20 \t

# Problem 1:

The expression is

$$\frac{(487{,}000)(12{,}027{,}300)+(9{,}621{,}001)(487{,}000)}{(19{,}367)(0.05)}$$

**Factor the numerator:** both terms share the factor 487,000.

$$487{,}000\,(12{,}027{,}300+9{,}621{,}001)=487{,}000\times 21{,}648{,}301$$

Estimate: $\approx (4.87\times10^5)(2.16\times10^7)\approx 1.05\times10^{13}$.

**Denominator:** $19{,}367\times0.05\approx 968$, i.e. about $9.7\times10^2$.

**Divide:**
$$\frac{1.05\times10^{13}}{9.7\times10^{2}}\approx 1.09\times10^{10}$$

That is about $10{,}900{,}000{,}000$, closest to $10{,}000{,}000{,}000$. (The next choice, $100{,}000{,}000{,}000$, is nearly ten times too big, and $1{,}000{,}000{,}000$ is nearly ten times too small.)

ANSWER 1: D

# Problem 2:

We need whole numbers with $\Diamond\times\triangle=36$, maximizing $\Diamond+\triangle$.

List factor pairs and sums:
- $1\cdot36 \Rightarrow 37$
- $2\cdot18 \Rightarrow 20$
- $3\cdot12 \Rightarrow 15$
- $4\cdot9 \Rightarrow 13$
- $6\cdot6 \Rightarrow 12$

The sum is largest when the factors are as far apart as possible: $1+36=37$. (Choices 12, 13, 15, 20 are the *other* pairs' sums — 12 is actually the smallest.)

ANSWER 2: E

# Problem 3:

Both sums have the same number of terms: from 1901 to 1993 there are $1993-1901+1=93$ terms, and from 101 to 193 there are also 93 terms.

Pair them term by term: $1901-101=1800$, $1902-102=1800$, …, $1993-193=1800$.

So the difference is
$$93\times1800=167{,}400.$$

(No need to compute either huge sum separately.)

ANSWER 3: A

# Problem 4:

We fill $5\ \_\ 4\ \_\ 6\ \_\ 3$ using $+,-,\times$ each exactly once. Try all 6 orders (remember $\times$ comes first):

- $+,-,\times$: $5+4-6\times3=5+4-18=-9$
- $+,\times,-$: $5+4\times6-3=5+24-3=26$
- $-,+,\times$: $5-4+6\times3=1+18=\mathbf{19}$
- $-,\times,+$: $5-4\times6+3=5-24+3=-16$
- $\times,+,-$: $5\times4+6-3=20+3=23$
- $\times,-,+$: $5\times4-6+3=20-6+3=17$

The only achievable value among the choices is $19$. (9, 10, 15, 16 never appear.)

ANSWER 4: E

# Problem 5:

Only one entry changed. The total was too large by
$$980{,}000-98{,}000=882{,}000\text{ dollars.}$$

The mean is the total divided by 1000 families, so the means differ by
$$\frac{882{,}000}{1000}=882\text{ dollars.}$$

(The $\$882{,}000$ choice is the error in the *total*, not the mean; $\$8200$ and the range are irrelevant.)

ANSWER 5: A

# Problem 6:

Savings:
- Jacket: $40\%$ of $80 = 0.40\times80=\$32$
- Shirt: $55\%$ of $40 = 0.55\times40=\$22$

Total saved $=32+22=\$54$. Total original price $=80+40=\$120$.

$$\frac{54}{120}=0.45=45\%$$

(Note it is **not** the average $\tfrac{40+55}{2}=47\tfrac12\%$, because the jacket costs more, pulling the percentage toward $40\%$.)

ANSWER 6: A

# Problem 7:

Sale price: $50\%$ below original means half price:
$$180\times0.50=\$90.$$

Saturday: an extra $20\%$ off that sale price, so you pay $80\%$ of \$90:
$$90\times0.80=\$72.$$

(\$54 would come from wrongly taking $70\%$ off the original; \$90 forgets the extra discount.)

ANSWER 7: B

# Problem 8:

Factor: $2020=2^2\cdot5\cdot101$, so it has $(2+1)(1+1)(1+1)=12$ divisors:
$$1,2,4,5,10,20,101,202,404,505,1010,2020.$$

A number has **more than 3** factors unless it is $1$ (one factor), a prime (two factors), or the square of a prime (three factors).

Among the list:
- $1$ → 1 factor (exclude)
- $2,5,101$ → primes, 2 factors (exclude)
- $4=2^2$ → 3 factors: $1,2,4$ (exclude)

All others ($10,20,202,404,505,1010,2020$) have more than 3 factors: that's $12-5=7$ numbers.

ANSWER 8: B

# Problem 9:

Count the cubes that touch **no** side and **not** the bottom, then subtract.

Set coordinates $1..4$ in each direction. A cube avoids all four vertical sides only if its horizontal position is $2$ or $3$ in both horizontal directions: $2\times2=4$ choices. It avoids the bottom if its level is $2,3,$ or $4$: 3 choices (the top is not mentioned, so touching the top is allowed).

Interior-and-not-bottom cubes: $2\times2\times3=12$.

Cubes touching a side or the bottom: $64-12=52$.

(48 would be the count of cubes on the outer shell of a $4\times4\times4$ block including the top face — but the top-face cubes not on a side don't touch a side or bottom, so 48 is wrong here.)

ANSWER 9: B

# Problem 10:

Each hop is up ($+1$) or down ($-1$). We need 6 hops summing to 0 (so 3 ups and 3 downs) and the height must never go below ground (he can't hop down from the ground).

These are exactly the "balanced sequences" counted by the Catalan number $C_3=\frac{1}{4}\binom{6}{3}=\frac{20}{4}=5$.

Listing them (U = up, D = down):
1. UUUDDD
2. UUDUDD
3. UUDDUD
4. UDUUDD
5. UDUDUD

Total: **5**. (The 20 sequences of three U's and three D's include 15 that dip below ground, like DUUUDD, which are impossible.)

ANSWER 10: B

Problem 1:

We are finding the value closest to the expression, which is intended to be

\[
\frac{(487{,}000)(12{,}027{,}300)+(9{,}621{,}001)(487{,}000)}
{(19{,}367)(0.05)}.
\]

Factor out \(487{,}000\) from the numerator:

\[
\frac{487{,}000(12{,}027{,}300+9{,}621{,}001)}
{19{,}367(0.05)}.
\]

Add inside the parentheses:

\[
12{,}027{,}300+9{,}621{,}001=21{,}648{,}301.
\]

Estimate the numerator:

\[
487{,}000\cdot 21{,}648{,}301\approx 4.87\times 10^5\cdot 2.165\times 10^7
\approx 1.05\times 10^{13}.
\]

Now estimate the denominator:

\[
19{,}367(0.05)=968.35\approx 10^3.
\]

Thus the whole value is approximately

\[
\frac{1.05\times 10^{13}}{10^3}\approx 1.05\times 10^{10}.
\]

So it is closest to \(10{,}000{,}000{,}000\).

\[
\boxed{\text{D}}
\]

ANSWER 1: D

Problem 2:

We need whole numbers \(\diamond\) and \(\triangle\) such that

\[
\diamond\cdot\triangle=36.
\]

To make their sum as large as possible, use the most unequal positive factor pair of 36:

\[
1\cdot 36=36.
\]

Then

\[
1+36=37.
\]

Other factor pairs give smaller sums:

\[
2+18=20,\qquad 3+12=15,\qquad 4+9=13,\qquad 6+6=12.
\]

The largest is \(37\).

\[
\boxed{\text{E}}
\]

ANSWER 2: E

Problem 3:

There are 93 terms in each sum:

\[
1901,1902,\dots,1993
\]

because

\[
1993-1901+1=93.
\]

Likewise,

\[
193-101+1=93.
\]

Match corresponding terms:

\[
1901-101=1800,
\]
\[
1902-102=1800,
\]

and so on. Every paired difference is \(1800\).

Therefore the total difference is

\[
93\cdot 1800=167{,}400.
\]

\[
\boxed{\text{A}}
\]

ANSWER 3: A

Problem 4:

We must use \(+\), \(-\), and \(\times\) exactly once in

\[
5\ \_\ 4\ \_\ 6\ \_\ 3.
\]

Remember that multiplication is done before addition and subtraction.

Try the possible placements:

\[
5+4-6\times3=9-18=-9
\]

\[
5+4\times6-3=5+24-3=26
\]

\[
5-4+6\times3=5-4+18=19
\]

This equals one of the choices.

For completeness, the remaining orders give:

\[
5-4\times6+3=5-24+3=-16,
\]

\[
5\times4+6-3=23,
\]

\[
5\times4-6+3=17.
\]

Only \(19\) appears among the answer choices.

\[
\boxed{\text{E}}
\]

ANSWER 4: E

Problem 5:

The actual largest income was

\[
98{,}000.
\]

It was incorrectly entered as

\[
980{,}000.
\]

So the total income in the computer data is too large by

\[
980{,}000-98{,}000=882{,}000.
\]

There are \(1000\) families, so the mean is too large by

\[
\frac{882{,}000}{1000}=882.
\]

Thus the difference between the two means is \(882\) dollars.

\[
\boxed{\text{A}}
\]

ANSWER 5: A

Problem 6:

The jacket originally cost \(80\) dollars and had a \(40\%\) discount.

Amount saved on jacket:

\[
0.40(80)=32.
\]

The shirt originally cost \(40\) dollars and had a \(55\%\) discount.

Amount saved on shirt:

\[
0.55(40)=22.
\]

Total saved:

\[
32+22=54.
\]

Total original price:

\[
80+40=120.
\]

The percent saved is

\[
\frac{54}{120}=0.45=45\%.
\]

\[
\boxed{\text{A}}
\]

ANSWER 6: A

Problem 7:

The original price is \(180\) dollars.

First, the store reduces the price by \(50\%\):

\[
180(0.50)=90.
\]

Then the Saturday discount takes another \(20\%\) off the sale price of \(90\):

\[
90(0.80)=72.
\]

The Saturday price is

\[
\$72.
\]

It is not \(\$54\), because the discounts are applied one after the other, not by adding \(50\%+20\%=70\%\) of the original price.

\[
\boxed{\text{B}}
\]

ANSWER 7: B

Problem 8:

First factor \(2020\):

\[
2020=2^2\cdot 5\cdot 101.
\]

So the number of factors of \(2020\) is

\[
(2+1)(1+1)(1+1)=12.
\]

We need to exclude factors that have at most 3 factors themselves.

- \(1\) has only 1 factor.
- The primes \(2,5,101\) each have 2 factors.
- \(4=2^2\) has 3 factors: \(1,2,4\).

Thus there are

\[
1+3+1=5
\]

factors of \(2020\) with at most 3 factors.

Therefore, the number with more than 3 factors is

\[
12-5=7.
\]

\[
\boxed{\text{B}}
\]

ANSWER 8: B

Problem 9:

There are

\[
4\cdot4\cdot4=64
\]

small cubes total.

It is easier to count the cubes that do **not** touch a side or the bottom.

To avoid touching a side, a cube must be in one of the middle \(2\times2\) positions horizontally:

\[
2\cdot2=4
\]

possible positions per layer.

To avoid touching the bottom, it can be in any of the upper 3 layers.

So the number that touch neither a side nor the bottom is

\[
2\cdot2\cdot3=12.
\]

Therefore the number that touch a side or the bottom is

\[
64-12=52.
\]

\[
\boxed{\text{B}}
\]

ANSWER 9: B

Problem 10:

Buzz makes 6 hops and ends where he started, on the ground. Therefore, he must make exactly 3 up-hops and 3 down-hops.

However, he cannot hop below the ground.

The valid sequences are:

\[
\text{up-up-up-down-down-down}
\]

\[
\text{up-up-down-up-down-down}
\]

\[
\text{up-up-down-down-up-down}
\]

\[
\text{up-down-up-up-down-down}
\]

\[
\text{up-down-up-down-up-down}
\]

There are \(5\) valid sequences.

For example, a sequence beginning with “down” is impossible, because Buzz would go below ground immediately.

\[
\boxed{\text{B}}
\]

ANSWER 10: B

Problem 1:
We are asked to estimate
[(487,000)(12,027,300) + (9,621,001)(487,000)] / [(19,367)(.05)]
to the nearest power of 10.

Factor the numerator:
487,000 × (12,027,300 + 9,621,001)
= 487,000 × 21,648,301.

Approximate for order of magnitude:
487,000 ≈ 5 × 10^5
21,648,301 ≈ 2.2 × 10^7
So numerator ≈ 5×10^5 × 2.2×10^7 ≈ 11×10^12 = 1.1×10^13.

Denominator:
19,367 × .05 ≈ 20,000 × .05 = 1,000 = 10^3.

Hence ratio ≈ 1.1×10^13 / 10^3 = 1.1×10^10 ≈ 11,000,000,000.

This is closest to 10,000,000,000, not 1,000,000,000 which is 10 times too small, nor 100,000,000,000 which is 10 times too large.
ANSWER 1: D

Problem 2:
We need whole numbers ◇, △ with product 36 to maximize the sum.

Whole numbers with product 36 must be positive factor pairs of 36. List them:
1 × 36, sum = 37
2 × 18, sum = 20
3 × 12, sum = 15
4 × 9, sum = 13
6 × 6, sum = 12

The sum gets smaller as the factors get closer together. The largest is obtained from the most extreme pair 1 and 36.

So 37 is largest, not 20 or 12.
ANSWER 2: E

Problem 3:
We are asked for
(1901+1902+...+1993) − (101+102+...+193).

Count terms: 1993−1901+1 = 93, and 193−101+1 = 93. So both sums have 93 terms.

Pair them in order:
1901−101 = 1800
1902−102 = 1800
...
1993−193 = 1800

There are 93 such differences, each 1800. So the total difference is
93 × 1800 = 93 × 18 × 100 = 1674 × 100 = 167,400.

This rules out e.g. adding instead of subtracting, or miscounting terms.
ANSWER 3: A

Problem 4:
We must place +, −, × each exactly once in 5 __ 4 __ 6 __ 3 and evaluate with normal precedence, × before + −.

List all 6 orderings:

1) 5+4−6×3 = 9−18 = −9
2) 5+4×6−3 = 5+24−3 = 26
3) 5−4+6×3 = 1+18 = 19
4) 5−4×6+3 = 5−24+3 = −16
5) 5×4+6−3 = 20+6−3 = 23
6) 5×4−6+3 = 20−6+3 = 17

The attainable values are −9, 26, 19, −16, 23, 17.

Of the choices 9,10,15,16,19, only 19 occurs, from 5−4+6×3.
ANSWER 4: E

Problem 5:
There are 1000 families. Mean = sum / 1000.

Only one income is wrong: true largest 98,000 entered as 980,000.
Increase in sum = 980,000 − 98,000 = 882,000.

Increase in mean = increase in sum divided by 1000:
882,000 / 1000 = 882.

So the incorrect mean is 882 dollars higher. This is not 980, which would ignore subtracting the correct value, nor hundreds of thousands, which forget to divide by 1000.
ANSWER 5: A

Problem 6:
Original prices total 80+40 = 120 dollars.

Saved on jacket: 40% of 80 = 0.40×80 = 32 dollars.
Saved on shirt: 55% of 40 = 0.55×40 = 22 dollars.
Total saved = 32+22 = 54 dollars.

As percent of original total:
54 / 120 = 9 / 20 = 45%.

Note 47½% is the simple average of 40% and 55%, tempting but wrong because the bases 80 and 40 are different.
ANSWER 6: A

Problem 7:
Original $180. Sale is 50% below original, so sale price is 50% of original:
0.50 × 180 = $90.

Saturday is an additional 20% off the sale price, so pay 100%−20% = 80% of sale price:
0.80 × 90 = $72.

$54 would be 70% off at once, which is wrong because successive discounts multiply, not add: 0.5×0.8 = 0.4 of original, not 0.3.
ANSWER 7: B

Problem 8:
Factor 2020 = 202×10 = 2×101×2×5 = 2^2 × 5^1 × 101^1.
Number of divisors = (2+1)(1+1)(1+1) = 12. They are:
1, 2, 4, 5, 10, 20, 101, 202, 404, 505, 1010, 2020.

A number has >3 factors unless it is 1 (1 factor), prime (2 factors), or square of a prime (3 factors).

Check:
1: 1 factor - no
2: prime - no
4 = 2^2: 3 factors - no, exactly 3 not more
5: prime - no
10 = 2×5: 4 factors - yes
20: 6 factors - yes
101: prime - no
202 = 2×101: 4 factors - yes
404 = 2^2×101: 6 factors - yes
505 = 5×101: 4 factors - yes
1010 = 2×5×101: 8 factors - yes
2020: 12 factors - yes

Yes count: 10,20,202,404,505,1010,2020 = 7.

Including 4 would give 8, but 4 has exactly 3, not more than 3.
ANSWER 8: B

Problem 9:
There are 4×4×4 = 64 small cubes. We want those that touch a vertical side or the bottom. The top does not count unless the cube also touches a side or bottom.

Count the complement: cubes touching neither a side nor the bottom.
To avoid the four sides, the cube must be interior in x and y: 2 by 2 = 4 per layer.
To avoid the bottom, it must not be in the bottom layer: so in the top 3 layers.

Complement = 2×2×3 = 12 cubes. These are the only ones that fail.

Hence desired = 64 − 12 = 52.

If all 6 faces counted, answer would be 64−8 = 56, which is not listed, confirming the top is excluded.
ANSWER 9: B

Problem 10:
6 hops, start and end at ground, one step up (U) or down (D) each time. To return, need 3 U and 3 D. We cannot go below ground, so height must stay ≥0.

In particular first hop must be U. This is counting Dyck paths of length 6.

Enumerate:
U U U D D D
U U D U D D
U U D D U D
U D U U D D
U D U D U D

That is 5 ways. This is the Catalan number C3 = 5.

Unrestricted 3U-3D would be C(6,3)=20, not listed, so the stay-above-ground condition is essential.
ANSWER 10: B

Here are the step-by-step solutions for each problem:

---

### Problem 1:

**Restatement of the problem:**  
We want to find the value closest to:
$$\frac{(487{,}000)(12{,}027{,}300) + (9{,}621{,}001)(487{,}000)}{(19{,}367)(0.05)}$$

**Step-by-step solution:**
1. Factor out the common term $487{,}000$ from the numerator:
   $$\text{Numerator} = 487{,}000 \times (12{,}027{,}300 + 9{,}621{,}001)$$
2. Add the terms inside the parentheses:
   $$12{,}027{,}300 + 9{,}621{,}001 = 21{,}648{,}301$$
3. Approximate the terms using scientific notation:
   - $487{,}000 \approx 5 \times 10^5$ (or more precisely, $4.87 \times 10^5$)
   - $21{,}648{,}301 \approx 2.16 \times 10^7$
   - Numerator $\approx (4.87 \times 10^5) \times (2.165 \times 10^7) \approx 1.05 \times 10^{13}$
4. Now calculate the denominator:
   $$\text{Denominator} = 19{,}367 \times 0.05 \approx 20{,}000 \times 0.05 = 1{,}000 = 10^3$$
5. Divide the numerator by the denominator:
   $$\frac{1.05 \times 10^{13}}{10^3} \approx 1.05 \times 10^{10} = 10{,}500{,}000{,}000$$
   This is closest to $10{,}000{,}000{,}000$ ($10^{10}$).

ANSWER 1: D

---

### Problem 2:

**Restatement of the problem:**  
Given that $\diamond$ and $\triangle$ are whole numbers and $\diamond \times \triangle = 36$, find the largest possible value of $\diamond + \triangle$.

**Step-by-step solution:**
1. The pairs of whole numbers (positive factors) that multiply to $36$ are:
   - $1 \times 36 = 36 \implies 1 + 36 = 37$
   - $2 \times 18 = 36 \implies 2 + 18 = 20$
   - $3 \times 12 = 36 \implies 3 + 12 = 15$
   - $4 \times 9 = 36 \implies 4 + 9 = 13$
   - $6 \times 6 = 36 \implies 6 + 6 = 12$
2. For two numbers with a fixed product, their sum is maximized when the two numbers are as far apart as possible.
3. The maximum sum occurs with the pair $1$ and $36$, which gives:
   $$1 + 36 = 37$$

ANSWER 2: E

---

### Problem 3:

**Restatement of the problem:**  
Calculate the difference:
$$(1901 + 1902 + 1903 + \dots + 1993) - (101 + 102 + 103 + \dots + 193)$$

**Step-by-step solution:**
1. Notice that both sums have the same number of terms: from $1$ to $93$, which is $93$ terms.
2. Group the terms into pairs by matching the $k$-th term of the first sum with the $k$-th term of the second sum:
   $$(1901 - 101) + (1902 - 102) + (1903 - 103) + \dots + (1993 - 193)$$
3. The difference between each corresponding pair is constant:
   $$1900 + k - (100 + k) = 1800$$
4. Since there are $93$ pairs, the total value is:
   $$93 \times 1800 = 93 \times 18 \times 100$$
5. Multiply $93 \times 18$:
   $$93 \times 18 = 93 \times (20 - 2) = 1860 - 186 = 1674$$
   $$1674 \times 100 = 167{,}400$$

ANSWER 3: A

---

### Problem 4:

**Restatement of the problem:**  
Each of the operations $+$, $-$, and $\times$ is used exactly once in the blanks of $5 \text{ \underline{ } } 4 \text{ \underline{ } } 6 \text{ \underline{ } } 3$. Determine which of the given choices could be the result.

**Step-by-step solution:**
1. Remember standard order of operations: multiplication ($\times$) is performed before addition and subtraction ($+$ and $-$).
2. Test all $6$ possible arrangements of the three signs:
   - $5 + 4 - 6 \times 3 = 5 + 4 - 18 = -9$
   - $5 + 4 \times 6 - 3 = 5 + 24 - 3 = 26$
   - $5 - 4 + 6 \times 3 = 5 - 4 + 18 = 19$
   - $5 - 4 \times 6 + 3 = 5 - 24 + 3 = -16$
   - $5 \times 4 + 6 - 3 = 20 + 6 - 3 = 23$
   - $5 \times 4 - 6 + 3 = 20 - 6 + 3 = 17$
3. Among the choices ($9, 10, 15, 16, 19$), the only value that appears in our list of possible outcomes is $19$.

ANSWER 4: E

---

### Problem 5:

**Restatement of the problem:**  
A set of $1{,}000$ incomes has one error: the largest income was entered as $\$980{,}000$ instead of $\$98{,}000$. Find the difference between the mean of the incorrect data and the mean of the actual data.

**Step-by-step solution:**
1. Let the correct sum of the $1{,}000$ incomes be $S_{\text{actual}}$.
2. The incorrect sum, $S_{\text{incorrect}}$, differs from the actual sum only by the error in the largest income:
   $$S_{\text{incorrect}} - S_{\text{actual}} = 980{,}000 - 98{,}000 = 882{,}000$$
3. The difference between the means is:
   $$\text{Mean}_{\text{incorrect}} - \text{Mean}_{\text{actual}} = \frac{S_{\text{incorrect}} - S_{\text{actual}}}{1{,}000} = \frac{882{,}000}{1{,}000} = 882\text{ dollars}$$

ANSWER 5: A

---

### Problem 6:

**Restatement of the problem:**  
A jacket originally cost $\$80$ and was discounted by $40\%$. A shirt originally cost $\$40$ and was discounted by $55\%$. Find the total amount saved as a percentage of the total original price.

**Step-by-step solution:**
1. Find the amount saved on the jacket:
   $$\text{Savings on jacket} = 40\% \text{ of } \$80 = 0.40 \times 80 = \$32$$
2. Find the amount saved on the shirt:
   $$\text{Savings on shirt} = 55\% \text{ of } \$40 = 0.55 \times 40 = \$22$$
3. Calculate the total savings:
   $$\text{Total savings} = \$32 + \$22 = \$54$$
4. Calculate the total original cost:
   $$\text{Total original price} = \$80 + \$40 = \$120$$
5. Find the percentage saved:
   $$\frac{54}{120} = \frac{9}{20} = \frac{45}{100} = 45\%$$

ANSWER 6: A

---

### Problem 7:

**Restatement of the problem:**  
A coat has an original price of $\$180$. The outlet sale price is $50\%$ off the original price, and on Saturday an additional $20\%$ is discounted from the sale price. Find the Saturday price.

**Step-by-step solution:**
1. Calculate the standard sale price:
   $$\text{Sale price} = \$180 - (0.50 \times \$180) = \$90$$
2. Calculate the Saturday price by taking an additional $20\%$ off the $\$90$ sale price:
   $$\text{Saturday discount} = 0.20 \times \$90 = \$18$$
   $$\text{Saturday price} = \$90 - \$18 = \$72$$
   *(Alternatively: $180 \times 0.50 \times 0.80 = 180 \times 0.40 = \$72$.)*

ANSWER 7: B

---

### Problem 8:

**Restatement of the problem:**  
Find how many factors of $2020$ have more than $3$ factors.

**Step-by-step solution:**
1. Find the prime factorization of $2020$:
   $$2020 = 20 \t

# Problem 1:

The expression is

$$\frac{(487{,}000)(12{,}027{,}300)+(9{,}621{,}001)(487{,}000)}{(19{,}367)(0.05)}$$

**Factor the numerator:** both terms share the factor 487,000.

$$487{,}000\,(12{,}027{,}300+9{,}621{,}001)=487{,}000\times 21{,}648{,}301$$

Estimate: $\approx (4.87\times10^5)(2.16\times10^7)\approx 1.05\times10^{13}$.

**Denominator:** $19{,}367\times0.05\approx 968$, i.e. about $9.7\times10^2$.

**Divide:**
$$\frac{1.05\times10^{13}}{9.7\times10^{2}}\approx 1.09\times10^{10}$$

That is about $10{,}900{,}000{,}000$, closest to $10{,}000{,}000{,}000$. (The next choice, $100{,}000{,}000{,}000$, is nearly ten times too big, and $1{,}000{,}000{,}000$ is nearly ten times too small.)

ANSWER 1: D

# Problem 2:

We need whole numbers with $\Diamond\times\triangle=36$, maximizing $\Diamond+\triangle$.

List factor pairs and sums:
- $1\cdot36 \Rightarrow 37$
- $2\cdot18 \Rightarrow 20$
- $3\cdot12 \Rightarrow 15$
- $4\cdot9 \Rightarrow 13$
- $6\cdot6 \Rightarrow 12$

The sum is largest when the factors are as far apart as possible: $1+36=37$. (Choices 12, 13, 15, 20 are the *other* pairs' sums — 12 is actually the smallest.)

ANSWER 2: E

# Problem 3:

Both sums have the same number of terms: from 1901 to 1993 there are $1993-1901+1=93$ terms, and from 101 to 193 there are also 93 terms.

Pair them term by term: $1901-101=1800$, $1902-102=1800$, …, $1993-193=1800$.

So the difference is
$$93\times1800=167{,}400.$$

(No need to compute either huge sum separately.)

ANSWER 3: A

# Problem 4:

We fill $5\ \_\ 4\ \_\ 6\ \_\ 3$ using $+,-,\times$ each exactly once. Try all 6 orders (remember $\times$ comes first):

- $+,-,\times$: $5+4-6\times3=5+4-18=-9$
- $+,\times,-$: $5+4\times6-3=5+24-3=26$
- $-,+,\times$: $5-4+6\times3=1+18=\mathbf{19}$
- $-,\times,+$: $5-4\times6+3=5-24+3=-16$
- $\times,+,-$: $5\times4+6-3=20+3=23$
- $\times,-,+$: $5\times4-6+3=20-6+3=17$

The only achievable value among the choices is $19$. (9, 10, 15, 16 never appear.)

ANSWER 4: E

# Problem 5:

Only one entry changed. The total was too large by
$$980{,}000-98{,}000=882{,}000\text{ dollars.}$$

The mean is the total divided by 1000 families, so the means differ by
$$\frac{882{,}000}{1000}=882\text{ dollars.}$$

(The $\$882{,}000$ choice is the error in the *total*, not the mean; $\$8200$ and the range are irrelevant.)

ANSWER 5: A

# Problem 6:

Savings:
- Jacket: $40\%$ of $80 = 0.40\times80=\$32$
- Shirt: $55\%$ of $40 = 0.55\times40=\$22$

Total saved $=32+22=\$54$. Total original price $=80+40=\$120$.

$$\frac{54}{120}=0.45=45\%$$

(Note it is **not** the average $\tfrac{40+55}{2}=47\tfrac12\%$, because the jacket costs more, pulling the percentage toward $40\%$.)

ANSWER 6: A

# Problem 7:

Sale price: $50\%$ below original means half price:
$$180\times0.50=\$90.$$

Saturday: an extra $20\%$ off that sale price, so you pay $80\%$ of \$90:
$$90\times0.80=\$72.$$

(\$54 would come from wrongly taking $70\%$ off the original; \$90 forgets the extra discount.)

ANSWER 7: B

# Problem 8:

Factor: $2020=2^2\cdot5\cdot101$, so it has $(2+1)(1+1)(1+1)=12$ divisors:
$$1,2,4,5,10,20,101,202,404,505,1010,2020.$$

A number has **more than 3** factors unless it is $1$ (one factor), a prime (two factors), or the square of a prime (three factors).

Among the list:
- $1$ → 1 factor (exclude)
- $2,5,101$ → primes, 2 factors (exclude)
- $4=2^2$ → 3 factors: $1,2,4$ (exclude)

All others ($10,20,202,404,505,1010,2020$) have more than 3 factors: that's $12-5=7$ numbers.

ANSWER 8: B

# Problem 9:

Count the cubes that touch **no** side and **not** the bottom, then subtract.

Set coordinates $1..4$ in each direction. A cube avoids all four vertical sides only if its horizontal position is $2$ or $3$ in both horizontal directions: $2\times2=4$ choices. It avoids the bottom if its level is $2,3,$ or $4$: 3 choices (the top is not mentioned, so touching the top is allowed).

Interior-and-not-bottom cubes: $2\times2\times3=12$.

Cubes touching a side or the bottom: $64-12=52$.

(48 would be the count of cubes on the outer shell of a $4\times4\times4$ block including the top face — but the top-face cubes not on a side don't touch a side or bottom, so 48 is wrong here.)

ANSWER 9: B

# Problem 10:

Each hop is up ($+1$) or down ($-1$). We need 6 hops summing to 0 (so 3 ups and 3 downs) and the height must never go below ground (he can't hop down from the ground).

These are exactly the "balanced sequences" counted by the Catalan number $C_3=\frac{1}{4}\binom{6}{3}=\frac{20}{4}=5$.

Listing them (U = up, D = down):
1. UUUDDD
2. UUDUDD
3. UUDDUD
4. UDUUDD
5. UDUDUD

Total: **5**. (The 20 sequences of three U's and three D's include 15 that dip below ground, like DUUUDD, which are impossible.)

ANSWER 10: B

Problem 1:

We are finding the value closest to the expression, which is intended to be

\[
\frac{(487{,}000)(12{,}027{,}300)+(9{,}621{,}001)(487{,}000)}
{(19{,}367)(0.05)}.
\]

Factor out \(487{,}000\) from the numerator:

\[
\frac{487{,}000(12{,}027{,}300+9{,}621{,}001)}
{19{,}367(0.05)}.
\]

Add inside the parentheses:

\[
12{,}027{,}300+9{,}621{,}001=21{,}648{,}301.
\]

Estimate the numerator:

\[
487{,}000\cdot 21{,}648{,}301\approx 4.87\times 10^5\cdot 2.165\times 10^7
\approx 1.05\times 10^{13}.
\]

Now estimate the denominator:

\[
19{,}367(0.05)=968.35\approx 10^3.
\]

Thus the whole value is approximately

\[
\frac{1.05\times 10^{13}}{10^3}\approx 1.05\times 10^{10}.
\]

So it is closest to \(10{,}000{,}000{,}000\).

\[
\boxed{\text{D}}
\]

ANSWER 1: D

Problem 2:

We need whole numbers \(\diamond\) and \(\triangle\) such that

\[
\diamond\cdot\triangle=36.
\]

To make their sum as large as possible, use the most unequal positive factor pair of 36:

\[
1\cdot 36=36.
\]

Then

\[
1+36=37.
\]

Other factor pairs give smaller sums:

\[
2+18=20,\qquad 3+12=15,\qquad 4+9=13,\qquad 6+6=12.
\]

The largest is \(37\).

\[
\boxed{\text{E}}
\]

ANSWER 2: E

Problem 3:

There are 93 terms in each sum:

\[
1901,1902,\dots,1993
\]

because

\[
1993-1901+1=93.
\]

Likewise,

\[
193-101+1=93.
\]

Match corresponding terms:

\[
1901-101=1800,
\]
\[
1902-102=1800,
\]

and so on. Every paired difference is \(1800\).

Therefore the total difference is

\[
93\cdot 1800=167{,}400.
\]

\[
\boxed{\text{A}}
\]

ANSWER 3: A

Problem 4:

We must use \(+\), \(-\), and \(\times\) exactly once in

\[
5\ \_\ 4\ \_\ 6\ \_\ 3.
\]

Remember that multiplication is done before addition and subtraction.

Try the possible placements:

\[
5+4-6\times3=9-18=-9
\]

\[
5+4\times6-3=5+24-3=26
\]

\[
5-4+6\times3=5-4+18=19
\]

This equals one of the choices.

For completeness, the remaining orders give:

\[
5-4\times6+3=5-24+3=-16,
\]

\[
5\times4+6-3=23,
\]

\[
5\times4-6+3=17.
\]

Only \(19\) appears among the answer choices.

\[
\boxed{\text{E}}
\]

ANSWER 4: E

Problem 5:

The actual largest income was

\[
98{,}000.
\]

It was incorrectly entered as

\[
980{,}000.
\]

So the total income in the computer data is too large by

\[
980{,}000-98{,}000=882{,}000.
\]

There are \(1000\) families, so the mean is too large by

\[
\frac{882{,}000}{1000}=882.
\]

Thus the difference between the two means is \(882\) dollars.

\[
\boxed{\text{A}}
\]

ANSWER 5: A

Problem 6:

The jacket originally cost \(80\) dollars and had a \(40\%\) discount.

Amount saved on jacket:

\[
0.40(80)=32.
\]

The shirt originally cost \(40\) dollars and had a \(55\%\) discount.

Amount saved on shirt:

\[
0.55(40)=22.
\]

Total saved:

\[
32+22=54.
\]

Total original price:

\[
80+40=120.
\]

The percent saved is

\[
\frac{54}{120}=0.45=45\%.
\]

\[
\boxed{\text{A}}
\]

ANSWER 6: A

Problem 7:

The original price is \(180\) dollars.

First, the store reduces the price by \(50\%\):

\[
180(0.50)=90.
\]

Then the Saturday discount takes another \(20\%\) off the sale price of \(90\):

\[
90(0.80)=72.
\]

The Saturday price is

\[
\$72.
\]

It is not \(\$54\), because the discounts are applied one after the other, not by adding \(50\%+20\%=70\%\) of the original price.

\[
\boxed{\text{B}}
\]

ANSWER 7: B

Problem 8:

First factor \(2020\):

\[
2020=2^2\cdot 5\cdot 101.
\]

So the number of factors of \(2020\) is

\[
(2+1)(1+1)(1+1)=12.
\]

We need to exclude factors that have at most 3 factors themselves.

- \(1\) has only 1 factor.
- The primes \(2,5,101\) each have 2 factors.
- \(4=2^2\) has 3 factors: \(1,2,4\).

Thus there are

\[
1+3+1=5
\]

factors of \(2020\) with at most 3 factors.

Therefore, the number with more than 3 factors is

\[
12-5=7.
\]

\[
\boxed{\text{B}}
\]

ANSWER 8: B

Problem 9:

There are

\[
4\cdot4\cdot4=64
\]

small cubes total.

It is easier to count the cubes that do **not** touch a side or the bottom.

To avoid touching a side, a cube must be in one of the middle \(2\times2\) positions horizontally:

\[
2\cdot2=4
\]

possible positions per layer.

To avoid touching the bottom, it can be in any of the upper 3 layers.

So the number that touch neither a side nor the bottom is

\[
2\cdot2\cdot3=12.
\]

Therefore the number that touch a side or the bottom is

\[
64-12=52.
\]

\[
\boxed{\text{B}}
\]

ANSWER 9: B

Problem 10:

Buzz makes 6 hops and ends where he started, on the ground. Therefore, he must make exactly 3 up-hops and 3 down-hops.

However, he cannot hop below the ground.

The valid sequences are:

\[
\text{up-up-up-down-down-down}
\]

\[
\text{up-up-down-up-down-down}
\]

\[
\text{up-up-down-down-up-down}
\]

\[
\text{up-down-up-up-down-down}
\]

\[
\text{up-down-up-down-up-down}
\]

There are \(5\) valid sequences.

For example, a sequence beginning with “down” is impossible, because Buzz would go below ground immediately.

\[
\boxed{\text{B}}
\]

ANSWER 10: B

Problem 1:
We are asked to estimate
[(487,000)(12,027,300) + (9,621,001)(487,000)] / [(19,367)(.05)]
to the nearest power of 10.

Factor the numerator:
487,000 × (12,027,300 + 9,621,001)
= 487,000 × 21,648,301.

Approximate for order of magnitude:
487,000 ≈ 5 × 10^5
21,648,301 ≈ 2.2 × 10^7
So numerator ≈ 5×10^5 × 2.2×10^7 ≈ 11×10^12 = 1.1×10^13.

Denominator:
19,367 × .05 ≈ 20,000 × .05 = 1,000 = 10^3.

Hence ratio ≈ 1.1×10^13 / 10^3 = 1.1×10^10 ≈ 11,000,000,000.

This is closest to 10,000,000,000, not 1,000,000,000 which is 10 times too small, nor 100,000,000,000 which is 10 times too large.
ANSWER 1: D

Problem 2:
We need whole numbers ◇, △ with product 36 to maximize the sum.

Whole numbers with product 36 must be positive factor pairs of 36. List them:
1 × 36, sum = 37
2 × 18, sum = 20
3 × 12, sum = 15
4 × 9, sum = 13
6 × 6, sum = 12

The sum gets smaller as the factors get closer together. The largest is obtained from the most extreme pair 1 and 36.

So 37 is largest, not 20 or 12.
ANSWER 2: E

Problem 3:
We are asked for
(1901+1902+...+1993) − (101+102+...+193).

Count terms: 1993−1901+1 = 93, and 193−101+1 = 93. So both sums have 93 terms.

Pair them in order:
1901−101 = 1800
1902−102 = 1800
...
1993−193 = 1800

There are 93 such differences, each 1800. So the total difference is
93 × 1800 = 93 × 18 × 100 = 1674 × 100 = 167,400.

This rules out e.g. adding instead of subtracting, or miscounting terms.
ANSWER 3: A

Problem 4:
We must place +, −, × each exactly once in 5 __ 4 __ 6 __ 3 and evaluate with normal precedence, × before + −.

List all 6 orderings:

1) 5+4−6×3 = 9−18 = −9
2) 5+4×6−3 = 5+24−3 = 26
3) 5−4+6×3 = 1+18 = 19
4) 5−4×6+3 = 5−24+3 = −16
5) 5×4+6−3 = 20+6−3 = 23
6) 5×4−6+3 = 20−6+3 = 17

The attainable values are −9, 26, 19, −16, 23, 17.

Of the choices 9,10,15,16,19, only 19 occurs, from 5−4+6×3.
ANSWER 4: E

Problem 5:
There are 1000 families. Mean = sum / 1000.

Only one income is wrong: true largest 98,000 entered as 980,000.
Increase in sum = 980,000 − 98,000 = 882,000.

Increase in mean = increase in sum divided by 1000:
882,000 / 1000 = 882.

So the incorrect mean is 882 dollars higher. This is not 980, which would ignore subtracting the correct value, nor hundreds of thousands, which forget to divide by 1000.
ANSWER 5: A

Problem 6:
Original prices total 80+40 = 120 dollars.

Saved on jacket: 40% of 80 = 0.40×80 = 32 dollars.
Saved on shirt: 55% of 40 = 0.55×40 = 22 dollars.
Total saved = 32+22 = 54 dollars.

As percent of original total:
54 / 120 = 9 / 20 = 45%.

Note 47½% is the simple average of 40% and 55%, tempting but wrong because the bases 80 and 40 are different.
ANSWER 6: A

Problem 7:
Original $180. Sale is 50% below original, so sale price is 50% of original:
0.50 × 180 = $90.

Saturday is an additional 20% off the sale price, so pay 100%−20% = 80% of sale price:
0.80 × 90 = $72.

$54 would be 70% off at once, which is wrong because successive discounts multiply, not add: 0.5×0.8 = 0.4 of original, not 0.3.
ANSWER 7: B

Problem 8:
Factor 2020 = 202×10 = 2×101×2×5 = 2^2 × 5^1 × 101^1.
Number of divisors = (2+1)(1+1)(1+1) = 12. They are:
1, 2, 4, 5, 10, 20, 101, 202, 404, 505, 1010, 2020.

A number has >3 factors unless it is 1 (1 factor), prime (2 factors), or square of a prime (3 factors).

Check:
1: 1 factor - no
2: prime - no
4 = 2^2: 3 factors - no, exactly 3 not more
5: prime - no
10 = 2×5: 4 factors - yes
20: 6 factors - yes
101: prime - no
202 = 2×101: 4 factors - yes
404 = 2^2×101: 6 factors - yes
505 = 5×101: 4 factors - yes
1010 = 2×5×101: 8 factors - yes
2020: 12 factors - yes

Yes count: 10,20,202,404,505,1010,2020 = 7.

Including 4 would give 8, but 4 has exactly 3, not more than 3.
ANSWER 8: B

Problem 9:
There are 4×4×4 = 64 small cubes. We want those that touch a vertical side or the bottom. The top does not count unless the cube also touches a side or bottom.

Count the complement: cubes touching neither a side nor the bottom.
To avoid the four sides, the cube must be interior in x and y: 2 by 2 = 4 per layer.
To avoid the bottom, it must not be in the bottom layer: so in the top 3 layers.

Complement = 2×2×3 = 12 cubes. These are the only ones that fail.

Hence desired = 64 − 12 = 52.

If all 6 faces counted, answer would be 64−8 = 56, which is not listed, confirming the top is excluded.
ANSWER 9: B

Problem 10:
6 hops, start and end at ground, one step up (U) or down (D) each time. To return, need 3 U and 3 D. We cannot go below ground, so height must stay ≥0.

In particular first hop must be U. This is counting Dyck paths of length 6.

Enumerate:
U U U D D D
U U D U D D
U U D D U D
U D U U D D
U D U D U D

That is 5 ways. This is the Catalan number C3 = 5.

Unrestricted 3U-3D would be C(6,3)=20, not listed, so the stay-above-ground condition is essential.
ANSWER 10: B

Here are the step-by-step solutions for each problem:

---

### Problem 1:

**Restatement of the problem:**  
We want to find the value closest to:
$$\frac{(487{,}000)(12{,}027{,}300) + (9{,}621{,}001)(487{,}000)}{(19{,}367)(0.05)}$$

**Step-by-step solution:**
1. Factor out the common term $487{,}000$ from the numerator:
   $$\text{Numerator} = 487{,}000 \times (12{,}027{,}300 + 9{,}621{,}001)$$
2. Add the terms inside the parentheses:
   $$12{,}027{,}300 + 9{,}621{,}001 = 21{,}648{,}301$$
3. Approximate the terms using scientific notation:
   - $487{,}000 \approx 5 \times 10^5$ (or more precisely, $4.87 \times 10^5$)
   - $21{,}648{,}301 \approx 2.16 \times 10^7$
   - Numerator $\approx (4.87 \times 10^5) \times (2.165 \times 10^7) \approx 1.05 \times 10^{13}$
4. Now calculate the denominator:
   $$\text{Denominator} = 19{,}367 \times 0.05 \approx 20{,}000 \times 0.05 = 1{,}000 = 10^3$$
5. Divide the numerator by the denominator:
   $$\frac{1.05 \times 10^{13}}{10^3} \approx 1.05 \times 10^{10} = 10{,}500{,}000{,}000$$
   This is closest to $10{,}000{,}000{,}000$ ($10^{10}$).

ANSWER 1: D

---

### Problem 2:

**Restatement of the problem:**  
Given that $\diamond$ and $\triangle$ are whole numbers and $\diamond \times \triangle = 36$, find the largest possible value of $\diamond + \triangle$.

**Step-by-step solution:**
1. The pairs of whole numbers (positive factors) that multiply to $36$ are:
   - $1 \times 36 = 36 \implies 1 + 36 = 37$
   - $2 \times 18 = 36 \implies 2 + 18 = 20$
   - $3 \times 12 = 36 \implies 3 + 12 = 15$
   - $4 \times 9 = 36 \implies 4 + 9 = 13$
   - $6 \times 6 = 36 \implies 6 + 6 = 12$
2. For two numbers with a fixed product, their sum is maximized when the two numbers are as far apart as possible.
3. The maximum sum occurs with the pair $1$ and $36$, which gives:
   $$1 + 36 = 37$$

ANSWER 2: E

---

### Problem 3:

**Restatement of the problem:**  
Calculate the difference:
$$(1901 + 1902 + 1903 + \dots + 1993) - (101 + 102 + 103 + \dots + 193)$$

**Step-by-step solution:**
1. Notice that both sums have the same number of terms: from $1$ to $93$, which is $93$ terms.
2. Group the terms into pairs by matching the $k$-th term of the first sum with the $k$-th term of the second sum:
   $$(1901 - 101) + (1902 - 102) + (1903 - 103) + \dots + (1993 - 193)$$
3. The difference between each corresponding pair is constant:
   $$1900 + k - (100 + k) = 1800$$
4. Since there are $93$ pairs, the total value is:
   $$93 \times 1800 = 93 \times 18 \times 100$$
5. Multiply $93 \times 18$:
   $$93 \times 18 = 93 \times (20 - 2) = 1860 - 186 = 1674$$
   $$1674 \times 100 = 167{,}400$$

ANSWER 3: A

---

### Problem 4:

**Restatement of the problem:**  
Each of the operations $+$, $-$, and $\times$ is used exactly once in the blanks of $5 \text{ \underline{ } } 4 \text{ \underline{ } } 6 \text{ \underline{ } } 3$. Determine which of the given choices could be the result.

**Step-by-step solution:**
1. Remember standard order of operations: multiplication ($\times$) is performed before addition and subtraction ($+$ and $-$).
2. Test all $6$ possible arrangements of the three signs:
   - $5 + 4 - 6 \times 3 = 5 + 4 - 18 = -9$
   - $5 + 4 \times 6 - 3 = 5 + 24 - 3 = 26$
   - $5 - 4 + 6 \times 3 = 5 - 4 + 18 = 19$
   - $5 - 4 \times 6 + 3 = 5 - 24 + 3 = -16$
   - $5 \times 4 + 6 - 3 = 20 + 6 - 3 = 23$
   - $5 \times 4 - 6 + 3 = 20 - 6 + 3 = 17$
3. Among the choices ($9, 10, 15, 16, 19$), the only value that appears in our list of possible outcomes is $19$.

ANSWER 4: E

---

### Problem 5:

**Restatement of the problem:**  
A set of $1{,}000$ incomes has one error: the largest income was entered as $\$980{,}000$ instead of $\$98{,}000$. Find the difference between the mean of the incorrect data and the mean of the actual data.

**Step-by-step solution:**
1. Let the correct sum of the $1{,}000$ incomes be $S_{\text{actual}}$.
2. The incorrect sum, $S_{\text{incorrect}}$, differs from the actual sum only by the error in the largest income:
   $$S_{\text{incorrect}} - S_{\text{actual}} = 980{,}000 - 98{,}000 = 882{,}000$$
3. The difference between the means is:
   $$\text{Mean}_{\text{incorrect}} - \text{Mean}_{\text{actual}} = \frac{S_{\text{incorrect}} - S_{\text{actual}}}{1{,}000} = \frac{882{,}000}{1{,}000} = 882\text{ dollars}$$

ANSWER 5: A

---

### Problem 6:

**Restatement of the problem:**  
A jacket originally cost $\$80$ and was discounted by $40\%$. A shirt originally cost $\$40$ and was discounted by $55\%$. Find the total amount saved as a percentage of the total original price.

**Step-by-step solution:**
1. Find the amount saved on the jacket:
   $$\text{Savings on jacket} = 40\% \text{ of } \$80 = 0.40 \times 80 = \$32$$
2. Find the amount saved on the shirt:
   $$\text{Savings on shirt} = 55\% \text{ of } \$40 = 0.55 \times 40 = \$22$$
3. Calculate the total savings:
   $$\text{Total savings} = \$32 + \$22 = \$54$$
4. Calculate the total original cost:
   $$\text{Total original price} = \$80 + \$40 = \$120$$
5. Find the percentage saved:
   $$\frac{54}{120} = \frac{9}{20} = \frac{45}{100} = 45\%$$

ANSWER 6: A

---

### Problem 7:

**Restatement of the problem:**  
A coat has an original price of $\$180$. The outlet sale price is $50\%$ off the original price, and on Saturday an additional $20\%$ is discounted from the sale price. Find the Saturday price.

**Step-by-step solution:**
1. Calculate the standard sale price:
   $$\text{Sale price} = \$180 - (0.50 \times \$180) = \$90$$
2. Calculate the Saturday price by taking an additional $20\%$ off the $\$90$ sale price:
   $$\text{Saturday discount} = 0.20 \times \$90 = \$18$$
   $$\text{Saturday price} = \$90 - \$18 = \$72$$
   *(Alternatively: $180 \times 0.50 \times 0.80 = 180 \times 0.40 = \$72$.)*

ANSWER 7: B

---

### Problem 8:

**Restatement of the problem:**  
Find how many factors of $2020$ have more than $3$ factors.

**Step-by-step solution:**
1. Find the prime factorization of $2020$:
   $$2020 = 20 \t

# Problem 1:

The expression is

$$\frac{(487{,}000)(12{,}027{,}300)+(9{,}621{,}001)(487{,}000)}{(19{,}367)(0.05)}$$

**Factor the numerator:** both terms share the factor 487,000.

$$487{,}000\,(12{,}027{,}300+9{,}621{,}001)=487{,}000\times 21{,}648{,}301$$

Estimate: $\approx (4.87\times10^5)(2.16\times10^7)\approx 1.05\times10^{13}$.

**Denominator:** $19{,}367\times0.05\approx 968$, i.e. about $9.7\times10^2$.

**Divide:**
$$\frac{1.05\times10^{13}}{9.7\times10^{2}}\approx 1.09\times10^{10}$$

That is about $10{,}900{,}000{,}000$, closest to $10{,}000{,}000{,}000$. (The next choice, $100{,}000{,}000{,}000$, is nearly ten times too big, and $1{,}000{,}000{,}000$ is nearly ten times too small.)

ANSWER 1: D

# Problem 2:

We need whole numbers with $\Diamond\times\triangle=36$, maximizing $\Diamond+\triangle$.

List factor pairs and sums:
- $1\cdot36 \Rightarrow 37$
- $2\cdot18 \Rightarrow 20$
- $3\cdot12 \Rightarrow 15$
- $4\cdot9 \Rightarrow 13$
- $6\cdot6 \Rightarrow 12$

The sum is largest when the factors are as far apart as possible: $1+36=37$. (Choices 12, 13, 15, 20 are the *other* pairs' sums — 12 is actually the smallest.)

ANSWER 2: E

# Problem 3:

Both sums have the same number of terms: from 1901 to 1993 there are $1993-1901+1=93$ terms, and from 101 to 193 there are also 93 terms.

Pair them term by term: $1901-101=1800$, $1902-102=1800$, …, $1993-193=1800$.

So the difference is
$$93\times1800=167{,}400.$$

(No need to compute either huge sum separately.)

ANSWER 3: A

# Problem 4:

We fill $5\ \_\ 4\ \_\ 6\ \_\ 3$ using $+,-,\times$ each exactly once. Try all 6 orders (remember $\times$ comes first):

- $+,-,\times$: $5+4-6\times3=5+4-18=-9$
- $+,\times,-$: $5+4\times6-3=5+24-3=26$
- $-,+,\times$: $5-4+6\times3=1+18=\mathbf{19}$
- $-,\times,+$: $5-4\times6+3=5-24+3=-16$
- $\times,+,-$: $5\times4+6-3=20+3=23$
- $\times,-,+$: $5\times4-6+3=20-6+3=17$

The only achievable value among the choices is $19$. (9, 10, 15, 16 never appear.)

ANSWER 4: E

# Problem 5:

Only one entry changed. The total was too large by
$$980{,}000-98{,}000=882{,}000\text{ dollars.}$$

The mean is the total divided by 1000 families, so the means differ by
$$\frac{882{,}000}{1000}=882\text{ dollars.}$$

(The $\$882{,}000$ choice is the error in the *total*, not the mean; $\$8200$ and the range are irrelevant.)

ANSWER 5: A

# Problem 6:

Savings:
- Jacket: $40\%$ of $80 = 0.40\times80=\$32$
- Shirt: $55\%$ of $40 = 0.55\times40=\$22$

Total saved $=32+22=\$54$. Total original price $=80+40=\$120$.

$$\frac{54}{120}=0.45=45\%$$

(Note it is **not** the average $\tfrac{40+55}{2}=47\tfrac12\%$, because the jacket costs more, pulling the percentage toward $40\%$.)

ANSWER 6: A

# Problem 7:

Sale price: $50\%$ below original means half price:
$$180\times0.50=\$90.$$

Saturday: an extra $20\%$ off that sale price, so you pay $80\%$ of \$90:
$$90\times0.80=\$72.$$

(\$54 would come from wrongly taking $70\%$ off the original; \$90 forgets the extra discount.)

ANSWER 7: B

# Problem 8:

Factor: $2020=2^2\cdot5\cdot101$, so it has $(2+1)(1+1)(1+1)=12$ divisors:
$$1,2,4,5,10,20,101,202,404,505,1010,2020.$$

A number has **more than 3** factors unless it is $1$ (one factor), a prime (two factors), or the square of a prime (three factors).

Among the list:
- $1$ → 1 factor (exclude)
- $2,5,101$ → primes, 2 factors (exclude)
- $4=2^2$ → 3 factors: $1,2,4$ (exclude)

All others ($10,20,202,404,505,1010,2020$) have more than 3 factors: that's $12-5=7$ numbers.

ANSWER 8: B

# Problem 9:

Count the cubes that touch **no** side and **not** the bottom, then subtract.

Set coordinates $1..4$ in each direction. A cube avoids all four vertical sides only if its horizontal position is $2$ or $3$ in both horizontal directions: $2\times2=4$ choices. It avoids the bottom if its level is $2,3,$ or $4$: 3 choices (the top is not mentioned, so touching the top is allowed).

Interior-and-not-bottom cubes: $2\times2\times3=12$.

Cubes touching a side or the bottom: $64-12=52$.

(48 would be the count of cubes on the outer shell of a $4\times4\times4$ block including the top face — but the top-face cubes not on a side don't touch a side or bottom, so 48 is wrong here.)

ANSWER 9: B

# Problem 10:

Each hop is up ($+1$) or down ($-1$). We need 6 hops summing to 0 (so 3 ups and 3 downs) and the height must never go below ground (he can't hop down from the ground).

These are exactly the "balanced sequences" counted by the Catalan number $C_3=\frac{1}{4}\binom{6}{3}=\frac{20}{4}=5$.

Listing them (U = up, D = down):
1. UUUDDD
2. UUDUDD
3. UUDDUD
4. UDUUDD
5. UDUDUD

Total: **5**. (The 20 sequences of three U's and three D's include 15 that dip below ground, like DUUUDD, which are impossible.)

ANSWER 10: B

Problem 1:

We are finding the value closest to the expression, which is intended to be

\[
\frac{(487{,}000)(12{,}027{,}300)+(9{,}621{,}001)(487{,}000)}
{(19{,}367)(0.05)}.
\]

Factor out \(487{,}000\) from the numerator:

\[
\frac{487{,}000(12{,}027{,}300+9{,}621{,}001)}
{19{,}367(0.05)}.
\]

Add inside the parentheses:

\[
12{,}027{,}300+9{,}621{,}001=21{,}648{,}301.
\]

Estimate the numerator:

\[
487{,}000\cdot 21{,}648{,}301\approx 4.87\times 10^5\cdot 2.165\times 10^7
\approx 1.05\times 10^{13}.
\]

Now estimate the denominator:

\[
19{,}367(0.05)=968.35\approx 10^3.
\]

Thus the whole value is approximately

\[
\frac{1.05\times 10^{13}}{10^3}\approx 1.05\times 10^{10}.
\]

So it is closest to \(10{,}000{,}000{,}000\).

\[
\boxed{\text{D}}
\]

ANSWER 1: D

Problem 2:

We need whole numbers \(\diamond\) and \(\triangle\) such that

\[
\diamond\cdot\triangle=36.
\]

To make their sum as large as possible, use the most unequal positive factor pair of 36:

\[
1\cdot 36=36.
\]

Then

\[
1+36=37.
\]

Other factor pairs give smaller sums:

\[
2+18=20,\qquad 3+12=15,\qquad 4+9=13,\qquad 6+6=12.
\]

The largest is \(37\).

\[
\boxed{\text{E}}
\]

ANSWER 2: E

Problem 3:

There are 93 terms in each sum:

\[
1901,1902,\dots,1993
\]

because

\[
1993-1901+1=93.
\]

Likewise,

\[
193-101+1=93.
\]

Match corresponding terms:

\[
1901-101=1800,
\]
\[
1902-102=1800,
\]

and so on. Every paired difference is \(1800\).

Therefore the total difference is

\[
93\cdot 1800=167{,}400.
\]

\[
\boxed{\text{A}}
\]

ANSWER 3: A

Problem 4:

We must use \(+\), \(-\), and \(\times\) exactly once in

\[
5\ \_\ 4\ \_\ 6\ \_\ 3.
\]

Remember that multiplication is done before addition and subtraction.

Try the possible placements:

\[
5+4-6\times3=9-18=-9
\]

\[
5+4\times6-3=5+24-3=26
\]

\[
5-4+6\times3=5-4+18=19
\]

This equals one of the choices.

For completeness, the remaining orders give:

\[
5-4\times6+3=5-24+3=-16,
\]

\[
5\times4+6-3=23,
\]

\[
5\times4-6+3=17.
\]

Only \(19\) appears among the answer choices.

\[
\boxed{\text{E}}
\]

ANSWER 4: E

Problem 5:

The actual largest income was

\[
98{,}000.
\]

It was incorrectly entered as

\[
980{,}000.
\]

So the total income in the computer data is too large by

\[
980{,}000-98{,}000=882{,}000.
\]

There are \(1000\) families, so the mean is too large by

\[
\frac{882{,}000}{1000}=882.
\]

Thus the difference between the two means is \(882\) dollars.

\[
\boxed{\text{A}}
\]

ANSWER 5: A

Problem 6:

The jacket originally cost \(80\) dollars and had a \(40\%\) discount.

Amount saved on jacket:

\[
0.40(80)=32.
\]

The shirt originally cost \(40\) dollars and had a \(55\%\) discount.

Amount saved on shirt:

\[
0.55(40)=22.
\]

Total saved:

\[
32+22=54.
\]

Total original price:

\[
80+40=120.
\]

The percent saved is

\[
\frac{54}{120}=0.45=45\%.
\]

\[
\boxed{\text{A}}
\]

ANSWER 6: A

Problem 7:

The original price is \(180\) dollars.

First, the store reduces the price by \(50\%\):

\[
180(0.50)=90.
\]

Then the Saturday discount takes another \(20\%\) off the sale price of \(90\):

\[
90(0.80)=72.
\]

The Saturday price is

\[
\$72.
\]

It is not \(\$54\), because the discounts are applied one after the other, not by adding \(50\%+20\%=70\%\) of the original price.

\[
\boxed{\text{B}}
\]

ANSWER 7: B

Problem 8:

First factor \(2020\):

\[
2020=2^2\cdot 5\cdot 101.
\]

So the number of factors of \(2020\) is

\[
(2+1)(1+1)(1+1)=12.
\]

We need to exclude factors that have at most 3 factors themselves.

- \(1\) has only 1 factor.
- The primes \(2,5,101\) each have 2 factors.
- \(4=2^2\) has 3 factors: \(1,2,4\).

Thus there are

\[
1+3+1=5
\]

factors of \(2020\) with at most 3 factors.

Therefore, the number with more than 3 factors is

\[
12-5=7.
\]

\[
\boxed{\text{B}}
\]

ANSWER 8: B

Problem 9:

There are

\[
4\cdot4\cdot4=64
\]

small cubes total.

It is easier to count the cubes that do **not** touch a side or the bottom.

To avoid touching a side, a cube must be in one of the middle \(2\times2\) positions horizontally:

\[
2\cdot2=4
\]

possible positions per layer.

To avoid touching the bottom, it can be in any of the upper 3 layers.

So the number that touch neither a side nor the bottom is

\[
2\cdot2\cdot3=12.
\]

Therefore the number that touch a side or the bottom is

\[
64-12=52.
\]

\[
\boxed{\text{B}}
\]

ANSWER 9: B

Problem 10:

Buzz makes 6 hops and ends where he started, on the ground. Therefore, he must make exactly 3 up-hops and 3 down-hops.

However, he cannot hop below the ground.

The valid sequences are:

\[
\text{up-up-up-down-down-down}
\]

\[
\text{up-up-down-up-down-down}
\]

\[
\text{up-up-down-down-up-down}
\]

\[
\text{up-down-up-up-down-down}
\]

\[
\text{up-down-up-down-up-down}
\]

There are \(5\) valid sequences.

For example, a sequence beginning with “down” is impossible, because Buzz would go below ground immediately.

\[
\boxed{\text{B}}
\]

ANSWER 10: B