🔬 Model Lab

Problem 1:
The problem asks us to compare the place value of the digits 9 and 3 in the number 74982.1035. 
First, we identify the positions of 9 and 3. The digit 9 is in the thousands place, and the digit 3 is in the thousandths place. 
The value of the place occupied by 9 is 1000, and the value of the place occupied by 3 is 0.001. 
To find how many times as great the value of the place occupied by 9 is compared to the value of the place occupied by 3, we calculate 1000 / 0.001. 
This is equal to 1000 * 1000 = 1,000,000. 
Let's examine the choices: A. 1,000, B. 10,000, C. 100,000, D. 1,000,000, E. 10,000,000. 
Our calculation matches choice D.

ANSWER 1: D

Problem 2:
The problem asks for the number halfway between 1/6 and 1/4. 
To find the number halfway between two numbers, we need to calculate their average. 
First, we need to find a common denominator for 1/6 and 1/4. The least common denominator is 12. 
1/6 = 2/12 and 1/4 = 3/12. 
The average is (2/12 + 3/12) / 2 = (5/12) / 2 = 5/24. 
Let's examine the choices: A. 1/10, B. 1/5, C. 5/24, D. 7/24, E. 5/12. 
Our calculation matches choice C.

ANSWER 2: C

Problem 3:
The numbers given are −2, 4, 6, 9, and 12. We need to rearrange them according to the given rules and find the average of the first and last numbers. 
First, let's arrange the numbers in ascending order: −2, 4, 6, 9, 12. 
The largest number is 12, the smallest is −2, and the median is 6. 
According to the rules: the largest isn't first but is in one of the first three places, the smallest isn't last but is in one of the last three places, and the median isn't first or last. 
Let's try to satisfy these conditions. 
Since the median (6) can't be first or last, it must be in the second, third, or fourth position. 
If 6 is in the third position, the arrangement could be: x, y, 6, z, w. 
Given that 12 is in one of the first three places and −2 is in one of the last three places, a possible arrangement is: 9, 12, 6, 4, −2 or 4, 12, 6, 9, −2. 
However, we need to check if these satisfy all conditions. 
For 9, 12, 6, 4, −2: 12 is in the second position (satisfies the first condition), and −2 is in the last position (does not satisfy the second condition because −2 is last). 
For 4, 12, 6, 9, −2: again, −2 is last, so it doesn't satisfy the condition. 
Let's consider 6 in the second position: x, 6, y, z, w. 
A possible arrangement could be: 9, 6, 12, −2, 4. 
Here, 12 is in the third position, and −2 is in the fourth, satisfying the conditions that 12 is in one of the first three places and −2 is in one of the last three. 
The first number is 9, and the last is 4. The average is (9 + 4) / 2 = 6.5. 
Let's examine the choices: A. 3.5, B. 5, C. 6.5, D. 7.5, E. 8. 
Our calculation matches choice C.

ANSWER 3: C

Problem 4:
The problem gives the equation 991 + 993 + 995 + 997 + 999 = 5000 − N and asks to find N. 
First, let's simplify the left-hand side of the equation. 
991 + 993 + 995 + 997 + 999 can be rewritten as (1000 - 9) + (1000 - 7) + (1000 - 5) + (1000 - 3) + (1000 - 1). 
This equals 5*1000 - (9 + 7 + 5 + 3 + 1) = 5000 - 25. 
So, 5000 - 25 = 5000 - N. 
This implies that N = 25. 
Let's examine the choices: A. 5, B. 10, C. 15, D. 20, E. 25. 
Our calculation matches choice E.

ANSWER 4: E

Problem 5:
The problem asks us to find how many 3-digit positive integers have digits whose product equals 24. 
To solve this, we need to list the factor combinations of 24 and see which ones can form 3-digit numbers. 
The prime factorization of 24 is 2^3 * 3. 
Possible combinations for the digits are derived from the factors: (1, 1, 24) - not valid since 24 is not a single digit; (1, 2, 12) - not valid; (1, 3, 8); (1, 4, 6); (2, 2, 6); (2, 3, 4); and (2, 2, 6). 
Let's count the permutations for valid combinations: 
- For (1, 3, 8), the permutations are 138, 183, 318, 381, 813, 831, which gives 6 numbers. 
- For (1, 4, 6), the permutations are 146, 164, 416, 461, 614, 641, which gives 6 numbers. 
- For (2, 2, 6), the permutations are 226, 262, 622, which gives 3 numbers. 
- For (2, 3, 4), the permutations are 234, 243, 324, 342, 423, 432, which gives 6 numbers. 
Adding these up: 6 + 6 + 3 + 6 = 21. 
Let's examine the choices: A. 12, B. 15, C. 18, D. 21, E. 24. 
Our calculation matches choice D.

ANSWER 5: D

Problem 6:
Brent has goldfish that quadruple every month, and Gretel has goldfish that double every month. 
Initially, Brent has 4 goldfish, and Gretel has 128 goldfish. 
Let's denote the number of months from the initial time as 'n'. 
After 'n' months, Brent will have 4 * (4^n) goldfish, and Gretel will have 128 * (2^n) goldfish. 
We need to find 'n' when 4 * (4^n) = 128 * (2^n). 
Simplifying, we get 4^(n+1) = 128 * 2^n. 
Since 4 = 2^2, we can rewrite this as (2^2)^(n+1) = 2^7 * 2^n. 
This simplifies to 2^(2n+2) = 2^(n+7). 
Equating the exponents, we get 2n + 2 = n + 7. 
Solving for 'n', we get n = 5. 
Let's examine the choices: A. 4, B. 5, C. 6, D. 7, E. 8. 
Our calculation matches choice B.

ANSWER 6: B

Problem 7:
In ▵ABC, AB = BC = 29, and AC = 42. 
Since AB = BC, ▵ABC is isosceles. 
To find the area, we need the height. 
Let's draw the altitude from B to AC, meeting AC at point D. 
Since ▵ABC is isosceles, D is the midpoint of AC. 
So, AD = DC = 42 / 2 = 21. 
Using the Pythagorean theorem in ▵ADB, we have BD^2 = AB^2 - AD^2 = 29^2 - 21^2. 
Calculating, BD^2 = 841 - 441 = 400, so BD = 20. 
The area of ▵ABC = 0.5 * base * height = 0.5 * 42 * 20 = 420. 
Let's examine the choices: A. 100, B. 420, C. 500, D. 609, E. 701. 
Our calculation matches choice B.

ANSWER 7: B

Problem 8:
The problem asks for the number of different isosceles triangles with integer side lengths and a perimeter of 23. 
For an isosceles triangle, two sides are equal. Let's denote the equal sides as 'a' and the base as 'b'. 
So, 2a + b = 23, and a + a > b (triangle inequality). 
This implies that 2a > b, or 2a > 23 - 2a, which simplifies to 4a > 23, or a > 5.75. 
Since 'a' is an integer, a ≥ 6. 
Also, a + b > a, which i

Problem 1:

We are asked about the **place** occupied by the digit 9 and the digit 3 in

\[
74982.1035
\]

The digit 9 is in the **hundreds** place, so the value of its place is

\[
100
\]

The digit 3 is in the **thousandths** place, so the value of its place is

\[
0.001=\frac{1}{1000}
\]

Now compare:

\[
100 \div 0.001 = 100 \div \frac{1}{1000}=100 \times 1000=100000
\]

So the place of the 9 is \(100{,}000\) times as great as the place of the 3.

Choices A and B are too small, and D and E are too large.

ANSWER 1: C

Problem 2:

We need the number halfway between

\[
\frac16 \quad \text{and} \quad \frac14
\]

The halfway point is the average:

\[
\frac{\frac16+\frac14}{2}
\]

Find a common denominator:

\[
\frac16=\frac{2}{12}, \qquad \frac14=\frac{3}{12}
\]

So

\[
\frac16+\frac14=\frac{2}{12}+\frac{3}{12}=\frac{5}{12}
\]

Now divide by 2:

\[
\frac{5}{12}\div 2=\frac{5}{12}\cdot \frac12=\frac{5}{24}
\]

So the number halfway between them is \(\frac{5}{24}\).

A tempting wrong answer is \(\frac15\), but that comes from averaging the denominators, which is not how fractions are averaged.

ANSWER 2: C

Problem 3:

The numbers are

\[
-2,\ 4,\ 6,\ 9,\ 12
\]

We are rearranging them using these rules:

- The largest number, \(12\), is not first, but is in one of the first three places.
- The smallest number, \(-2\), is not last, but is in one of the last three places.
- The median number, \(6\), is not first or last.

The largest number \(12\) must be in position 2 or 3.

The smallest number \(-2\) must be in position 3 or 4.

The median \(6\) must be in position 2, 3, or 4.

So the numbers \(12\), \(-2\), and \(6\) occupy the middle three positions in some order.

That means the first and last positions must be filled by the remaining two numbers:

\[
4 \quad \text{and} \quad 9
\]

Their average is

\[
\frac{4+9}{2}=\frac{13}{2}=6.5
\]

So the average of the first and last numbers is \(6.5\).

ANSWER 3: C

Problem 4:

We are given

\[
991+993+995+997+999=5000-N
\]

Add the five numbers. Since they are evenly spaced, their average is the middle number:

\[
995
\]

There are 5 numbers, so the sum is

\[
5 \cdot 995=4975
\]

Thus

\[
4975=5000-N
\]

Solve for \(N\):

\[
N=5000-4975=25
\]

So \(N=25\).

ANSWER 4: E

Problem 5:

We need to count 3-digit positive integers whose digits multiply to 24.

Since the product is 24, no digit can be 0.

Factor:

\[
24=2^3\cdot 3
\]

Now list digit triples whose product is 24:

\[
1,3,8
\]

\[
1,4,6
\]

\[
2,2,6
\]

\[
2,3,4
\]

Now count rearrangements.

For \(1,3,8\), all digits are different:

\[
3!=6
\]

For \(1,4,6\), all digits are different:

\[
3!=6
\]

For \(2,2,6\), two digits are the same, so there are

\[
\frac{3!}{2!}=3
\]

For \(2,3,4\), all digits are different:

\[
3!=6
\]

Total:

\[
6+6+3+6=21
\]

So there are 21 such integers.

ANSWER 5: D

Problem 6:

Brent starts with 4 goldfish and quadruples every month.

After \(m\) months, Brent has

\[
4\cdot 4^m
\]

Gretel starts with 128 goldfish and doubles every month.

After \(m\) months, Gretel has

\[
128\cdot 2^m
\]

We want them equal:

\[
4\cdot 4^m=128\cdot 2^m
\]

Rewrite everything as powers of 2:

\[
4=2^2,\quad 4^m=2^{2m},\quad 128=2^7
\]

So

\[
2^2\cdot 2^{2m}=2^7\cdot 2^m
\]

\[
2^{2+2m}=2^{7+m}
\]

Therefore,

\[
2+2m=7+m
\]

\[
m=5
\]

So in 5 months, they will have the same number of goldfish.

ANSWER 6: B

Problem 7:

We have an isosceles triangle with

\[
AB=BC=29
\]

and base

\[
AC=42
\]

The altitude from \(B\) to \(AC\) bisects the base, so each half is

\[
\frac{42}{2}=21
\]

Now use the Pythagorean Theorem to find the height \(h\):

\[
h^2+21^2=29^2
\]

\[
h^2+441=841
\]

\[
h^2=400
\]

\[
h=20
\]

Now find the area:

\[
\text{Area}=\frac12 \cdot \text{base} \cdot \text{height}
\]

\[
\text{Area}=\frac12 \cdot 42 \cdot 20=420
\]

So the area is 420.

ANSWER 7: B

Problem 8:

We need isosceles triangles with integer side lengths and perimeter 23.

Let the equal sides each have length \(a\), and let the base have length \(b\).

Then

\[
2a+b=23
\]

So

\[
b=23-2a
\]

For a triangle, the sum of the two equal sides must be greater than the base:

\[
2a>b
\]

Substitute \(b=23-2a\):

\[
2a>23-2a
\]

\[
4a>23
\]

So

\[
a>5.75
\]

Thus

\[
a\ge 6
\]

Also, \(b\) must be positive:

\[
23-2a>0
\]

\[
2a<23
\]

\[
a<11.5
\]

Thus

\[
a\le 11
\]

So possible values of \(a\) are

\[
6,7,8,9,10,11
\]

That gives 6 different triangles.

ANSWER 8: C

Problem 9:

We draw three digits from

\[
1,2,3,4
\]

and form a three-digit number.

A number is divisible by 3 if the sum of its digits is divisible by 3.

The sum of all four digits is

\[
1+2+3+4=10
\]

We choose three digits, which is the same as leaving out one digit.

If we leave out digit \(d\), then the chosen digit sum is

\[
10-d
\]

We need

\[
10-d
\]

to be divisible by 3.

Check the possible omitted digits:

- Omit 1: sum is \(9\), divisible by 3.
- Omit 2: sum is \(8\), not divisible by 3.
- Omit 3: sum is \(7\), not divisible by 3.
- Omit 4: sum is \(6\), divisible by 3.

So 2 of the 4 possible sets of digits work.

For each chosen set, all rearrangements are also multiples of 3, since digit sum does not change.

Thus the probability is

\[
\frac{2}{4}=\frac12
\]

ANSWER 9: C

Problem 10:

If the phone is on but not being used, the battery lasts 24 hours, so it uses

\[
\frac{1}{24}
\]

of the battery per hour.

If the phone is being used constantly, the battery lasts 3 hours, so it uses

\[
\frac{1}{3}
\]

of the battery per hour.

Since the last recharge, the phone has been on for 9 hours, and Niki used it for 60 minutes, which is 1 hour.

So the phone was:

- used for 1 hour
- on but not used for 8 hours

Battery used while talking:

\[
1\cdot \frac13=\frac13
\]

Battery used while on but idle:

\[
8\cdot \frac{1}{24}=\frac{8}{24}=\frac13
\]

Total battery used:

\[
\frac13+\frac13=\frac23
\]

Battery remaining:

\[
1-\frac23=\frac13
\]

Now she will leave the phone on but not talk, so it uses \(\frac{1}{24}\)

Problem 1:
**Restatement:** We are asked to find how many times greater the value of the place occupied by the digit 9 is compared to the value of the place occupied by the digit 3 in the number 74982.1035.

**Step-by-step Working:**
1. Identify the place value of the digit 9: In 74982.1035, the 9 is in the hundreds place. Its value is $9 \times 100$, so the place value is $100$.
2. Identify the place value of the digit 3: The 3 is in the thousandths place (three places to the right of the decimal point). Its value is $3 \times 0.001$, so the place value is $0.001$ (or $\frac{1}{1000}$).
3. Calculate the ratio: To find out how many times as great the hundreds place is compared to the thousandths place, we divide the two place values:
$$\frac{100}{0.001} = 100 \times 1000 = 100,000$$

**Ruling out wrong choices:** 
- A (1,000) and B (10,000) are tempting if you miscount the decimal places between the two digits.
- D (1,000,000) would be the answer if the 3 were in the ten-thousandths place.

ANSWER 1: C

Problem 2:
**Restatement:** We need to find the number that is exactly halfway between the fractions $\frac{1}{6}$ and $\frac{1}{4}$.

**Step-by-step Working:**
1. The number halfway between two numbers is their average. We need to calculate $\frac{\frac{1}{6} + \frac{1}{4}}{2}$.
2. Find a common denominator to add the fractions. The least common multiple of 6 and 4 is 12.
$$\frac{1}{6} = \frac{2}{12}$$
$$\frac{1}{4} = \frac{3}{12}$$
3. Add the fractions:
$$\frac{2}{12} + \frac{3}{12} = \frac{5}{12}$$
4. Divide the sum by 2 to find the average:
$$\frac{\frac{5}{12}}{2} = \frac{5}{12} \times \frac{1}{2} = \frac{5}{24}$$

**Ruling out wrong choices:**
- B ($\frac{1}{5}$) is a common mistake made by incorrectly "averaging" the denominators (halfway between 4 and 6).
- E ($\frac{5}{12}$) is the sum of the two fractions, forgetting to divide by 2.

ANSWER 2: C

Problem 3:
**Restatement:** We are given the numbers $-2, 4, 6, 9,$ and $12$. They are placed in five positions based on specific rules. We need to find the average of the numbers in the first and last positions.

**Step-by-step Working:**
1. Identify the key numbers:
   - Largest: $12$
   - Smallest: $-2$
   - Median (middle value when sorted): $6$
   - Remaining numbers: $4$ and $9$
2. Analyze the rules for the 5 positions (1, 2, 3, 4, 5):
   - The largest ($12$) isn't first, but is in the first three places. So, $12$ must be in position 2 or 3. This means $12$ is NOT in position 1 or 5.
   - The smallest ($-2$) isn't last, but is in the last three places. So, $-2$ must be in position 3 or 4. This means $-2$ is NOT in position 1 or 5.
   - The median ($6$) isn't first or last. So, $6$ is NOT in position 1 or 5.
3. Determine the first and last numbers:
   Since $12$, $-2$, and $6$ are all forbidden from being in position 1 and position 5, the only numbers left to fill the first and last spots are $4$ and $9$.
4. Calculate the average of the first and last numbers:
   $$\frac{4 + 9}{2} = \frac{13}{2} = 6.5$$

**Ruling out wrong choices:**
- If you mistakenly thought the median was part of the ends, you might get 5 (average of 4 and 6) or 7.5 (average of 6 and 9). 

ANSWER 3: C

Problem 4:
**Restatement:** We are given the equation $991 + 993 + 995 + 997 + 999 = 5000 - N$ and need to solve for $N$.

**Step-by-step Working:**
1. Notice that each number on the left side is slightly less than 1000. We can rewrite each term:
   $991 = 1000 - 9$
   $993 = 1000 - 7$
   $995 = 1000 - 5$
   $997 = 1000 - 3$
   $999 = 1000 - 1$
2. Substitute these into the sum:
   $(1000 - 9) + (1000 - 7) + (1000 - 5) + (1000 - 3) + (1000 - 1)$
3. Group the 1000s and the subtracted amounts:
   $= 5000 - (9 + 7 + 5 + 3 + 1)$
4. Add the subtracted amounts:
   $9 + 7 + 5 + 3 + 1 = 25$
5. The sum is $5000 - 25$. Therefore, $N = 25$.

**Ruling out wrong choices:**
- A (5) is tempting if you just count the number of terms being added.
- C (15) or D (20) might result from an arithmetic error when adding the odd numbers.

ANSWER 4: E

Problem 5:
**Restatement:** We need to find how many 3-digit positive integers exist such that the product of their three digits is exactly 24.

**Step-by-step Working:**
1. Find all combinations of 3 single-digit numbers (from 1 to 9) that multiply to 24.
   - The prime factorization of 24 is $2^3 \times 3$.
   - Possible sets of digits:
     - $\{1, 3, 8\}$
     - $\{1, 4, 6\}$
     - $\{2, 2, 6\}$
     - $\{2, 3, 4\}$
   (No other combinations work. For example, we cannot use 9, 7, or 5 because 24 is not divisible by them.)
2. Count the number of permutations (arrangements) for each set to form 3-digit numbers:
   - $\{1, 3, 8\}$: 3 distinct digits $\rightarrow 3! = 6$ numbers.
   - $\{1, 4, 6\}$: 3 distinct digits $\rightarrow 3! = 6$ numbers.
   - $\{2, 3, 4\}$: 3 distinct digits $\rightarrow 3! = 6$ numbers.
   - $\{2, 2, 6\}$: 2 identical digits $\rightarrow \frac{3!}{2!} = 3$ numbers.
3. Add them up:
   $6 + 6 + 6 + 3 = 21$

**Ruling out wrong choices:**
- E (24) is tempting if one assumes all 4 sets have 6 permutations ($4 \times 6 = 24$).
- C (18) is tempting if one misses the $\{2, 2, 6\}$ set entirely.

ANSWER 5: D

Problem 6:
**Restatement:** Brent starts with 4 goldfish that quadruple every month. Gretel starts with 128 goldfish that double every month. We need to find the number of months until they have the same amount.

**Step-by-step Working:**
1. Let $m$ be the number of months.
2. Write an expression for Brent's goldfish:
   $B(m) = 4 \times 4^m$
3. Write an expression for Gretel's goldfish:
   $G(m) = 128 \times 2^m$
4. Set them equal to each other:
   $4 \times 4^m = 128 \times 2^m$
5. Express all bases as powers of 2 to easily solve for $m$:
   $4 = 2^2$
   $128 = 2^7$
   $2^2 \times (2^2)^m = 2^7 \times 2^m$
   $2^2 \times 2^{2m} = 2^{m+7}$
   $2^{2m+2} = 2^{m+7}$
6. Equate the exponents:
   $2m + 2 = m + 7$
   $m = 5$

**Ruling out wrong choices:**
- A (4) or C (6) could be obtained if one makes an arithmetic error 

Problem 1:
The problem asks us to compare the place value of the digits 9 and 3 in the number 74982.1035. 
First, we identify the positions of 9 and 3. The digit 9 is in the thousands place, and the digit 3 is in the thousandths place. 
The value of the place occupied by 9 is 1000, and the value of the place occupied by 3 is 0.001. 
To find how many times as great the value of the place occupied by 9 is compared to the value of the place occupied by 3, we calculate 1000 / 0.001. 
This is equal to 1000 * 1000 = 1,000,000. 
Let's examine the choices: A. 1,000, B. 10,000, C. 100,000, D. 1,000,000, E. 10,000,000. 
Our calculation matches choice D.

ANSWER 1: D

Problem 2:
The problem asks for the number halfway between 1/6 and 1/4. 
To find the number halfway between two numbers, we need to calculate their average. 
First, we need to find a common denominator for 1/6 and 1/4. The least common denominator is 12. 
1/6 = 2/12 and 1/4 = 3/12. 
The average is (2/12 + 3/12) / 2 = (5/12) / 2 = 5/24. 
Let's examine the choices: A. 1/10, B. 1/5, C. 5/24, D. 7/24, E. 5/12. 
Our calculation matches choice C.

ANSWER 2: C

Problem 3:
The numbers given are −2, 4, 6, 9, and 12. We need to rearrange them according to the given rules and find the average of the first and last numbers. 
First, let's arrange the numbers in ascending order: −2, 4, 6, 9, 12. 
The largest number is 12, the smallest is −2, and the median is 6. 
According to the rules: the largest isn't first but is in one of the first three places, the smallest isn't last but is in one of the last three places, and the median isn't first or last. 
Let's try to satisfy these conditions. 
Since the median (6) can't be first or last, it must be in the second, third, or fourth position. 
If 6 is in the third position, the arrangement could be: x, y, 6, z, w. 
Given that 12 is in one of the first three places and −2 is in one of the last three places, a possible arrangement is: 9, 12, 6, 4, −2 or 4, 12, 6, 9, −2. 
However, we need to check if these satisfy all conditions. 
For 9, 12, 6, 4, −2: 12 is in the second position (satisfies the first condition), and −2 is in the last position (does not satisfy the second condition because −2 is last). 
For 4, 12, 6, 9, −2: again, −2 is last, so it doesn't satisfy the condition. 
Let's consider 6 in the second position: x, 6, y, z, w. 
A possible arrangement could be: 9, 6, 12, −2, 4. 
Here, 12 is in the third position, and −2 is in the fourth, satisfying the conditions that 12 is in one of the first three places and −2 is in one of the last three. 
The first number is 9, and the last is 4. The average is (9 + 4) / 2 = 6.5. 
Let's examine the choices: A. 3.5, B. 5, C. 6.5, D. 7.5, E. 8. 
Our calculation matches choice C.

ANSWER 3: C

Problem 4:
The problem gives the equation 991 + 993 + 995 + 997 + 999 = 5000 − N and asks to find N. 
First, let's simplify the left-hand side of the equation. 
991 + 993 + 995 + 997 + 999 can be rewritten as (1000 - 9) + (1000 - 7) + (1000 - 5) + (1000 - 3) + (1000 - 1). 
This equals 5*1000 - (9 + 7 + 5 + 3 + 1) = 5000 - 25. 
So, 5000 - 25 = 5000 - N. 
This implies that N = 25. 
Let's examine the choices: A. 5, B. 10, C. 15, D. 20, E. 25. 
Our calculation matches choice E.

ANSWER 4: E

Problem 5:
The problem asks us to find how many 3-digit positive integers have digits whose product equals 24. 
To solve this, we need to list the factor combinations of 24 and see which ones can form 3-digit numbers. 
The prime factorization of 24 is 2^3 * 3. 
Possible combinations for the digits are derived from the factors: (1, 1, 24) - not valid since 24 is not a single digit; (1, 2, 12) - not valid; (1, 3, 8); (1, 4, 6); (2, 2, 6); (2, 3, 4); and (2, 2, 6). 
Let's count the permutations for valid combinations: 
- For (1, 3, 8), the permutations are 138, 183, 318, 381, 813, 831, which gives 6 numbers. 
- For (1, 4, 6), the permutations are 146, 164, 416, 461, 614, 641, which gives 6 numbers. 
- For (2, 2, 6), the permutations are 226, 262, 622, which gives 3 numbers. 
- For (2, 3, 4), the permutations are 234, 243, 324, 342, 423, 432, which gives 6 numbers. 
Adding these up: 6 + 6 + 3 + 6 = 21. 
Let's examine the choices: A. 12, B. 15, C. 18, D. 21, E. 24. 
Our calculation matches choice D.

ANSWER 5: D

Problem 6:
Brent has goldfish that quadruple every month, and Gretel has goldfish that double every month. 
Initially, Brent has 4 goldfish, and Gretel has 128 goldfish. 
Let's denote the number of months from the initial time as 'n'. 
After 'n' months, Brent will have 4 * (4^n) goldfish, and Gretel will have 128 * (2^n) goldfish. 
We need to find 'n' when 4 * (4^n) = 128 * (2^n). 
Simplifying, we get 4^(n+1) = 128 * 2^n. 
Since 4 = 2^2, we can rewrite this as (2^2)^(n+1) = 2^7 * 2^n. 
This simplifies to 2^(2n+2) = 2^(n+7). 
Equating the exponents, we get 2n + 2 = n + 7. 
Solving for 'n', we get n = 5. 
Let's examine the choices: A. 4, B. 5, C. 6, D. 7, E. 8. 
Our calculation matches choice B.

ANSWER 6: B

Problem 7:
In ▵ABC, AB = BC = 29, and AC = 42. 
Since AB = BC, ▵ABC is isosceles. 
To find the area, we need the height. 
Let's draw the altitude from B to AC, meeting AC at point D. 
Since ▵ABC is isosceles, D is the midpoint of AC. 
So, AD = DC = 42 / 2 = 21. 
Using the Pythagorean theorem in ▵ADB, we have BD^2 = AB^2 - AD^2 = 29^2 - 21^2. 
Calculating, BD^2 = 841 - 441 = 400, so BD = 20. 
The area of ▵ABC = 0.5 * base * height = 0.5 * 42 * 20 = 420. 
Let's examine the choices: A. 100, B. 420, C. 500, D. 609, E. 701. 
Our calculation matches choice B.

ANSWER 7: B

Problem 8:
The problem asks for the number of different isosceles triangles with integer side lengths and a perimeter of 23. 
For an isosceles triangle, two sides are equal. Let's denote the equal sides as 'a' and the base as 'b'. 
So, 2a + b = 23, and a + a > b (triangle inequality). 
This implies that 2a > b, or 2a > 23 - 2a, which simplifies to 4a > 23, or a > 5.75. 
Since 'a' is an integer, a ≥ 6. 
Also, a + b > a, which i

Problem 1:

We are asked about the **place** occupied by the digit 9 and the digit 3 in

\[
74982.1035
\]

The digit 9 is in the **hundreds** place, so the value of its place is

\[
100
\]

The digit 3 is in the **thousandths** place, so the value of its place is

\[
0.001=\frac{1}{1000}
\]

Now compare:

\[
100 \div 0.001 = 100 \div \frac{1}{1000}=100 \times 1000=100000
\]

So the place of the 9 is \(100{,}000\) times as great as the place of the 3.

Choices A and B are too small, and D and E are too large.

ANSWER 1: C

Problem 2:

We need the number halfway between

\[
\frac16 \quad \text{and} \quad \frac14
\]

The halfway point is the average:

\[
\frac{\frac16+\frac14}{2}
\]

Find a common denominator:

\[
\frac16=\frac{2}{12}, \qquad \frac14=\frac{3}{12}
\]

So

\[
\frac16+\frac14=\frac{2}{12}+\frac{3}{12}=\frac{5}{12}
\]

Now divide by 2:

\[
\frac{5}{12}\div 2=\frac{5}{12}\cdot \frac12=\frac{5}{24}
\]

So the number halfway between them is \(\frac{5}{24}\).

A tempting wrong answer is \(\frac15\), but that comes from averaging the denominators, which is not how fractions are averaged.

ANSWER 2: C

Problem 3:

The numbers are

\[
-2,\ 4,\ 6,\ 9,\ 12
\]

We are rearranging them using these rules:

- The largest number, \(12\), is not first, but is in one of the first three places.
- The smallest number, \(-2\), is not last, but is in one of the last three places.
- The median number, \(6\), is not first or last.

The largest number \(12\) must be in position 2 or 3.

The smallest number \(-2\) must be in position 3 or 4.

The median \(6\) must be in position 2, 3, or 4.

So the numbers \(12\), \(-2\), and \(6\) occupy the middle three positions in some order.

That means the first and last positions must be filled by the remaining two numbers:

\[
4 \quad \text{and} \quad 9
\]

Their average is

\[
\frac{4+9}{2}=\frac{13}{2}=6.5
\]

So the average of the first and last numbers is \(6.5\).

ANSWER 3: C

Problem 4:

We are given

\[
991+993+995+997+999=5000-N
\]

Add the five numbers. Since they are evenly spaced, their average is the middle number:

\[
995
\]

There are 5 numbers, so the sum is

\[
5 \cdot 995=4975
\]

Thus

\[
4975=5000-N
\]

Solve for \(N\):

\[
N=5000-4975=25
\]

So \(N=25\).

ANSWER 4: E

Problem 5:

We need to count 3-digit positive integers whose digits multiply to 24.

Since the product is 24, no digit can be 0.

Factor:

\[
24=2^3\cdot 3
\]

Now list digit triples whose product is 24:

\[
1,3,8
\]

\[
1,4,6
\]

\[
2,2,6
\]

\[
2,3,4
\]

Now count rearrangements.

For \(1,3,8\), all digits are different:

\[
3!=6
\]

For \(1,4,6\), all digits are different:

\[
3!=6
\]

For \(2,2,6\), two digits are the same, so there are

\[
\frac{3!}{2!}=3
\]

For \(2,3,4\), all digits are different:

\[
3!=6
\]

Total:

\[
6+6+3+6=21
\]

So there are 21 such integers.

ANSWER 5: D

Problem 6:

Brent starts with 4 goldfish and quadruples every month.

After \(m\) months, Brent has

\[
4\cdot 4^m
\]

Gretel starts with 128 goldfish and doubles every month.

After \(m\) months, Gretel has

\[
128\cdot 2^m
\]

We want them equal:

\[
4\cdot 4^m=128\cdot 2^m
\]

Rewrite everything as powers of 2:

\[
4=2^2,\quad 4^m=2^{2m},\quad 128=2^7
\]

So

\[
2^2\cdot 2^{2m}=2^7\cdot 2^m
\]

\[
2^{2+2m}=2^{7+m}
\]

Therefore,

\[
2+2m=7+m
\]

\[
m=5
\]

So in 5 months, they will have the same number of goldfish.

ANSWER 6: B

Problem 7:

We have an isosceles triangle with

\[
AB=BC=29
\]

and base

\[
AC=42
\]

The altitude from \(B\) to \(AC\) bisects the base, so each half is

\[
\frac{42}{2}=21
\]

Now use the Pythagorean Theorem to find the height \(h\):

\[
h^2+21^2=29^2
\]

\[
h^2+441=841
\]

\[
h^2=400
\]

\[
h=20
\]

Now find the area:

\[
\text{Area}=\frac12 \cdot \text{base} \cdot \text{height}
\]

\[
\text{Area}=\frac12 \cdot 42 \cdot 20=420
\]

So the area is 420.

ANSWER 7: B

Problem 8:

We need isosceles triangles with integer side lengths and perimeter 23.

Let the equal sides each have length \(a\), and let the base have length \(b\).

Then

\[
2a+b=23
\]

So

\[
b=23-2a
\]

For a triangle, the sum of the two equal sides must be greater than the base:

\[
2a>b
\]

Substitute \(b=23-2a\):

\[
2a>23-2a
\]

\[
4a>23
\]

So

\[
a>5.75
\]

Thus

\[
a\ge 6
\]

Also, \(b\) must be positive:

\[
23-2a>0
\]

\[
2a<23
\]

\[
a<11.5
\]

Thus

\[
a\le 11
\]

So possible values of \(a\) are

\[
6,7,8,9,10,11
\]

That gives 6 different triangles.

ANSWER 8: C

Problem 9:

We draw three digits from

\[
1,2,3,4
\]

and form a three-digit number.

A number is divisible by 3 if the sum of its digits is divisible by 3.

The sum of all four digits is

\[
1+2+3+4=10
\]

We choose three digits, which is the same as leaving out one digit.

If we leave out digit \(d\), then the chosen digit sum is

\[
10-d
\]

We need

\[
10-d
\]

to be divisible by 3.

Check the possible omitted digits:

- Omit 1: sum is \(9\), divisible by 3.
- Omit 2: sum is \(8\), not divisible by 3.
- Omit 3: sum is \(7\), not divisible by 3.
- Omit 4: sum is \(6\), divisible by 3.

So 2 of the 4 possible sets of digits work.

For each chosen set, all rearrangements are also multiples of 3, since digit sum does not change.

Thus the probability is

\[
\frac{2}{4}=\frac12
\]

ANSWER 9: C

Problem 10:

If the phone is on but not being used, the battery lasts 24 hours, so it uses

\[
\frac{1}{24}
\]

of the battery per hour.

If the phone is being used constantly, the battery lasts 3 hours, so it uses

\[
\frac{1}{3}
\]

of the battery per hour.

Since the last recharge, the phone has been on for 9 hours, and Niki used it for 60 minutes, which is 1 hour.

So the phone was:

- used for 1 hour
- on but not used for 8 hours

Battery used while talking:

\[
1\cdot \frac13=\frac13
\]

Battery used while on but idle:

\[
8\cdot \frac{1}{24}=\frac{8}{24}=\frac13
\]

Total battery used:

\[
\frac13+\frac13=\frac23
\]

Battery remaining:

\[
1-\frac23=\frac13
\]

Now she will leave the phone on but not talk, so it uses \(\frac{1}{24}\)

Problem 1:
**Restatement:** We are asked to find how many times greater the value of the place occupied by the digit 9 is compared to the value of the place occupied by the digit 3 in the number 74982.1035.

**Step-by-step Working:**
1. Identify the place value of the digit 9: In 74982.1035, the 9 is in the hundreds place. Its value is $9 \times 100$, so the place value is $100$.
2. Identify the place value of the digit 3: The 3 is in the thousandths place (three places to the right of the decimal point). Its value is $3 \times 0.001$, so the place value is $0.001$ (or $\frac{1}{1000}$).
3. Calculate the ratio: To find out how many times as great the hundreds place is compared to the thousandths place, we divide the two place values:
$$\frac{100}{0.001} = 100 \times 1000 = 100,000$$

**Ruling out wrong choices:** 
- A (1,000) and B (10,000) are tempting if you miscount the decimal places between the two digits.
- D (1,000,000) would be the answer if the 3 were in the ten-thousandths place.

ANSWER 1: C

Problem 2:
**Restatement:** We need to find the number that is exactly halfway between the fractions $\frac{1}{6}$ and $\frac{1}{4}$.

**Step-by-step Working:**
1. The number halfway between two numbers is their average. We need to calculate $\frac{\frac{1}{6} + \frac{1}{4}}{2}$.
2. Find a common denominator to add the fractions. The least common multiple of 6 and 4 is 12.
$$\frac{1}{6} = \frac{2}{12}$$
$$\frac{1}{4} = \frac{3}{12}$$
3. Add the fractions:
$$\frac{2}{12} + \frac{3}{12} = \frac{5}{12}$$
4. Divide the sum by 2 to find the average:
$$\frac{\frac{5}{12}}{2} = \frac{5}{12} \times \frac{1}{2} = \frac{5}{24}$$

**Ruling out wrong choices:**
- B ($\frac{1}{5}$) is a common mistake made by incorrectly "averaging" the denominators (halfway between 4 and 6).
- E ($\frac{5}{12}$) is the sum of the two fractions, forgetting to divide by 2.

ANSWER 2: C

Problem 3:
**Restatement:** We are given the numbers $-2, 4, 6, 9,$ and $12$. They are placed in five positions based on specific rules. We need to find the average of the numbers in the first and last positions.

**Step-by-step Working:**
1. Identify the key numbers:
   - Largest: $12$
   - Smallest: $-2$
   - Median (middle value when sorted): $6$
   - Remaining numbers: $4$ and $9$
2. Analyze the rules for the 5 positions (1, 2, 3, 4, 5):
   - The largest ($12$) isn't first, but is in the first three places. So, $12$ must be in position 2 or 3. This means $12$ is NOT in position 1 or 5.
   - The smallest ($-2$) isn't last, but is in the last three places. So, $-2$ must be in position 3 or 4. This means $-2$ is NOT in position 1 or 5.
   - The median ($6$) isn't first or last. So, $6$ is NOT in position 1 or 5.
3. Determine the first and last numbers:
   Since $12$, $-2$, and $6$ are all forbidden from being in position 1 and position 5, the only numbers left to fill the first and last spots are $4$ and $9$.
4. Calculate the average of the first and last numbers:
   $$\frac{4 + 9}{2} = \frac{13}{2} = 6.5$$

**Ruling out wrong choices:**
- If you mistakenly thought the median was part of the ends, you might get 5 (average of 4 and 6) or 7.5 (average of 6 and 9). 

ANSWER 3: C

Problem 4:
**Restatement:** We are given the equation $991 + 993 + 995 + 997 + 999 = 5000 - N$ and need to solve for $N$.

**Step-by-step Working:**
1. Notice that each number on the left side is slightly less than 1000. We can rewrite each term:
   $991 = 1000 - 9$
   $993 = 1000 - 7$
   $995 = 1000 - 5$
   $997 = 1000 - 3$
   $999 = 1000 - 1$
2. Substitute these into the sum:
   $(1000 - 9) + (1000 - 7) + (1000 - 5) + (1000 - 3) + (1000 - 1)$
3. Group the 1000s and the subtracted amounts:
   $= 5000 - (9 + 7 + 5 + 3 + 1)$
4. Add the subtracted amounts:
   $9 + 7 + 5 + 3 + 1 = 25$
5. The sum is $5000 - 25$. Therefore, $N = 25$.

**Ruling out wrong choices:**
- A (5) is tempting if you just count the number of terms being added.
- C (15) or D (20) might result from an arithmetic error when adding the odd numbers.

ANSWER 4: E

Problem 5:
**Restatement:** We need to find how many 3-digit positive integers exist such that the product of their three digits is exactly 24.

**Step-by-step Working:**
1. Find all combinations of 3 single-digit numbers (from 1 to 9) that multiply to 24.
   - The prime factorization of 24 is $2^3 \times 3$.
   - Possible sets of digits:
     - $\{1, 3, 8\}$
     - $\{1, 4, 6\}$
     - $\{2, 2, 6\}$
     - $\{2, 3, 4\}$
   (No other combinations work. For example, we cannot use 9, 7, or 5 because 24 is not divisible by them.)
2. Count the number of permutations (arrangements) for each set to form 3-digit numbers:
   - $\{1, 3, 8\}$: 3 distinct digits $\rightarrow 3! = 6$ numbers.
   - $\{1, 4, 6\}$: 3 distinct digits $\rightarrow 3! = 6$ numbers.
   - $\{2, 3, 4\}$: 3 distinct digits $\rightarrow 3! = 6$ numbers.
   - $\{2, 2, 6\}$: 2 identical digits $\rightarrow \frac{3!}{2!} = 3$ numbers.
3. Add them up:
   $6 + 6 + 6 + 3 = 21$

**Ruling out wrong choices:**
- E (24) is tempting if one assumes all 4 sets have 6 permutations ($4 \times 6 = 24$).
- C (18) is tempting if one misses the $\{2, 2, 6\}$ set entirely.

ANSWER 5: D

Problem 6:
**Restatement:** Brent starts with 4 goldfish that quadruple every month. Gretel starts with 128 goldfish that double every month. We need to find the number of months until they have the same amount.

**Step-by-step Working:**
1. Let $m$ be the number of months.
2. Write an expression for Brent's goldfish:
   $B(m) = 4 \times 4^m$
3. Write an expression for Gretel's goldfish:
   $G(m) = 128 \times 2^m$
4. Set them equal to each other:
   $4 \times 4^m = 128 \times 2^m$
5. Express all bases as powers of 2 to easily solve for $m$:
   $4 = 2^2$
   $128 = 2^7$
   $2^2 \times (2^2)^m = 2^7 \times 2^m$
   $2^2 \times 2^{2m} = 2^{m+7}$
   $2^{2m+2} = 2^{m+7}$
6. Equate the exponents:
   $2m + 2 = m + 7$
   $m = 5$

**Ruling out wrong choices:**
- A (4) or C (6) could be obtained if one makes an arithmetic error 

Problem 1:
The problem asks us to compare the place value of the digits 9 and 3 in the number 74982.1035. 
First, we identify the positions of 9 and 3. The digit 9 is in the thousands place, and the digit 3 is in the thousandths place. 
The value of the place occupied by 9 is 1000, and the value of the place occupied by 3 is 0.001. 
To find how many times as great the value of the place occupied by 9 is compared to the value of the place occupied by 3, we calculate 1000 / 0.001. 
This is equal to 1000 * 1000 = 1,000,000. 
Let's examine the choices: A. 1,000, B. 10,000, C. 100,000, D. 1,000,000, E. 10,000,000. 
Our calculation matches choice D.

ANSWER 1: D

Problem 2:
The problem asks for the number halfway between 1/6 and 1/4. 
To find the number halfway between two numbers, we need to calculate their average. 
First, we need to find a common denominator for 1/6 and 1/4. The least common denominator is 12. 
1/6 = 2/12 and 1/4 = 3/12. 
The average is (2/12 + 3/12) / 2 = (5/12) / 2 = 5/24. 
Let's examine the choices: A. 1/10, B. 1/5, C. 5/24, D. 7/24, E. 5/12. 
Our calculation matches choice C.

ANSWER 2: C

Problem 3:
The numbers given are −2, 4, 6, 9, and 12. We need to rearrange them according to the given rules and find the average of the first and last numbers. 
First, let's arrange the numbers in ascending order: −2, 4, 6, 9, 12. 
The largest number is 12, the smallest is −2, and the median is 6. 
According to the rules: the largest isn't first but is in one of the first three places, the smallest isn't last but is in one of the last three places, and the median isn't first or last. 
Let's try to satisfy these conditions. 
Since the median (6) can't be first or last, it must be in the second, third, or fourth position. 
If 6 is in the third position, the arrangement could be: x, y, 6, z, w. 
Given that 12 is in one of the first three places and −2 is in one of the last three places, a possible arrangement is: 9, 12, 6, 4, −2 or 4, 12, 6, 9, −2. 
However, we need to check if these satisfy all conditions. 
For 9, 12, 6, 4, −2: 12 is in the second position (satisfies the first condition), and −2 is in the last position (does not satisfy the second condition because −2 is last). 
For 4, 12, 6, 9, −2: again, −2 is last, so it doesn't satisfy the condition. 
Let's consider 6 in the second position: x, 6, y, z, w. 
A possible arrangement could be: 9, 6, 12, −2, 4. 
Here, 12 is in the third position, and −2 is in the fourth, satisfying the conditions that 12 is in one of the first three places and −2 is in one of the last three. 
The first number is 9, and the last is 4. The average is (9 + 4) / 2 = 6.5. 
Let's examine the choices: A. 3.5, B. 5, C. 6.5, D. 7.5, E. 8. 
Our calculation matches choice C.

ANSWER 3: C

Problem 4:
The problem gives the equation 991 + 993 + 995 + 997 + 999 = 5000 − N and asks to find N. 
First, let's simplify the left-hand side of the equation. 
991 + 993 + 995 + 997 + 999 can be rewritten as (1000 - 9) + (1000 - 7) + (1000 - 5) + (1000 - 3) + (1000 - 1). 
This equals 5*1000 - (9 + 7 + 5 + 3 + 1) = 5000 - 25. 
So, 5000 - 25 = 5000 - N. 
This implies that N = 25. 
Let's examine the choices: A. 5, B. 10, C. 15, D. 20, E. 25. 
Our calculation matches choice E.

ANSWER 4: E

Problem 5:
The problem asks us to find how many 3-digit positive integers have digits whose product equals 24. 
To solve this, we need to list the factor combinations of 24 and see which ones can form 3-digit numbers. 
The prime factorization of 24 is 2^3 * 3. 
Possible combinations for the digits are derived from the factors: (1, 1, 24) - not valid since 24 is not a single digit; (1, 2, 12) - not valid; (1, 3, 8); (1, 4, 6); (2, 2, 6); (2, 3, 4); and (2, 2, 6). 
Let's count the permutations for valid combinations: 
- For (1, 3, 8), the permutations are 138, 183, 318, 381, 813, 831, which gives 6 numbers. 
- For (1, 4, 6), the permutations are 146, 164, 416, 461, 614, 641, which gives 6 numbers. 
- For (2, 2, 6), the permutations are 226, 262, 622, which gives 3 numbers. 
- For (2, 3, 4), the permutations are 234, 243, 324, 342, 423, 432, which gives 6 numbers. 
Adding these up: 6 + 6 + 3 + 6 = 21. 
Let's examine the choices: A. 12, B. 15, C. 18, D. 21, E. 24. 
Our calculation matches choice D.

ANSWER 5: D

Problem 6:
Brent has goldfish that quadruple every month, and Gretel has goldfish that double every month. 
Initially, Brent has 4 goldfish, and Gretel has 128 goldfish. 
Let's denote the number of months from the initial time as 'n'. 
After 'n' months, Brent will have 4 * (4^n) goldfish, and Gretel will have 128 * (2^n) goldfish. 
We need to find 'n' when 4 * (4^n) = 128 * (2^n). 
Simplifying, we get 4^(n+1) = 128 * 2^n. 
Since 4 = 2^2, we can rewrite this as (2^2)^(n+1) = 2^7 * 2^n. 
This simplifies to 2^(2n+2) = 2^(n+7). 
Equating the exponents, we get 2n + 2 = n + 7. 
Solving for 'n', we get n = 5. 
Let's examine the choices: A. 4, B. 5, C. 6, D. 7, E. 8. 
Our calculation matches choice B.

ANSWER 6: B

Problem 7:
In ▵ABC, AB = BC = 29, and AC = 42. 
Since AB = BC, ▵ABC is isosceles. 
To find the area, we need the height. 
Let's draw the altitude from B to AC, meeting AC at point D. 
Since ▵ABC is isosceles, D is the midpoint of AC. 
So, AD = DC = 42 / 2 = 21. 
Using the Pythagorean theorem in ▵ADB, we have BD^2 = AB^2 - AD^2 = 29^2 - 21^2. 
Calculating, BD^2 = 841 - 441 = 400, so BD = 20. 
The area of ▵ABC = 0.5 * base * height = 0.5 * 42 * 20 = 420. 
Let's examine the choices: A. 100, B. 420, C. 500, D. 609, E. 701. 
Our calculation matches choice B.

ANSWER 7: B

Problem 8:
The problem asks for the number of different isosceles triangles with integer side lengths and a perimeter of 23. 
For an isosceles triangle, two sides are equal. Let's denote the equal sides as 'a' and the base as 'b'. 
So, 2a + b = 23, and a + a > b (triangle inequality). 
This implies that 2a > b, or 2a > 23 - 2a, which simplifies to 4a > 23, or a > 5.75. 
Since 'a' is an integer, a ≥ 6. 
Also, a + b > a, which i

Problem 1:

We are asked about the **place** occupied by the digit 9 and the digit 3 in

\[
74982.1035
\]

The digit 9 is in the **hundreds** place, so the value of its place is

\[
100
\]

The digit 3 is in the **thousandths** place, so the value of its place is

\[
0.001=\frac{1}{1000}
\]

Now compare:

\[
100 \div 0.001 = 100 \div \frac{1}{1000}=100 \times 1000=100000
\]

So the place of the 9 is \(100{,}000\) times as great as the place of the 3.

Choices A and B are too small, and D and E are too large.

ANSWER 1: C

Problem 2:

We need the number halfway between

\[
\frac16 \quad \text{and} \quad \frac14
\]

The halfway point is the average:

\[
\frac{\frac16+\frac14}{2}
\]

Find a common denominator:

\[
\frac16=\frac{2}{12}, \qquad \frac14=\frac{3}{12}
\]

So

\[
\frac16+\frac14=\frac{2}{12}+\frac{3}{12}=\frac{5}{12}
\]

Now divide by 2:

\[
\frac{5}{12}\div 2=\frac{5}{12}\cdot \frac12=\frac{5}{24}
\]

So the number halfway between them is \(\frac{5}{24}\).

A tempting wrong answer is \(\frac15\), but that comes from averaging the denominators, which is not how fractions are averaged.

ANSWER 2: C

Problem 3:

The numbers are

\[
-2,\ 4,\ 6,\ 9,\ 12
\]

We are rearranging them using these rules:

- The largest number, \(12\), is not first, but is in one of the first three places.
- The smallest number, \(-2\), is not last, but is in one of the last three places.
- The median number, \(6\), is not first or last.

The largest number \(12\) must be in position 2 or 3.

The smallest number \(-2\) must be in position 3 or 4.

The median \(6\) must be in position 2, 3, or 4.

So the numbers \(12\), \(-2\), and \(6\) occupy the middle three positions in some order.

That means the first and last positions must be filled by the remaining two numbers:

\[
4 \quad \text{and} \quad 9
\]

Their average is

\[
\frac{4+9}{2}=\frac{13}{2}=6.5
\]

So the average of the first and last numbers is \(6.5\).

ANSWER 3: C

Problem 4:

We are given

\[
991+993+995+997+999=5000-N
\]

Add the five numbers. Since they are evenly spaced, their average is the middle number:

\[
995
\]

There are 5 numbers, so the sum is

\[
5 \cdot 995=4975
\]

Thus

\[
4975=5000-N
\]

Solve for \(N\):

\[
N=5000-4975=25
\]

So \(N=25\).

ANSWER 4: E

Problem 5:

We need to count 3-digit positive integers whose digits multiply to 24.

Since the product is 24, no digit can be 0.

Factor:

\[
24=2^3\cdot 3
\]

Now list digit triples whose product is 24:

\[
1,3,8
\]

\[
1,4,6
\]

\[
2,2,6
\]

\[
2,3,4
\]

Now count rearrangements.

For \(1,3,8\), all digits are different:

\[
3!=6
\]

For \(1,4,6\), all digits are different:

\[
3!=6
\]

For \(2,2,6\), two digits are the same, so there are

\[
\frac{3!}{2!}=3
\]

For \(2,3,4\), all digits are different:

\[
3!=6
\]

Total:

\[
6+6+3+6=21
\]

So there are 21 such integers.

ANSWER 5: D

Problem 6:

Brent starts with 4 goldfish and quadruples every month.

After \(m\) months, Brent has

\[
4\cdot 4^m
\]

Gretel starts with 128 goldfish and doubles every month.

After \(m\) months, Gretel has

\[
128\cdot 2^m
\]

We want them equal:

\[
4\cdot 4^m=128\cdot 2^m
\]

Rewrite everything as powers of 2:

\[
4=2^2,\quad 4^m=2^{2m},\quad 128=2^7
\]

So

\[
2^2\cdot 2^{2m}=2^7\cdot 2^m
\]

\[
2^{2+2m}=2^{7+m}
\]

Therefore,

\[
2+2m=7+m
\]

\[
m=5
\]

So in 5 months, they will have the same number of goldfish.

ANSWER 6: B

Problem 7:

We have an isosceles triangle with

\[
AB=BC=29
\]

and base

\[
AC=42
\]

The altitude from \(B\) to \(AC\) bisects the base, so each half is

\[
\frac{42}{2}=21
\]

Now use the Pythagorean Theorem to find the height \(h\):

\[
h^2+21^2=29^2
\]

\[
h^2+441=841
\]

\[
h^2=400
\]

\[
h=20
\]

Now find the area:

\[
\text{Area}=\frac12 \cdot \text{base} \cdot \text{height}
\]

\[
\text{Area}=\frac12 \cdot 42 \cdot 20=420
\]

So the area is 420.

ANSWER 7: B

Problem 8:

We need isosceles triangles with integer side lengths and perimeter 23.

Let the equal sides each have length \(a\), and let the base have length \(b\).

Then

\[
2a+b=23
\]

So

\[
b=23-2a
\]

For a triangle, the sum of the two equal sides must be greater than the base:

\[
2a>b
\]

Substitute \(b=23-2a\):

\[
2a>23-2a
\]

\[
4a>23
\]

So

\[
a>5.75
\]

Thus

\[
a\ge 6
\]

Also, \(b\) must be positive:

\[
23-2a>0
\]

\[
2a<23
\]

\[
a<11.5
\]

Thus

\[
a\le 11
\]

So possible values of \(a\) are

\[
6,7,8,9,10,11
\]

That gives 6 different triangles.

ANSWER 8: C

Problem 9:

We draw three digits from

\[
1,2,3,4
\]

and form a three-digit number.

A number is divisible by 3 if the sum of its digits is divisible by 3.

The sum of all four digits is

\[
1+2+3+4=10
\]

We choose three digits, which is the same as leaving out one digit.

If we leave out digit \(d\), then the chosen digit sum is

\[
10-d
\]

We need

\[
10-d
\]

to be divisible by 3.

Check the possible omitted digits:

- Omit 1: sum is \(9\), divisible by 3.
- Omit 2: sum is \(8\), not divisible by 3.
- Omit 3: sum is \(7\), not divisible by 3.
- Omit 4: sum is \(6\), divisible by 3.

So 2 of the 4 possible sets of digits work.

For each chosen set, all rearrangements are also multiples of 3, since digit sum does not change.

Thus the probability is

\[
\frac{2}{4}=\frac12
\]

ANSWER 9: C

Problem 10:

If the phone is on but not being used, the battery lasts 24 hours, so it uses

\[
\frac{1}{24}
\]

of the battery per hour.

If the phone is being used constantly, the battery lasts 3 hours, so it uses

\[
\frac{1}{3}
\]

of the battery per hour.

Since the last recharge, the phone has been on for 9 hours, and Niki used it for 60 minutes, which is 1 hour.

So the phone was:

- used for 1 hour
- on but not used for 8 hours

Battery used while talking:

\[
1\cdot \frac13=\frac13
\]

Battery used while on but idle:

\[
8\cdot \frac{1}{24}=\frac{8}{24}=\frac13
\]

Total battery used:

\[
\frac13+\frac13=\frac23
\]

Battery remaining:

\[
1-\frac23=\frac13
\]

Now she will leave the phone on but not talk, so it uses \(\frac{1}{24}\)

Problem 1:
**Restatement:** We are asked to find how many times greater the value of the place occupied by the digit 9 is compared to the value of the place occupied by the digit 3 in the number 74982.1035.

**Step-by-step Working:**
1. Identify the place value of the digit 9: In 74982.1035, the 9 is in the hundreds place. Its value is $9 \times 100$, so the place value is $100$.
2. Identify the place value of the digit 3: The 3 is in the thousandths place (three places to the right of the decimal point). Its value is $3 \times 0.001$, so the place value is $0.001$ (or $\frac{1}{1000}$).
3. Calculate the ratio: To find out how many times as great the hundreds place is compared to the thousandths place, we divide the two place values:
$$\frac{100}{0.001} = 100 \times 1000 = 100,000$$

**Ruling out wrong choices:** 
- A (1,000) and B (10,000) are tempting if you miscount the decimal places between the two digits.
- D (1,000,000) would be the answer if the 3 were in the ten-thousandths place.

ANSWER 1: C

Problem 2:
**Restatement:** We need to find the number that is exactly halfway between the fractions $\frac{1}{6}$ and $\frac{1}{4}$.

**Step-by-step Working:**
1. The number halfway between two numbers is their average. We need to calculate $\frac{\frac{1}{6} + \frac{1}{4}}{2}$.
2. Find a common denominator to add the fractions. The least common multiple of 6 and 4 is 12.
$$\frac{1}{6} = \frac{2}{12}$$
$$\frac{1}{4} = \frac{3}{12}$$
3. Add the fractions:
$$\frac{2}{12} + \frac{3}{12} = \frac{5}{12}$$
4. Divide the sum by 2 to find the average:
$$\frac{\frac{5}{12}}{2} = \frac{5}{12} \times \frac{1}{2} = \frac{5}{24}$$

**Ruling out wrong choices:**
- B ($\frac{1}{5}$) is a common mistake made by incorrectly "averaging" the denominators (halfway between 4 and 6).
- E ($\frac{5}{12}$) is the sum of the two fractions, forgetting to divide by 2.

ANSWER 2: C

Problem 3:
**Restatement:** We are given the numbers $-2, 4, 6, 9,$ and $12$. They are placed in five positions based on specific rules. We need to find the average of the numbers in the first and last positions.

**Step-by-step Working:**
1. Identify the key numbers:
   - Largest: $12$
   - Smallest: $-2$
   - Median (middle value when sorted): $6$
   - Remaining numbers: $4$ and $9$
2. Analyze the rules for the 5 positions (1, 2, 3, 4, 5):
   - The largest ($12$) isn't first, but is in the first three places. So, $12$ must be in position 2 or 3. This means $12$ is NOT in position 1 or 5.
   - The smallest ($-2$) isn't last, but is in the last three places. So, $-2$ must be in position 3 or 4. This means $-2$ is NOT in position 1 or 5.
   - The median ($6$) isn't first or last. So, $6$ is NOT in position 1 or 5.
3. Determine the first and last numbers:
   Since $12$, $-2$, and $6$ are all forbidden from being in position 1 and position 5, the only numbers left to fill the first and last spots are $4$ and $9$.
4. Calculate the average of the first and last numbers:
   $$\frac{4 + 9}{2} = \frac{13}{2} = 6.5$$

**Ruling out wrong choices:**
- If you mistakenly thought the median was part of the ends, you might get 5 (average of 4 and 6) or 7.5 (average of 6 and 9). 

ANSWER 3: C

Problem 4:
**Restatement:** We are given the equation $991 + 993 + 995 + 997 + 999 = 5000 - N$ and need to solve for $N$.

**Step-by-step Working:**
1. Notice that each number on the left side is slightly less than 1000. We can rewrite each term:
   $991 = 1000 - 9$
   $993 = 1000 - 7$
   $995 = 1000 - 5$
   $997 = 1000 - 3$
   $999 = 1000 - 1$
2. Substitute these into the sum:
   $(1000 - 9) + (1000 - 7) + (1000 - 5) + (1000 - 3) + (1000 - 1)$
3. Group the 1000s and the subtracted amounts:
   $= 5000 - (9 + 7 + 5 + 3 + 1)$
4. Add the subtracted amounts:
   $9 + 7 + 5 + 3 + 1 = 25$
5. The sum is $5000 - 25$. Therefore, $N = 25$.

**Ruling out wrong choices:**
- A (5) is tempting if you just count the number of terms being added.
- C (15) or D (20) might result from an arithmetic error when adding the odd numbers.

ANSWER 4: E

Problem 5:
**Restatement:** We need to find how many 3-digit positive integers exist such that the product of their three digits is exactly 24.

**Step-by-step Working:**
1. Find all combinations of 3 single-digit numbers (from 1 to 9) that multiply to 24.
   - The prime factorization of 24 is $2^3 \times 3$.
   - Possible sets of digits:
     - $\{1, 3, 8\}$
     - $\{1, 4, 6\}$
     - $\{2, 2, 6\}$
     - $\{2, 3, 4\}$
   (No other combinations work. For example, we cannot use 9, 7, or 5 because 24 is not divisible by them.)
2. Count the number of permutations (arrangements) for each set to form 3-digit numbers:
   - $\{1, 3, 8\}$: 3 distinct digits $\rightarrow 3! = 6$ numbers.
   - $\{1, 4, 6\}$: 3 distinct digits $\rightarrow 3! = 6$ numbers.
   - $\{2, 3, 4\}$: 3 distinct digits $\rightarrow 3! = 6$ numbers.
   - $\{2, 2, 6\}$: 2 identical digits $\rightarrow \frac{3!}{2!} = 3$ numbers.
3. Add them up:
   $6 + 6 + 6 + 3 = 21$

**Ruling out wrong choices:**
- E (24) is tempting if one assumes all 4 sets have 6 permutations ($4 \times 6 = 24$).
- C (18) is tempting if one misses the $\{2, 2, 6\}$ set entirely.

ANSWER 5: D

Problem 6:
**Restatement:** Brent starts with 4 goldfish that quadruple every month. Gretel starts with 128 goldfish that double every month. We need to find the number of months until they have the same amount.

**Step-by-step Working:**
1. Let $m$ be the number of months.
2. Write an expression for Brent's goldfish:
   $B(m) = 4 \times 4^m$
3. Write an expression for Gretel's goldfish:
   $G(m) = 128 \times 2^m$
4. Set them equal to each other:
   $4 \times 4^m = 128 \times 2^m$
5. Express all bases as powers of 2 to easily solve for $m$:
   $4 = 2^2$
   $128 = 2^7$
   $2^2 \times (2^2)^m = 2^7 \times 2^m$
   $2^2 \times 2^{2m} = 2^{m+7}$
   $2^{2m+2} = 2^{m+7}$
6. Equate the exponents:
   $2m + 2 = m + 7$
   $m = 5$

**Ruling out wrong choices:**
- A (4) or C (6) could be obtained if one makes an arithmetic error 

Problem 1:
The problem asks us to compare the place value of the digits 9 and 3 in the number 74982.1035. 
First, we identify the positions of 9 and 3. The digit 9 is in the thousands place, and the digit 3 is in the thousandths place. 
The value of the place occupied by 9 is 1000, and the value of the place occupied by 3 is 0.001. 
To find how many times as great the value of the place occupied by 9 is compared to the value of the place occupied by 3, we calculate 1000 / 0.001. 
This is equal to 1000 * 1000 = 1,000,000. 
Let's examine the choices: A. 1,000, B. 10,000, C. 100,000, D. 1,000,000, E. 10,000,000. 
Our calculation matches choice D.

ANSWER 1: D

Problem 2:
The problem asks for the number halfway between 1/6 and 1/4. 
To find the number halfway between two numbers, we need to calculate their average. 
First, we need to find a common denominator for 1/6 and 1/4. The least common denominator is 12. 
1/6 = 2/12 and 1/4 = 3/12. 
The average is (2/12 + 3/12) / 2 = (5/12) / 2 = 5/24. 
Let's examine the choices: A. 1/10, B. 1/5, C. 5/24, D. 7/24, E. 5/12. 
Our calculation matches choice C.

ANSWER 2: C

Problem 3:
The numbers given are −2, 4, 6, 9, and 12. We need to rearrange them according to the given rules and find the average of the first and last numbers. 
First, let's arrange the numbers in ascending order: −2, 4, 6, 9, 12. 
The largest number is 12, the smallest is −2, and the median is 6. 
According to the rules: the largest isn't first but is in one of the first three places, the smallest isn't last but is in one of the last three places, and the median isn't first or last. 
Let's try to satisfy these conditions. 
Since the median (6) can't be first or last, it must be in the second, third, or fourth position. 
If 6 is in the third position, the arrangement could be: x, y, 6, z, w. 
Given that 12 is in one of the first three places and −2 is in one of the last three places, a possible arrangement is: 9, 12, 6, 4, −2 or 4, 12, 6, 9, −2. 
However, we need to check if these satisfy all conditions. 
For 9, 12, 6, 4, −2: 12 is in the second position (satisfies the first condition), and −2 is in the last position (does not satisfy the second condition because −2 is last). 
For 4, 12, 6, 9, −2: again, −2 is last, so it doesn't satisfy the condition. 
Let's consider 6 in the second position: x, 6, y, z, w. 
A possible arrangement could be: 9, 6, 12, −2, 4. 
Here, 12 is in the third position, and −2 is in the fourth, satisfying the conditions that 12 is in one of the first three places and −2 is in one of the last three. 
The first number is 9, and the last is 4. The average is (9 + 4) / 2 = 6.5. 
Let's examine the choices: A. 3.5, B. 5, C. 6.5, D. 7.5, E. 8. 
Our calculation matches choice C.

ANSWER 3: C

Problem 4:
The problem gives the equation 991 + 993 + 995 + 997 + 999 = 5000 − N and asks to find N. 
First, let's simplify the left-hand side of the equation. 
991 + 993 + 995 + 997 + 999 can be rewritten as (1000 - 9) + (1000 - 7) + (1000 - 5) + (1000 - 3) + (1000 - 1). 
This equals 5*1000 - (9 + 7 + 5 + 3 + 1) = 5000 - 25. 
So, 5000 - 25 = 5000 - N. 
This implies that N = 25. 
Let's examine the choices: A. 5, B. 10, C. 15, D. 20, E. 25. 
Our calculation matches choice E.

ANSWER 4: E

Problem 5:
The problem asks us to find how many 3-digit positive integers have digits whose product equals 24. 
To solve this, we need to list the factor combinations of 24 and see which ones can form 3-digit numbers. 
The prime factorization of 24 is 2^3 * 3. 
Possible combinations for the digits are derived from the factors: (1, 1, 24) - not valid since 24 is not a single digit; (1, 2, 12) - not valid; (1, 3, 8); (1, 4, 6); (2, 2, 6); (2, 3, 4); and (2, 2, 6). 
Let's count the permutations for valid combinations: 
- For (1, 3, 8), the permutations are 138, 183, 318, 381, 813, 831, which gives 6 numbers. 
- For (1, 4, 6), the permutations are 146, 164, 416, 461, 614, 641, which gives 6 numbers. 
- For (2, 2, 6), the permutations are 226, 262, 622, which gives 3 numbers. 
- For (2, 3, 4), the permutations are 234, 243, 324, 342, 423, 432, which gives 6 numbers. 
Adding these up: 6 + 6 + 3 + 6 = 21. 
Let's examine the choices: A. 12, B. 15, C. 18, D. 21, E. 24. 
Our calculation matches choice D.

ANSWER 5: D

Problem 6:
Brent has goldfish that quadruple every month, and Gretel has goldfish that double every month. 
Initially, Brent has 4 goldfish, and Gretel has 128 goldfish. 
Let's denote the number of months from the initial time as 'n'. 
After 'n' months, Brent will have 4 * (4^n) goldfish, and Gretel will have 128 * (2^n) goldfish. 
We need to find 'n' when 4 * (4^n) = 128 * (2^n). 
Simplifying, we get 4^(n+1) = 128 * 2^n. 
Since 4 = 2^2, we can rewrite this as (2^2)^(n+1) = 2^7 * 2^n. 
This simplifies to 2^(2n+2) = 2^(n+7). 
Equating the exponents, we get 2n + 2 = n + 7. 
Solving for 'n', we get n = 5. 
Let's examine the choices: A. 4, B. 5, C. 6, D. 7, E. 8. 
Our calculation matches choice B.

ANSWER 6: B

Problem 7:
In ▵ABC, AB = BC = 29, and AC = 42. 
Since AB = BC, ▵ABC is isosceles. 
To find the area, we need the height. 
Let's draw the altitude from B to AC, meeting AC at point D. 
Since ▵ABC is isosceles, D is the midpoint of AC. 
So, AD = DC = 42 / 2 = 21. 
Using the Pythagorean theorem in ▵ADB, we have BD^2 = AB^2 - AD^2 = 29^2 - 21^2. 
Calculating, BD^2 = 841 - 441 = 400, so BD = 20. 
The area of ▵ABC = 0.5 * base * height = 0.5 * 42 * 20 = 420. 
Let's examine the choices: A. 100, B. 420, C. 500, D. 609, E. 701. 
Our calculation matches choice B.

ANSWER 7: B

Problem 8:
The problem asks for the number of different isosceles triangles with integer side lengths and a perimeter of 23. 
For an isosceles triangle, two sides are equal. Let's denote the equal sides as 'a' and the base as 'b'. 
So, 2a + b = 23, and a + a > b (triangle inequality). 
This implies that 2a > b, or 2a > 23 - 2a, which simplifies to 4a > 23, or a > 5.75. 
Since 'a' is an integer, a ≥ 6. 
Also, a + b > a, which i

Problem 1:

We are asked about the **place** occupied by the digit 9 and the digit 3 in

\[
74982.1035
\]

The digit 9 is in the **hundreds** place, so the value of its place is

\[
100
\]

The digit 3 is in the **thousandths** place, so the value of its place is

\[
0.001=\frac{1}{1000}
\]

Now compare:

\[
100 \div 0.001 = 100 \div \frac{1}{1000}=100 \times 1000=100000
\]

So the place of the 9 is \(100{,}000\) times as great as the place of the 3.

Choices A and B are too small, and D and E are too large.

ANSWER 1: C

Problem 2:

We need the number halfway between

\[
\frac16 \quad \text{and} \quad \frac14
\]

The halfway point is the average:

\[
\frac{\frac16+\frac14}{2}
\]

Find a common denominator:

\[
\frac16=\frac{2}{12}, \qquad \frac14=\frac{3}{12}
\]

So

\[
\frac16+\frac14=\frac{2}{12}+\frac{3}{12}=\frac{5}{12}
\]

Now divide by 2:

\[
\frac{5}{12}\div 2=\frac{5}{12}\cdot \frac12=\frac{5}{24}
\]

So the number halfway between them is \(\frac{5}{24}\).

A tempting wrong answer is \(\frac15\), but that comes from averaging the denominators, which is not how fractions are averaged.

ANSWER 2: C

Problem 3:

The numbers are

\[
-2,\ 4,\ 6,\ 9,\ 12
\]

We are rearranging them using these rules:

- The largest number, \(12\), is not first, but is in one of the first three places.
- The smallest number, \(-2\), is not last, but is in one of the last three places.
- The median number, \(6\), is not first or last.

The largest number \(12\) must be in position 2 or 3.

The smallest number \(-2\) must be in position 3 or 4.

The median \(6\) must be in position 2, 3, or 4.

So the numbers \(12\), \(-2\), and \(6\) occupy the middle three positions in some order.

That means the first and last positions must be filled by the remaining two numbers:

\[
4 \quad \text{and} \quad 9
\]

Their average is

\[
\frac{4+9}{2}=\frac{13}{2}=6.5
\]

So the average of the first and last numbers is \(6.5\).

ANSWER 3: C

Problem 4:

We are given

\[
991+993+995+997+999=5000-N
\]

Add the five numbers. Since they are evenly spaced, their average is the middle number:

\[
995
\]

There are 5 numbers, so the sum is

\[
5 \cdot 995=4975
\]

Thus

\[
4975=5000-N
\]

Solve for \(N\):

\[
N=5000-4975=25
\]

So \(N=25\).

ANSWER 4: E

Problem 5:

We need to count 3-digit positive integers whose digits multiply to 24.

Since the product is 24, no digit can be 0.

Factor:

\[
24=2^3\cdot 3
\]

Now list digit triples whose product is 24:

\[
1,3,8
\]

\[
1,4,6
\]

\[
2,2,6
\]

\[
2,3,4
\]

Now count rearrangements.

For \(1,3,8\), all digits are different:

\[
3!=6
\]

For \(1,4,6\), all digits are different:

\[
3!=6
\]

For \(2,2,6\), two digits are the same, so there are

\[
\frac{3!}{2!}=3
\]

For \(2,3,4\), all digits are different:

\[
3!=6
\]

Total:

\[
6+6+3+6=21
\]

So there are 21 such integers.

ANSWER 5: D

Problem 6:

Brent starts with 4 goldfish and quadruples every month.

After \(m\) months, Brent has

\[
4\cdot 4^m
\]

Gretel starts with 128 goldfish and doubles every month.

After \(m\) months, Gretel has

\[
128\cdot 2^m
\]

We want them equal:

\[
4\cdot 4^m=128\cdot 2^m
\]

Rewrite everything as powers of 2:

\[
4=2^2,\quad 4^m=2^{2m},\quad 128=2^7
\]

So

\[
2^2\cdot 2^{2m}=2^7\cdot 2^m
\]

\[
2^{2+2m}=2^{7+m}
\]

Therefore,

\[
2+2m=7+m
\]

\[
m=5
\]

So in 5 months, they will have the same number of goldfish.

ANSWER 6: B

Problem 7:

We have an isosceles triangle with

\[
AB=BC=29
\]

and base

\[
AC=42
\]

The altitude from \(B\) to \(AC\) bisects the base, so each half is

\[
\frac{42}{2}=21
\]

Now use the Pythagorean Theorem to find the height \(h\):

\[
h^2+21^2=29^2
\]

\[
h^2+441=841
\]

\[
h^2=400
\]

\[
h=20
\]

Now find the area:

\[
\text{Area}=\frac12 \cdot \text{base} \cdot \text{height}
\]

\[
\text{Area}=\frac12 \cdot 42 \cdot 20=420
\]

So the area is 420.

ANSWER 7: B

Problem 8:

We need isosceles triangles with integer side lengths and perimeter 23.

Let the equal sides each have length \(a\), and let the base have length \(b\).

Then

\[
2a+b=23
\]

So

\[
b=23-2a
\]

For a triangle, the sum of the two equal sides must be greater than the base:

\[
2a>b
\]

Substitute \(b=23-2a\):

\[
2a>23-2a
\]

\[
4a>23
\]

So

\[
a>5.75
\]

Thus

\[
a\ge 6
\]

Also, \(b\) must be positive:

\[
23-2a>0
\]

\[
2a<23
\]

\[
a<11.5
\]

Thus

\[
a\le 11
\]

So possible values of \(a\) are

\[
6,7,8,9,10,11
\]

That gives 6 different triangles.

ANSWER 8: C

Problem 9:

We draw three digits from

\[
1,2,3,4
\]

and form a three-digit number.

A number is divisible by 3 if the sum of its digits is divisible by 3.

The sum of all four digits is

\[
1+2+3+4=10
\]

We choose three digits, which is the same as leaving out one digit.

If we leave out digit \(d\), then the chosen digit sum is

\[
10-d
\]

We need

\[
10-d
\]

to be divisible by 3.

Check the possible omitted digits:

- Omit 1: sum is \(9\), divisible by 3.
- Omit 2: sum is \(8\), not divisible by 3.
- Omit 3: sum is \(7\), not divisible by 3.
- Omit 4: sum is \(6\), divisible by 3.

So 2 of the 4 possible sets of digits work.

For each chosen set, all rearrangements are also multiples of 3, since digit sum does not change.

Thus the probability is

\[
\frac{2}{4}=\frac12
\]

ANSWER 9: C

Problem 10:

If the phone is on but not being used, the battery lasts 24 hours, so it uses

\[
\frac{1}{24}
\]

of the battery per hour.

If the phone is being used constantly, the battery lasts 3 hours, so it uses

\[
\frac{1}{3}
\]

of the battery per hour.

Since the last recharge, the phone has been on for 9 hours, and Niki used it for 60 minutes, which is 1 hour.

So the phone was:

- used for 1 hour
- on but not used for 8 hours

Battery used while talking:

\[
1\cdot \frac13=\frac13
\]

Battery used while on but idle:

\[
8\cdot \frac{1}{24}=\frac{8}{24}=\frac13
\]

Total battery used:

\[
\frac13+\frac13=\frac23
\]

Battery remaining:

\[
1-\frac23=\frac13
\]

Now she will leave the phone on but not talk, so it uses \(\frac{1}{24}\)

Problem 1:
**Restatement:** We are asked to find how many times greater the value of the place occupied by the digit 9 is compared to the value of the place occupied by the digit 3 in the number 74982.1035.

**Step-by-step Working:**
1. Identify the place value of the digit 9: In 74982.1035, the 9 is in the hundreds place. Its value is $9 \times 100$, so the place value is $100$.
2. Identify the place value of the digit 3: The 3 is in the thousandths place (three places to the right of the decimal point). Its value is $3 \times 0.001$, so the place value is $0.001$ (or $\frac{1}{1000}$).
3. Calculate the ratio: To find out how many times as great the hundreds place is compared to the thousandths place, we divide the two place values:
$$\frac{100}{0.001} = 100 \times 1000 = 100,000$$

**Ruling out wrong choices:** 
- A (1,000) and B (10,000) are tempting if you miscount the decimal places between the two digits.
- D (1,000,000) would be the answer if the 3 were in the ten-thousandths place.

ANSWER 1: C

Problem 2:
**Restatement:** We need to find the number that is exactly halfway between the fractions $\frac{1}{6}$ and $\frac{1}{4}$.

**Step-by-step Working:**
1. The number halfway between two numbers is their average. We need to calculate $\frac{\frac{1}{6} + \frac{1}{4}}{2}$.
2. Find a common denominator to add the fractions. The least common multiple of 6 and 4 is 12.
$$\frac{1}{6} = \frac{2}{12}$$
$$\frac{1}{4} = \frac{3}{12}$$
3. Add the fractions:
$$\frac{2}{12} + \frac{3}{12} = \frac{5}{12}$$
4. Divide the sum by 2 to find the average:
$$\frac{\frac{5}{12}}{2} = \frac{5}{12} \times \frac{1}{2} = \frac{5}{24}$$

**Ruling out wrong choices:**
- B ($\frac{1}{5}$) is a common mistake made by incorrectly "averaging" the denominators (halfway between 4 and 6).
- E ($\frac{5}{12}$) is the sum of the two fractions, forgetting to divide by 2.

ANSWER 2: C

Problem 3:
**Restatement:** We are given the numbers $-2, 4, 6, 9,$ and $12$. They are placed in five positions based on specific rules. We need to find the average of the numbers in the first and last positions.

**Step-by-step Working:**
1. Identify the key numbers:
   - Largest: $12$
   - Smallest: $-2$
   - Median (middle value when sorted): $6$
   - Remaining numbers: $4$ and $9$
2. Analyze the rules for the 5 positions (1, 2, 3, 4, 5):
   - The largest ($12$) isn't first, but is in the first three places. So, $12$ must be in position 2 or 3. This means $12$ is NOT in position 1 or 5.
   - The smallest ($-2$) isn't last, but is in the last three places. So, $-2$ must be in position 3 or 4. This means $-2$ is NOT in position 1 or 5.
   - The median ($6$) isn't first or last. So, $6$ is NOT in position 1 or 5.
3. Determine the first and last numbers:
   Since $12$, $-2$, and $6$ are all forbidden from being in position 1 and position 5, the only numbers left to fill the first and last spots are $4$ and $9$.
4. Calculate the average of the first and last numbers:
   $$\frac{4 + 9}{2} = \frac{13}{2} = 6.5$$

**Ruling out wrong choices:**
- If you mistakenly thought the median was part of the ends, you might get 5 (average of 4 and 6) or 7.5 (average of 6 and 9). 

ANSWER 3: C

Problem 4:
**Restatement:** We are given the equation $991 + 993 + 995 + 997 + 999 = 5000 - N$ and need to solve for $N$.

**Step-by-step Working:**
1. Notice that each number on the left side is slightly less than 1000. We can rewrite each term:
   $991 = 1000 - 9$
   $993 = 1000 - 7$
   $995 = 1000 - 5$
   $997 = 1000 - 3$
   $999 = 1000 - 1$
2. Substitute these into the sum:
   $(1000 - 9) + (1000 - 7) + (1000 - 5) + (1000 - 3) + (1000 - 1)$
3. Group the 1000s and the subtracted amounts:
   $= 5000 - (9 + 7 + 5 + 3 + 1)$
4. Add the subtracted amounts:
   $9 + 7 + 5 + 3 + 1 = 25$
5. The sum is $5000 - 25$. Therefore, $N = 25$.

**Ruling out wrong choices:**
- A (5) is tempting if you just count the number of terms being added.
- C (15) or D (20) might result from an arithmetic error when adding the odd numbers.

ANSWER 4: E

Problem 5:
**Restatement:** We need to find how many 3-digit positive integers exist such that the product of their three digits is exactly 24.

**Step-by-step Working:**
1. Find all combinations of 3 single-digit numbers (from 1 to 9) that multiply to 24.
   - The prime factorization of 24 is $2^3 \times 3$.
   - Possible sets of digits:
     - $\{1, 3, 8\}$
     - $\{1, 4, 6\}$
     - $\{2, 2, 6\}$
     - $\{2, 3, 4\}$
   (No other combinations work. For example, we cannot use 9, 7, or 5 because 24 is not divisible by them.)
2. Count the number of permutations (arrangements) for each set to form 3-digit numbers:
   - $\{1, 3, 8\}$: 3 distinct digits $\rightarrow 3! = 6$ numbers.
   - $\{1, 4, 6\}$: 3 distinct digits $\rightarrow 3! = 6$ numbers.
   - $\{2, 3, 4\}$: 3 distinct digits $\rightarrow 3! = 6$ numbers.
   - $\{2, 2, 6\}$: 2 identical digits $\rightarrow \frac{3!}{2!} = 3$ numbers.
3. Add them up:
   $6 + 6 + 6 + 3 = 21$

**Ruling out wrong choices:**
- E (24) is tempting if one assumes all 4 sets have 6 permutations ($4 \times 6 = 24$).
- C (18) is tempting if one misses the $\{2, 2, 6\}$ set entirely.

ANSWER 5: D

Problem 6:
**Restatement:** Brent starts with 4 goldfish that quadruple every month. Gretel starts with 128 goldfish that double every month. We need to find the number of months until they have the same amount.

**Step-by-step Working:**
1. Let $m$ be the number of months.
2. Write an expression for Brent's goldfish:
   $B(m) = 4 \times 4^m$
3. Write an expression for Gretel's goldfish:
   $G(m) = 128 \times 2^m$
4. Set them equal to each other:
   $4 \times 4^m = 128 \times 2^m$
5. Express all bases as powers of 2 to easily solve for $m$:
   $4 = 2^2$
   $128 = 2^7$
   $2^2 \times (2^2)^m = 2^7 \times 2^m$
   $2^2 \times 2^{2m} = 2^{m+7}$
   $2^{2m+2} = 2^{m+7}$
6. Equate the exponents:
   $2m + 2 = m + 7$
   $m = 5$

**Ruling out wrong choices:**
- A (4) or C (6) could be obtained if one makes an arithmetic error 

Problem 1:
The problem asks us to compare the place value of the digits 9 and 3 in the number 74982.1035. 
First, we identify the positions of 9 and 3. The digit 9 is in the thousands place, and the digit 3 is in the thousandths place. 
The value of the place occupied by 9 is 1000, and the value of the place occupied by 3 is 0.001. 
To find how many times as great the value of the place occupied by 9 is compared to the value of the place occupied by 3, we calculate 1000 / 0.001. 
This is equal to 1000 * 1000 = 1,000,000. 
Let's examine the choices: A. 1,000, B. 10,000, C. 100,000, D. 1,000,000, E. 10,000,000. 
Our calculation matches choice D.

ANSWER 1: D

Problem 2:
The problem asks for the number halfway between 1/6 and 1/4. 
To find the number halfway between two numbers, we need to calculate their average. 
First, we need to find a common denominator for 1/6 and 1/4. The least common denominator is 12. 
1/6 = 2/12 and 1/4 = 3/12. 
The average is (2/12 + 3/12) / 2 = (5/12) / 2 = 5/24. 
Let's examine the choices: A. 1/10, B. 1/5, C. 5/24, D. 7/24, E. 5/12. 
Our calculation matches choice C.

ANSWER 2: C

Problem 3:
The numbers given are −2, 4, 6, 9, and 12. We need to rearrange them according to the given rules and find the average of the first and last numbers. 
First, let's arrange the numbers in ascending order: −2, 4, 6, 9, 12. 
The largest number is 12, the smallest is −2, and the median is 6. 
According to the rules: the largest isn't first but is in one of the first three places, the smallest isn't last but is in one of the last three places, and the median isn't first or last. 
Let's try to satisfy these conditions. 
Since the median (6) can't be first or last, it must be in the second, third, or fourth position. 
If 6 is in the third position, the arrangement could be: x, y, 6, z, w. 
Given that 12 is in one of the first three places and −2 is in one of the last three places, a possible arrangement is: 9, 12, 6, 4, −2 or 4, 12, 6, 9, −2. 
However, we need to check if these satisfy all conditions. 
For 9, 12, 6, 4, −2: 12 is in the second position (satisfies the first condition), and −2 is in the last position (does not satisfy the second condition because −2 is last). 
For 4, 12, 6, 9, −2: again, −2 is last, so it doesn't satisfy the condition. 
Let's consider 6 in the second position: x, 6, y, z, w. 
A possible arrangement could be: 9, 6, 12, −2, 4. 
Here, 12 is in the third position, and −2 is in the fourth, satisfying the conditions that 12 is in one of the first three places and −2 is in one of the last three. 
The first number is 9, and the last is 4. The average is (9 + 4) / 2 = 6.5. 
Let's examine the choices: A. 3.5, B. 5, C. 6.5, D. 7.5, E. 8. 
Our calculation matches choice C.

ANSWER 3: C

Problem 4:
The problem gives the equation 991 + 993 + 995 + 997 + 999 = 5000 − N and asks to find N. 
First, let's simplify the left-hand side of the equation. 
991 + 993 + 995 + 997 + 999 can be rewritten as (1000 - 9) + (1000 - 7) + (1000 - 5) + (1000 - 3) + (1000 - 1). 
This equals 5*1000 - (9 + 7 + 5 + 3 + 1) = 5000 - 25. 
So, 5000 - 25 = 5000 - N. 
This implies that N = 25. 
Let's examine the choices: A. 5, B. 10, C. 15, D. 20, E. 25. 
Our calculation matches choice E.

ANSWER 4: E

Problem 5:
The problem asks us to find how many 3-digit positive integers have digits whose product equals 24. 
To solve this, we need to list the factor combinations of 24 and see which ones can form 3-digit numbers. 
The prime factorization of 24 is 2^3 * 3. 
Possible combinations for the digits are derived from the factors: (1, 1, 24) - not valid since 24 is not a single digit; (1, 2, 12) - not valid; (1, 3, 8); (1, 4, 6); (2, 2, 6); (2, 3, 4); and (2, 2, 6). 
Let's count the permutations for valid combinations: 
- For (1, 3, 8), the permutations are 138, 183, 318, 381, 813, 831, which gives 6 numbers. 
- For (1, 4, 6), the permutations are 146, 164, 416, 461, 614, 641, which gives 6 numbers. 
- For (2, 2, 6), the permutations are 226, 262, 622, which gives 3 numbers. 
- For (2, 3, 4), the permutations are 234, 243, 324, 342, 423, 432, which gives 6 numbers. 
Adding these up: 6 + 6 + 3 + 6 = 21. 
Let's examine the choices: A. 12, B. 15, C. 18, D. 21, E. 24. 
Our calculation matches choice D.

ANSWER 5: D

Problem 6:
Brent has goldfish that quadruple every month, and Gretel has goldfish that double every month. 
Initially, Brent has 4 goldfish, and Gretel has 128 goldfish. 
Let's denote the number of months from the initial time as 'n'. 
After 'n' months, Brent will have 4 * (4^n) goldfish, and Gretel will have 128 * (2^n) goldfish. 
We need to find 'n' when 4 * (4^n) = 128 * (2^n). 
Simplifying, we get 4^(n+1) = 128 * 2^n. 
Since 4 = 2^2, we can rewrite this as (2^2)^(n+1) = 2^7 * 2^n. 
This simplifies to 2^(2n+2) = 2^(n+7). 
Equating the exponents, we get 2n + 2 = n + 7. 
Solving for 'n', we get n = 5. 
Let's examine the choices: A. 4, B. 5, C. 6, D. 7, E. 8. 
Our calculation matches choice B.

ANSWER 6: B

Problem 7:
In ▵ABC, AB = BC = 29, and AC = 42. 
Since AB = BC, ▵ABC is isosceles. 
To find the area, we need the height. 
Let's draw the altitude from B to AC, meeting AC at point D. 
Since ▵ABC is isosceles, D is the midpoint of AC. 
So, AD = DC = 42 / 2 = 21. 
Using the Pythagorean theorem in ▵ADB, we have BD^2 = AB^2 - AD^2 = 29^2 - 21^2. 
Calculating, BD^2 = 841 - 441 = 400, so BD = 20. 
The area of ▵ABC = 0.5 * base * height = 0.5 * 42 * 20 = 420. 
Let's examine the choices: A. 100, B. 420, C. 500, D. 609, E. 701. 
Our calculation matches choice B.

ANSWER 7: B

Problem 8:
The problem asks for the number of different isosceles triangles with integer side lengths and a perimeter of 23. 
For an isosceles triangle, two sides are equal. Let's denote the equal sides as 'a' and the base as 'b'. 
So, 2a + b = 23, and a + a > b (triangle inequality). 
This implies that 2a > b, or 2a > 23 - 2a, which simplifies to 4a > 23, or a > 5.75. 
Since 'a' is an integer, a ≥ 6. 
Also, a + b > a, which i

Problem 1:

We are asked about the **place** occupied by the digit 9 and the digit 3 in

\[
74982.1035
\]

The digit 9 is in the **hundreds** place, so the value of its place is

\[
100
\]

The digit 3 is in the **thousandths** place, so the value of its place is

\[
0.001=\frac{1}{1000}
\]

Now compare:

\[
100 \div 0.001 = 100 \div \frac{1}{1000}=100 \times 1000=100000
\]

So the place of the 9 is \(100{,}000\) times as great as the place of the 3.

Choices A and B are too small, and D and E are too large.

ANSWER 1: C

Problem 2:

We need the number halfway between

\[
\frac16 \quad \text{and} \quad \frac14
\]

The halfway point is the average:

\[
\frac{\frac16+\frac14}{2}
\]

Find a common denominator:

\[
\frac16=\frac{2}{12}, \qquad \frac14=\frac{3}{12}
\]

So

\[
\frac16+\frac14=\frac{2}{12}+\frac{3}{12}=\frac{5}{12}
\]

Now divide by 2:

\[
\frac{5}{12}\div 2=\frac{5}{12}\cdot \frac12=\frac{5}{24}
\]

So the number halfway between them is \(\frac{5}{24}\).

A tempting wrong answer is \(\frac15\), but that comes from averaging the denominators, which is not how fractions are averaged.

ANSWER 2: C

Problem 3:

The numbers are

\[
-2,\ 4,\ 6,\ 9,\ 12
\]

We are rearranging them using these rules:

- The largest number, \(12\), is not first, but is in one of the first three places.
- The smallest number, \(-2\), is not last, but is in one of the last three places.
- The median number, \(6\), is not first or last.

The largest number \(12\) must be in position 2 or 3.

The smallest number \(-2\) must be in position 3 or 4.

The median \(6\) must be in position 2, 3, or 4.

So the numbers \(12\), \(-2\), and \(6\) occupy the middle three positions in some order.

That means the first and last positions must be filled by the remaining two numbers:

\[
4 \quad \text{and} \quad 9
\]

Their average is

\[
\frac{4+9}{2}=\frac{13}{2}=6.5
\]

So the average of the first and last numbers is \(6.5\).

ANSWER 3: C

Problem 4:

We are given

\[
991+993+995+997+999=5000-N
\]

Add the five numbers. Since they are evenly spaced, their average is the middle number:

\[
995
\]

There are 5 numbers, so the sum is

\[
5 \cdot 995=4975
\]

Thus

\[
4975=5000-N
\]

Solve for \(N\):

\[
N=5000-4975=25
\]

So \(N=25\).

ANSWER 4: E

Problem 5:

We need to count 3-digit positive integers whose digits multiply to 24.

Since the product is 24, no digit can be 0.

Factor:

\[
24=2^3\cdot 3
\]

Now list digit triples whose product is 24:

\[
1,3,8
\]

\[
1,4,6
\]

\[
2,2,6
\]

\[
2,3,4
\]

Now count rearrangements.

For \(1,3,8\), all digits are different:

\[
3!=6
\]

For \(1,4,6\), all digits are different:

\[
3!=6
\]

For \(2,2,6\), two digits are the same, so there are

\[
\frac{3!}{2!}=3
\]

For \(2,3,4\), all digits are different:

\[
3!=6
\]

Total:

\[
6+6+3+6=21
\]

So there are 21 such integers.

ANSWER 5: D

Problem 6:

Brent starts with 4 goldfish and quadruples every month.

After \(m\) months, Brent has

\[
4\cdot 4^m
\]

Gretel starts with 128 goldfish and doubles every month.

After \(m\) months, Gretel has

\[
128\cdot 2^m
\]

We want them equal:

\[
4\cdot 4^m=128\cdot 2^m
\]

Rewrite everything as powers of 2:

\[
4=2^2,\quad 4^m=2^{2m},\quad 128=2^7
\]

So

\[
2^2\cdot 2^{2m}=2^7\cdot 2^m
\]

\[
2^{2+2m}=2^{7+m}
\]

Therefore,

\[
2+2m=7+m
\]

\[
m=5
\]

So in 5 months, they will have the same number of goldfish.

ANSWER 6: B

Problem 7:

We have an isosceles triangle with

\[
AB=BC=29
\]

and base

\[
AC=42
\]

The altitude from \(B\) to \(AC\) bisects the base, so each half is

\[
\frac{42}{2}=21
\]

Now use the Pythagorean Theorem to find the height \(h\):

\[
h^2+21^2=29^2
\]

\[
h^2+441=841
\]

\[
h^2=400
\]

\[
h=20
\]

Now find the area:

\[
\text{Area}=\frac12 \cdot \text{base} \cdot \text{height}
\]

\[
\text{Area}=\frac12 \cdot 42 \cdot 20=420
\]

So the area is 420.

ANSWER 7: B

Problem 8:

We need isosceles triangles with integer side lengths and perimeter 23.

Let the equal sides each have length \(a\), and let the base have length \(b\).

Then

\[
2a+b=23
\]

So

\[
b=23-2a
\]

For a triangle, the sum of the two equal sides must be greater than the base:

\[
2a>b
\]

Substitute \(b=23-2a\):

\[
2a>23-2a
\]

\[
4a>23
\]

So

\[
a>5.75
\]

Thus

\[
a\ge 6
\]

Also, \(b\) must be positive:

\[
23-2a>0
\]

\[
2a<23
\]

\[
a<11.5
\]

Thus

\[
a\le 11
\]

So possible values of \(a\) are

\[
6,7,8,9,10,11
\]

That gives 6 different triangles.

ANSWER 8: C

Problem 9:

We draw three digits from

\[
1,2,3,4
\]

and form a three-digit number.

A number is divisible by 3 if the sum of its digits is divisible by 3.

The sum of all four digits is

\[
1+2+3+4=10
\]

We choose three digits, which is the same as leaving out one digit.

If we leave out digit \(d\), then the chosen digit sum is

\[
10-d
\]

We need

\[
10-d
\]

to be divisible by 3.

Check the possible omitted digits:

- Omit 1: sum is \(9\), divisible by 3.
- Omit 2: sum is \(8\), not divisible by 3.
- Omit 3: sum is \(7\), not divisible by 3.
- Omit 4: sum is \(6\), divisible by 3.

So 2 of the 4 possible sets of digits work.

For each chosen set, all rearrangements are also multiples of 3, since digit sum does not change.

Thus the probability is

\[
\frac{2}{4}=\frac12
\]

ANSWER 9: C

Problem 10:

If the phone is on but not being used, the battery lasts 24 hours, so it uses

\[
\frac{1}{24}
\]

of the battery per hour.

If the phone is being used constantly, the battery lasts 3 hours, so it uses

\[
\frac{1}{3}
\]

of the battery per hour.

Since the last recharge, the phone has been on for 9 hours, and Niki used it for 60 minutes, which is 1 hour.

So the phone was:

- used for 1 hour
- on but not used for 8 hours

Battery used while talking:

\[
1\cdot \frac13=\frac13
\]

Battery used while on but idle:

\[
8\cdot \frac{1}{24}=\frac{8}{24}=\frac13
\]

Total battery used:

\[
\frac13+\frac13=\frac23
\]

Battery remaining:

\[
1-\frac23=\frac13
\]

Now she will leave the phone on but not talk, so it uses \(\frac{1}{24}\)

Problem 1:
**Restatement:** We are asked to find how many times greater the value of the place occupied by the digit 9 is compared to the value of the place occupied by the digit 3 in the number 74982.1035.

**Step-by-step Working:**
1. Identify the place value of the digit 9: In 74982.1035, the 9 is in the hundreds place. Its value is $9 \times 100$, so the place value is $100$.
2. Identify the place value of the digit 3: The 3 is in the thousandths place (three places to the right of the decimal point). Its value is $3 \times 0.001$, so the place value is $0.001$ (or $\frac{1}{1000}$).
3. Calculate the ratio: To find out how many times as great the hundreds place is compared to the thousandths place, we divide the two place values:
$$\frac{100}{0.001} = 100 \times 1000 = 100,000$$

**Ruling out wrong choices:** 
- A (1,000) and B (10,000) are tempting if you miscount the decimal places between the two digits.
- D (1,000,000) would be the answer if the 3 were in the ten-thousandths place.

ANSWER 1: C

Problem 2:
**Restatement:** We need to find the number that is exactly halfway between the fractions $\frac{1}{6}$ and $\frac{1}{4}$.

**Step-by-step Working:**
1. The number halfway between two numbers is their average. We need to calculate $\frac{\frac{1}{6} + \frac{1}{4}}{2}$.
2. Find a common denominator to add the fractions. The least common multiple of 6 and 4 is 12.
$$\frac{1}{6} = \frac{2}{12}$$
$$\frac{1}{4} = \frac{3}{12}$$
3. Add the fractions:
$$\frac{2}{12} + \frac{3}{12} = \frac{5}{12}$$
4. Divide the sum by 2 to find the average:
$$\frac{\frac{5}{12}}{2} = \frac{5}{12} \times \frac{1}{2} = \frac{5}{24}$$

**Ruling out wrong choices:**
- B ($\frac{1}{5}$) is a common mistake made by incorrectly "averaging" the denominators (halfway between 4 and 6).
- E ($\frac{5}{12}$) is the sum of the two fractions, forgetting to divide by 2.

ANSWER 2: C

Problem 3:
**Restatement:** We are given the numbers $-2, 4, 6, 9,$ and $12$. They are placed in five positions based on specific rules. We need to find the average of the numbers in the first and last positions.

**Step-by-step Working:**
1. Identify the key numbers:
   - Largest: $12$
   - Smallest: $-2$
   - Median (middle value when sorted): $6$
   - Remaining numbers: $4$ and $9$
2. Analyze the rules for the 5 positions (1, 2, 3, 4, 5):
   - The largest ($12$) isn't first, but is in the first three places. So, $12$ must be in position 2 or 3. This means $12$ is NOT in position 1 or 5.
   - The smallest ($-2$) isn't last, but is in the last three places. So, $-2$ must be in position 3 or 4. This means $-2$ is NOT in position 1 or 5.
   - The median ($6$) isn't first or last. So, $6$ is NOT in position 1 or 5.
3. Determine the first and last numbers:
   Since $12$, $-2$, and $6$ are all forbidden from being in position 1 and position 5, the only numbers left to fill the first and last spots are $4$ and $9$.
4. Calculate the average of the first and last numbers:
   $$\frac{4 + 9}{2} = \frac{13}{2} = 6.5$$

**Ruling out wrong choices:**
- If you mistakenly thought the median was part of the ends, you might get 5 (average of 4 and 6) or 7.5 (average of 6 and 9). 

ANSWER 3: C

Problem 4:
**Restatement:** We are given the equation $991 + 993 + 995 + 997 + 999 = 5000 - N$ and need to solve for $N$.

**Step-by-step Working:**
1. Notice that each number on the left side is slightly less than 1000. We can rewrite each term:
   $991 = 1000 - 9$
   $993 = 1000 - 7$
   $995 = 1000 - 5$
   $997 = 1000 - 3$
   $999 = 1000 - 1$
2. Substitute these into the sum:
   $(1000 - 9) + (1000 - 7) + (1000 - 5) + (1000 - 3) + (1000 - 1)$
3. Group the 1000s and the subtracted amounts:
   $= 5000 - (9 + 7 + 5 + 3 + 1)$
4. Add the subtracted amounts:
   $9 + 7 + 5 + 3 + 1 = 25$
5. The sum is $5000 - 25$. Therefore, $N = 25$.

**Ruling out wrong choices:**
- A (5) is tempting if you just count the number of terms being added.
- C (15) or D (20) might result from an arithmetic error when adding the odd numbers.

ANSWER 4: E

Problem 5:
**Restatement:** We need to find how many 3-digit positive integers exist such that the product of their three digits is exactly 24.

**Step-by-step Working:**
1. Find all combinations of 3 single-digit numbers (from 1 to 9) that multiply to 24.
   - The prime factorization of 24 is $2^3 \times 3$.
   - Possible sets of digits:
     - $\{1, 3, 8\}$
     - $\{1, 4, 6\}$
     - $\{2, 2, 6\}$
     - $\{2, 3, 4\}$
   (No other combinations work. For example, we cannot use 9, 7, or 5 because 24 is not divisible by them.)
2. Count the number of permutations (arrangements) for each set to form 3-digit numbers:
   - $\{1, 3, 8\}$: 3 distinct digits $\rightarrow 3! = 6$ numbers.
   - $\{1, 4, 6\}$: 3 distinct digits $\rightarrow 3! = 6$ numbers.
   - $\{2, 3, 4\}$: 3 distinct digits $\rightarrow 3! = 6$ numbers.
   - $\{2, 2, 6\}$: 2 identical digits $\rightarrow \frac{3!}{2!} = 3$ numbers.
3. Add them up:
   $6 + 6 + 6 + 3 = 21$

**Ruling out wrong choices:**
- E (24) is tempting if one assumes all 4 sets have 6 permutations ($4 \times 6 = 24$).
- C (18) is tempting if one misses the $\{2, 2, 6\}$ set entirely.

ANSWER 5: D

Problem 6:
**Restatement:** Brent starts with 4 goldfish that quadruple every month. Gretel starts with 128 goldfish that double every month. We need to find the number of months until they have the same amount.

**Step-by-step Working:**
1. Let $m$ be the number of months.
2. Write an expression for Brent's goldfish:
   $B(m) = 4 \times 4^m$
3. Write an expression for Gretel's goldfish:
   $G(m) = 128 \times 2^m$
4. Set them equal to each other:
   $4 \times 4^m = 128 \times 2^m$
5. Express all bases as powers of 2 to easily solve for $m$:
   $4 = 2^2$
   $128 = 2^7$
   $2^2 \times (2^2)^m = 2^7 \times 2^m$
   $2^2 \times 2^{2m} = 2^{m+7}$
   $2^{2m+2} = 2^{m+7}$
6. Equate the exponents:
   $2m + 2 = m + 7$
   $m = 5$

**Ruling out wrong choices:**
- A (4) or C (6) could be obtained if one makes an arithmetic error 

Problem 1:
The problem asks us to compare the place value of the digits 9 and 3 in the number 74982.1035. 
First, we identify the positions of 9 and 3. The digit 9 is in the thousands place, and the digit 3 is in the thousandths place. 
The value of the place occupied by 9 is 1000, and the value of the place occupied by 3 is 0.001. 
To find how many times as great the value of the place occupied by 9 is compared to the value of the place occupied by 3, we calculate 1000 / 0.001. 
This is equal to 1000 * 1000 = 1,000,000. 
Let's examine the choices: A. 1,000, B. 10,000, C. 100,000, D. 1,000,000, E. 10,000,000. 
Our calculation matches choice D.

ANSWER 1: D

Problem 2:
The problem asks for the number halfway between 1/6 and 1/4. 
To find the number halfway between two numbers, we need to calculate their average. 
First, we need to find a common denominator for 1/6 and 1/4. The least common denominator is 12. 
1/6 = 2/12 and 1/4 = 3/12. 
The average is (2/12 + 3/12) / 2 = (5/12) / 2 = 5/24. 
Let's examine the choices: A. 1/10, B. 1/5, C. 5/24, D. 7/24, E. 5/12. 
Our calculation matches choice C.

ANSWER 2: C

Problem 3:
The numbers given are −2, 4, 6, 9, and 12. We need to rearrange them according to the given rules and find the average of the first and last numbers. 
First, let's arrange the numbers in ascending order: −2, 4, 6, 9, 12. 
The largest number is 12, the smallest is −2, and the median is 6. 
According to the rules: the largest isn't first but is in one of the first three places, the smallest isn't last but is in one of the last three places, and the median isn't first or last. 
Let's try to satisfy these conditions. 
Since the median (6) can't be first or last, it must be in the second, third, or fourth position. 
If 6 is in the third position, the arrangement could be: x, y, 6, z, w. 
Given that 12 is in one of the first three places and −2 is in one of the last three places, a possible arrangement is: 9, 12, 6, 4, −2 or 4, 12, 6, 9, −2. 
However, we need to check if these satisfy all conditions. 
For 9, 12, 6, 4, −2: 12 is in the second position (satisfies the first condition), and −2 is in the last position (does not satisfy the second condition because −2 is last). 
For 4, 12, 6, 9, −2: again, −2 is last, so it doesn't satisfy the condition. 
Let's consider 6 in the second position: x, 6, y, z, w. 
A possible arrangement could be: 9, 6, 12, −2, 4. 
Here, 12 is in the third position, and −2 is in the fourth, satisfying the conditions that 12 is in one of the first three places and −2 is in one of the last three. 
The first number is 9, and the last is 4. The average is (9 + 4) / 2 = 6.5. 
Let's examine the choices: A. 3.5, B. 5, C. 6.5, D. 7.5, E. 8. 
Our calculation matches choice C.

ANSWER 3: C

Problem 4:
The problem gives the equation 991 + 993 + 995 + 997 + 999 = 5000 − N and asks to find N. 
First, let's simplify the left-hand side of the equation. 
991 + 993 + 995 + 997 + 999 can be rewritten as (1000 - 9) + (1000 - 7) + (1000 - 5) + (1000 - 3) + (1000 - 1). 
This equals 5*1000 - (9 + 7 + 5 + 3 + 1) = 5000 - 25. 
So, 5000 - 25 = 5000 - N. 
This implies that N = 25. 
Let's examine the choices: A. 5, B. 10, C. 15, D. 20, E. 25. 
Our calculation matches choice E.

ANSWER 4: E

Problem 5:
The problem asks us to find how many 3-digit positive integers have digits whose product equals 24. 
To solve this, we need to list the factor combinations of 24 and see which ones can form 3-digit numbers. 
The prime factorization of 24 is 2^3 * 3. 
Possible combinations for the digits are derived from the factors: (1, 1, 24) - not valid since 24 is not a single digit; (1, 2, 12) - not valid; (1, 3, 8); (1, 4, 6); (2, 2, 6); (2, 3, 4); and (2, 2, 6). 
Let's count the permutations for valid combinations: 
- For (1, 3, 8), the permutations are 138, 183, 318, 381, 813, 831, which gives 6 numbers. 
- For (1, 4, 6), the permutations are 146, 164, 416, 461, 614, 641, which gives 6 numbers. 
- For (2, 2, 6), the permutations are 226, 262, 622, which gives 3 numbers. 
- For (2, 3, 4), the permutations are 234, 243, 324, 342, 423, 432, which gives 6 numbers. 
Adding these up: 6 + 6 + 3 + 6 = 21. 
Let's examine the choices: A. 12, B. 15, C. 18, D. 21, E. 24. 
Our calculation matches choice D.

ANSWER 5: D

Problem 6:
Brent has goldfish that quadruple every month, and Gretel has goldfish that double every month. 
Initially, Brent has 4 goldfish, and Gretel has 128 goldfish. 
Let's denote the number of months from the initial time as 'n'. 
After 'n' months, Brent will have 4 * (4^n) goldfish, and Gretel will have 128 * (2^n) goldfish. 
We need to find 'n' when 4 * (4^n) = 128 * (2^n). 
Simplifying, we get 4^(n+1) = 128 * 2^n. 
Since 4 = 2^2, we can rewrite this as (2^2)^(n+1) = 2^7 * 2^n. 
This simplifies to 2^(2n+2) = 2^(n+7). 
Equating the exponents, we get 2n + 2 = n + 7. 
Solving for 'n', we get n = 5. 
Let's examine the choices: A. 4, B. 5, C. 6, D. 7, E. 8. 
Our calculation matches choice B.

ANSWER 6: B

Problem 7:
In ▵ABC, AB = BC = 29, and AC = 42. 
Since AB = BC, ▵ABC is isosceles. 
To find the area, we need the height. 
Let's draw the altitude from B to AC, meeting AC at point D. 
Since ▵ABC is isosceles, D is the midpoint of AC. 
So, AD = DC = 42 / 2 = 21. 
Using the Pythagorean theorem in ▵ADB, we have BD^2 = AB^2 - AD^2 = 29^2 - 21^2. 
Calculating, BD^2 = 841 - 441 = 400, so BD = 20. 
The area of ▵ABC = 0.5 * base * height = 0.5 * 42 * 20 = 420. 
Let's examine the choices: A. 100, B. 420, C. 500, D. 609, E. 701. 
Our calculation matches choice B.

ANSWER 7: B

Problem 8:
The problem asks for the number of different isosceles triangles with integer side lengths and a perimeter of 23. 
For an isosceles triangle, two sides are equal. Let's denote the equal sides as 'a' and the base as 'b'. 
So, 2a + b = 23, and a + a > b (triangle inequality). 
This implies that 2a > b, or 2a > 23 - 2a, which simplifies to 4a > 23, or a > 5.75. 
Since 'a' is an integer, a ≥ 6. 
Also, a + b > a, which i

Problem 1:

We are asked about the **place** occupied by the digit 9 and the digit 3 in

\[
74982.1035
\]

The digit 9 is in the **hundreds** place, so the value of its place is

\[
100
\]

The digit 3 is in the **thousandths** place, so the value of its place is

\[
0.001=\frac{1}{1000}
\]

Now compare:

\[
100 \div 0.001 = 100 \div \frac{1}{1000}=100 \times 1000=100000
\]

So the place of the 9 is \(100{,}000\) times as great as the place of the 3.

Choices A and B are too small, and D and E are too large.

ANSWER 1: C

Problem 2:

We need the number halfway between

\[
\frac16 \quad \text{and} \quad \frac14
\]

The halfway point is the average:

\[
\frac{\frac16+\frac14}{2}
\]

Find a common denominator:

\[
\frac16=\frac{2}{12}, \qquad \frac14=\frac{3}{12}
\]

So

\[
\frac16+\frac14=\frac{2}{12}+\frac{3}{12}=\frac{5}{12}
\]

Now divide by 2:

\[
\frac{5}{12}\div 2=\frac{5}{12}\cdot \frac12=\frac{5}{24}
\]

So the number halfway between them is \(\frac{5}{24}\).

A tempting wrong answer is \(\frac15\), but that comes from averaging the denominators, which is not how fractions are averaged.

ANSWER 2: C

Problem 3:

The numbers are

\[
-2,\ 4,\ 6,\ 9,\ 12
\]

We are rearranging them using these rules:

- The largest number, \(12\), is not first, but is in one of the first three places.
- The smallest number, \(-2\), is not last, but is in one of the last three places.
- The median number, \(6\), is not first or last.

The largest number \(12\) must be in position 2 or 3.

The smallest number \(-2\) must be in position 3 or 4.

The median \(6\) must be in position 2, 3, or 4.

So the numbers \(12\), \(-2\), and \(6\) occupy the middle three positions in some order.

That means the first and last positions must be filled by the remaining two numbers:

\[
4 \quad \text{and} \quad 9
\]

Their average is

\[
\frac{4+9}{2}=\frac{13}{2}=6.5
\]

So the average of the first and last numbers is \(6.5\).

ANSWER 3: C

Problem 4:

We are given

\[
991+993+995+997+999=5000-N
\]

Add the five numbers. Since they are evenly spaced, their average is the middle number:

\[
995
\]

There are 5 numbers, so the sum is

\[
5 \cdot 995=4975
\]

Thus

\[
4975=5000-N
\]

Solve for \(N\):

\[
N=5000-4975=25
\]

So \(N=25\).

ANSWER 4: E

Problem 5:

We need to count 3-digit positive integers whose digits multiply to 24.

Since the product is 24, no digit can be 0.

Factor:

\[
24=2^3\cdot 3
\]

Now list digit triples whose product is 24:

\[
1,3,8
\]

\[
1,4,6
\]

\[
2,2,6
\]

\[
2,3,4
\]

Now count rearrangements.

For \(1,3,8\), all digits are different:

\[
3!=6
\]

For \(1,4,6\), all digits are different:

\[
3!=6
\]

For \(2,2,6\), two digits are the same, so there are

\[
\frac{3!}{2!}=3
\]

For \(2,3,4\), all digits are different:

\[
3!=6
\]

Total:

\[
6+6+3+6=21
\]

So there are 21 such integers.

ANSWER 5: D

Problem 6:

Brent starts with 4 goldfish and quadruples every month.

After \(m\) months, Brent has

\[
4\cdot 4^m
\]

Gretel starts with 128 goldfish and doubles every month.

After \(m\) months, Gretel has

\[
128\cdot 2^m
\]

We want them equal:

\[
4\cdot 4^m=128\cdot 2^m
\]

Rewrite everything as powers of 2:

\[
4=2^2,\quad 4^m=2^{2m},\quad 128=2^7
\]

So

\[
2^2\cdot 2^{2m}=2^7\cdot 2^m
\]

\[
2^{2+2m}=2^{7+m}
\]

Therefore,

\[
2+2m=7+m
\]

\[
m=5
\]

So in 5 months, they will have the same number of goldfish.

ANSWER 6: B

Problem 7:

We have an isosceles triangle with

\[
AB=BC=29
\]

and base

\[
AC=42
\]

The altitude from \(B\) to \(AC\) bisects the base, so each half is

\[
\frac{42}{2}=21
\]

Now use the Pythagorean Theorem to find the height \(h\):

\[
h^2+21^2=29^2
\]

\[
h^2+441=841
\]

\[
h^2=400
\]

\[
h=20
\]

Now find the area:

\[
\text{Area}=\frac12 \cdot \text{base} \cdot \text{height}
\]

\[
\text{Area}=\frac12 \cdot 42 \cdot 20=420
\]

So the area is 420.

ANSWER 7: B

Problem 8:

We need isosceles triangles with integer side lengths and perimeter 23.

Let the equal sides each have length \(a\), and let the base have length \(b\).

Then

\[
2a+b=23
\]

So

\[
b=23-2a
\]

For a triangle, the sum of the two equal sides must be greater than the base:

\[
2a>b
\]

Substitute \(b=23-2a\):

\[
2a>23-2a
\]

\[
4a>23
\]

So

\[
a>5.75
\]

Thus

\[
a\ge 6
\]

Also, \(b\) must be positive:

\[
23-2a>0
\]

\[
2a<23
\]

\[
a<11.5
\]

Thus

\[
a\le 11
\]

So possible values of \(a\) are

\[
6,7,8,9,10,11
\]

That gives 6 different triangles.

ANSWER 8: C

Problem 9:

We draw three digits from

\[
1,2,3,4
\]

and form a three-digit number.

A number is divisible by 3 if the sum of its digits is divisible by 3.

The sum of all four digits is

\[
1+2+3+4=10
\]

We choose three digits, which is the same as leaving out one digit.

If we leave out digit \(d\), then the chosen digit sum is

\[
10-d
\]

We need

\[
10-d
\]

to be divisible by 3.

Check the possible omitted digits:

- Omit 1: sum is \(9\), divisible by 3.
- Omit 2: sum is \(8\), not divisible by 3.
- Omit 3: sum is \(7\), not divisible by 3.
- Omit 4: sum is \(6\), divisible by 3.

So 2 of the 4 possible sets of digits work.

For each chosen set, all rearrangements are also multiples of 3, since digit sum does not change.

Thus the probability is

\[
\frac{2}{4}=\frac12
\]

ANSWER 9: C

Problem 10:

If the phone is on but not being used, the battery lasts 24 hours, so it uses

\[
\frac{1}{24}
\]

of the battery per hour.

If the phone is being used constantly, the battery lasts 3 hours, so it uses

\[
\frac{1}{3}
\]

of the battery per hour.

Since the last recharge, the phone has been on for 9 hours, and Niki used it for 60 minutes, which is 1 hour.

So the phone was:

- used for 1 hour
- on but not used for 8 hours

Battery used while talking:

\[
1\cdot \frac13=\frac13
\]

Battery used while on but idle:

\[
8\cdot \frac{1}{24}=\frac{8}{24}=\frac13
\]

Total battery used:

\[
\frac13+\frac13=\frac23
\]

Battery remaining:

\[
1-\frac23=\frac13
\]

Now she will leave the phone on but not talk, so it uses \(\frac{1}{24}\)

Problem 1:
**Restatement:** We are asked to find how many times greater the value of the place occupied by the digit 9 is compared to the value of the place occupied by the digit 3 in the number 74982.1035.

**Step-by-step Working:**
1. Identify the place value of the digit 9: In 74982.1035, the 9 is in the hundreds place. Its value is $9 \times 100$, so the place value is $100$.
2. Identify the place value of the digit 3: The 3 is in the thousandths place (three places to the right of the decimal point). Its value is $3 \times 0.001$, so the place value is $0.001$ (or $\frac{1}{1000}$).
3. Calculate the ratio: To find out how many times as great the hundreds place is compared to the thousandths place, we divide the two place values:
$$\frac{100}{0.001} = 100 \times 1000 = 100,000$$

**Ruling out wrong choices:** 
- A (1,000) and B (10,000) are tempting if you miscount the decimal places between the two digits.
- D (1,000,000) would be the answer if the 3 were in the ten-thousandths place.

ANSWER 1: C

Problem 2:
**Restatement:** We need to find the number that is exactly halfway between the fractions $\frac{1}{6}$ and $\frac{1}{4}$.

**Step-by-step Working:**
1. The number halfway between two numbers is their average. We need to calculate $\frac{\frac{1}{6} + \frac{1}{4}}{2}$.
2. Find a common denominator to add the fractions. The least common multiple of 6 and 4 is 12.
$$\frac{1}{6} = \frac{2}{12}$$
$$\frac{1}{4} = \frac{3}{12}$$
3. Add the fractions:
$$\frac{2}{12} + \frac{3}{12} = \frac{5}{12}$$
4. Divide the sum by 2 to find the average:
$$\frac{\frac{5}{12}}{2} = \frac{5}{12} \times \frac{1}{2} = \frac{5}{24}$$

**Ruling out wrong choices:**
- B ($\frac{1}{5}$) is a common mistake made by incorrectly "averaging" the denominators (halfway between 4 and 6).
- E ($\frac{5}{12}$) is the sum of the two fractions, forgetting to divide by 2.

ANSWER 2: C

Problem 3:
**Restatement:** We are given the numbers $-2, 4, 6, 9,$ and $12$. They are placed in five positions based on specific rules. We need to find the average of the numbers in the first and last positions.

**Step-by-step Working:**
1. Identify the key numbers:
   - Largest: $12$
   - Smallest: $-2$
   - Median (middle value when sorted): $6$
   - Remaining numbers: $4$ and $9$
2. Analyze the rules for the 5 positions (1, 2, 3, 4, 5):
   - The largest ($12$) isn't first, but is in the first three places. So, $12$ must be in position 2 or 3. This means $12$ is NOT in position 1 or 5.
   - The smallest ($-2$) isn't last, but is in the last three places. So, $-2$ must be in position 3 or 4. This means $-2$ is NOT in position 1 or 5.
   - The median ($6$) isn't first or last. So, $6$ is NOT in position 1 or 5.
3. Determine the first and last numbers:
   Since $12$, $-2$, and $6$ are all forbidden from being in position 1 and position 5, the only numbers left to fill the first and last spots are $4$ and $9$.
4. Calculate the average of the first and last numbers:
   $$\frac{4 + 9}{2} = \frac{13}{2} = 6.5$$

**Ruling out wrong choices:**
- If you mistakenly thought the median was part of the ends, you might get 5 (average of 4 and 6) or 7.5 (average of 6 and 9). 

ANSWER 3: C

Problem 4:
**Restatement:** We are given the equation $991 + 993 + 995 + 997 + 999 = 5000 - N$ and need to solve for $N$.

**Step-by-step Working:**
1. Notice that each number on the left side is slightly less than 1000. We can rewrite each term:
   $991 = 1000 - 9$
   $993 = 1000 - 7$
   $995 = 1000 - 5$
   $997 = 1000 - 3$
   $999 = 1000 - 1$
2. Substitute these into the sum:
   $(1000 - 9) + (1000 - 7) + (1000 - 5) + (1000 - 3) + (1000 - 1)$
3. Group the 1000s and the subtracted amounts:
   $= 5000 - (9 + 7 + 5 + 3 + 1)$
4. Add the subtracted amounts:
   $9 + 7 + 5 + 3 + 1 = 25$
5. The sum is $5000 - 25$. Therefore, $N = 25$.

**Ruling out wrong choices:**
- A (5) is tempting if you just count the number of terms being added.
- C (15) or D (20) might result from an arithmetic error when adding the odd numbers.

ANSWER 4: E

Problem 5:
**Restatement:** We need to find how many 3-digit positive integers exist such that the product of their three digits is exactly 24.

**Step-by-step Working:**
1. Find all combinations of 3 single-digit numbers (from 1 to 9) that multiply to 24.
   - The prime factorization of 24 is $2^3 \times 3$.
   - Possible sets of digits:
     - $\{1, 3, 8\}$
     - $\{1, 4, 6\}$
     - $\{2, 2, 6\}$
     - $\{2, 3, 4\}$
   (No other combinations work. For example, we cannot use 9, 7, or 5 because 24 is not divisible by them.)
2. Count the number of permutations (arrangements) for each set to form 3-digit numbers:
   - $\{1, 3, 8\}$: 3 distinct digits $\rightarrow 3! = 6$ numbers.
   - $\{1, 4, 6\}$: 3 distinct digits $\rightarrow 3! = 6$ numbers.
   - $\{2, 3, 4\}$: 3 distinct digits $\rightarrow 3! = 6$ numbers.
   - $\{2, 2, 6\}$: 2 identical digits $\rightarrow \frac{3!}{2!} = 3$ numbers.
3. Add them up:
   $6 + 6 + 6 + 3 = 21$

**Ruling out wrong choices:**
- E (24) is tempting if one assumes all 4 sets have 6 permutations ($4 \times 6 = 24$).
- C (18) is tempting if one misses the $\{2, 2, 6\}$ set entirely.

ANSWER 5: D

Problem 6:
**Restatement:** Brent starts with 4 goldfish that quadruple every month. Gretel starts with 128 goldfish that double every month. We need to find the number of months until they have the same amount.

**Step-by-step Working:**
1. Let $m$ be the number of months.
2. Write an expression for Brent's goldfish:
   $B(m) = 4 \times 4^m$
3. Write an expression for Gretel's goldfish:
   $G(m) = 128 \times 2^m$
4. Set them equal to each other:
   $4 \times 4^m = 128 \times 2^m$
5. Express all bases as powers of 2 to easily solve for $m$:
   $4 = 2^2$
   $128 = 2^7$
   $2^2 \times (2^2)^m = 2^7 \times 2^m$
   $2^2 \times 2^{2m} = 2^{m+7}$
   $2^{2m+2} = 2^{m+7}$
6. Equate the exponents:
   $2m + 2 = m + 7$
   $m = 5$

**Ruling out wrong choices:**
- A (4) or C (6) could be obtained if one makes an arithmetic error 

Problem 1:
The problem asks us to compare the place value of the digits 9 and 3 in the number 74982.1035. 
First, we identify the positions of 9 and 3. The digit 9 is in the thousands place, and the digit 3 is in the thousandths place. 
The value of the place occupied by 9 is 1000, and the value of the place occupied by 3 is 0.001. 
To find how many times as great the value of the place occupied by 9 is compared to the value of the place occupied by 3, we calculate 1000 / 0.001. 
This is equal to 1000 * 1000 = 1,000,000. 
Let's examine the choices: A. 1,000, B. 10,000, C. 100,000, D. 1,000,000, E. 10,000,000. 
Our calculation matches choice D.

ANSWER 1: D

Problem 2:
The problem asks for the number halfway between 1/6 and 1/4. 
To find the number halfway between two numbers, we need to calculate their average. 
First, we need to find a common denominator for 1/6 and 1/4. The least common denominator is 12. 
1/6 = 2/12 and 1/4 = 3/12. 
The average is (2/12 + 3/12) / 2 = (5/12) / 2 = 5/24. 
Let's examine the choices: A. 1/10, B. 1/5, C. 5/24, D. 7/24, E. 5/12. 
Our calculation matches choice C.

ANSWER 2: C

Problem 3:
The numbers given are −2, 4, 6, 9, and 12. We need to rearrange them according to the given rules and find the average of the first and last numbers. 
First, let's arrange the numbers in ascending order: −2, 4, 6, 9, 12. 
The largest number is 12, the smallest is −2, and the median is 6. 
According to the rules: the largest isn't first but is in one of the first three places, the smallest isn't last but is in one of the last three places, and the median isn't first or last. 
Let's try to satisfy these conditions. 
Since the median (6) can't be first or last, it must be in the second, third, or fourth position. 
If 6 is in the third position, the arrangement could be: x, y, 6, z, w. 
Given that 12 is in one of the first three places and −2 is in one of the last three places, a possible arrangement is: 9, 12, 6, 4, −2 or 4, 12, 6, 9, −2. 
However, we need to check if these satisfy all conditions. 
For 9, 12, 6, 4, −2: 12 is in the second position (satisfies the first condition), and −2 is in the last position (does not satisfy the second condition because −2 is last). 
For 4, 12, 6, 9, −2: again, −2 is last, so it doesn't satisfy the condition. 
Let's consider 6 in the second position: x, 6, y, z, w. 
A possible arrangement could be: 9, 6, 12, −2, 4. 
Here, 12 is in the third position, and −2 is in the fourth, satisfying the conditions that 12 is in one of the first three places and −2 is in one of the last three. 
The first number is 9, and the last is 4. The average is (9 + 4) / 2 = 6.5. 
Let's examine the choices: A. 3.5, B. 5, C. 6.5, D. 7.5, E. 8. 
Our calculation matches choice C.

ANSWER 3: C

Problem 4:
The problem gives the equation 991 + 993 + 995 + 997 + 999 = 5000 − N and asks to find N. 
First, let's simplify the left-hand side of the equation. 
991 + 993 + 995 + 997 + 999 can be rewritten as (1000 - 9) + (1000 - 7) + (1000 - 5) + (1000 - 3) + (1000 - 1). 
This equals 5*1000 - (9 + 7 + 5 + 3 + 1) = 5000 - 25. 
So, 5000 - 25 = 5000 - N. 
This implies that N = 25. 
Let's examine the choices: A. 5, B. 10, C. 15, D. 20, E. 25. 
Our calculation matches choice E.

ANSWER 4: E

Problem 5:
The problem asks us to find how many 3-digit positive integers have digits whose product equals 24. 
To solve this, we need to list the factor combinations of 24 and see which ones can form 3-digit numbers. 
The prime factorization of 24 is 2^3 * 3. 
Possible combinations for the digits are derived from the factors: (1, 1, 24) - not valid since 24 is not a single digit; (1, 2, 12) - not valid; (1, 3, 8); (1, 4, 6); (2, 2, 6); (2, 3, 4); and (2, 2, 6). 
Let's count the permutations for valid combinations: 
- For (1, 3, 8), the permutations are 138, 183, 318, 381, 813, 831, which gives 6 numbers. 
- For (1, 4, 6), the permutations are 146, 164, 416, 461, 614, 641, which gives 6 numbers. 
- For (2, 2, 6), the permutations are 226, 262, 622, which gives 3 numbers. 
- For (2, 3, 4), the permutations are 234, 243, 324, 342, 423, 432, which gives 6 numbers. 
Adding these up: 6 + 6 + 3 + 6 = 21. 
Let's examine the choices: A. 12, B. 15, C. 18, D. 21, E. 24. 
Our calculation matches choice D.

ANSWER 5: D

Problem 6:
Brent has goldfish that quadruple every month, and Gretel has goldfish that double every month. 
Initially, Brent has 4 goldfish, and Gretel has 128 goldfish. 
Let's denote the number of months from the initial time as 'n'. 
After 'n' months, Brent will have 4 * (4^n) goldfish, and Gretel will have 128 * (2^n) goldfish. 
We need to find 'n' when 4 * (4^n) = 128 * (2^n). 
Simplifying, we get 4^(n+1) = 128 * 2^n. 
Since 4 = 2^2, we can rewrite this as (2^2)^(n+1) = 2^7 * 2^n. 
This simplifies to 2^(2n+2) = 2^(n+7). 
Equating the exponents, we get 2n + 2 = n + 7. 
Solving for 'n', we get n = 5. 
Let's examine the choices: A. 4, B. 5, C. 6, D. 7, E. 8. 
Our calculation matches choice B.

ANSWER 6: B

Problem 7:
In ▵ABC, AB = BC = 29, and AC = 42. 
Since AB = BC, ▵ABC is isosceles. 
To find the area, we need the height. 
Let's draw the altitude from B to AC, meeting AC at point D. 
Since ▵ABC is isosceles, D is the midpoint of AC. 
So, AD = DC = 42 / 2 = 21. 
Using the Pythagorean theorem in ▵ADB, we have BD^2 = AB^2 - AD^2 = 29^2 - 21^2. 
Calculating, BD^2 = 841 - 441 = 400, so BD = 20. 
The area of ▵ABC = 0.5 * base * height = 0.5 * 42 * 20 = 420. 
Let's examine the choices: A. 100, B. 420, C. 500, D. 609, E. 701. 
Our calculation matches choice B.

ANSWER 7: B

Problem 8:
The problem asks for the number of different isosceles triangles with integer side lengths and a perimeter of 23. 
For an isosceles triangle, two sides are equal. Let's denote the equal sides as 'a' and the base as 'b'. 
So, 2a + b = 23, and a + a > b (triangle inequality). 
This implies that 2a > b, or 2a > 23 - 2a, which simplifies to 4a > 23, or a > 5.75. 
Since 'a' is an integer, a ≥ 6. 
Also, a + b > a, which i

Problem 1:

We are asked about the **place** occupied by the digit 9 and the digit 3 in

\[
74982.1035
\]

The digit 9 is in the **hundreds** place, so the value of its place is

\[
100
\]

The digit 3 is in the **thousandths** place, so the value of its place is

\[
0.001=\frac{1}{1000}
\]

Now compare:

\[
100 \div 0.001 = 100 \div \frac{1}{1000}=100 \times 1000=100000
\]

So the place of the 9 is \(100{,}000\) times as great as the place of the 3.

Choices A and B are too small, and D and E are too large.

ANSWER 1: C

Problem 2:

We need the number halfway between

\[
\frac16 \quad \text{and} \quad \frac14
\]

The halfway point is the average:

\[
\frac{\frac16+\frac14}{2}
\]

Find a common denominator:

\[
\frac16=\frac{2}{12}, \qquad \frac14=\frac{3}{12}
\]

So

\[
\frac16+\frac14=\frac{2}{12}+\frac{3}{12}=\frac{5}{12}
\]

Now divide by 2:

\[
\frac{5}{12}\div 2=\frac{5}{12}\cdot \frac12=\frac{5}{24}
\]

So the number halfway between them is \(\frac{5}{24}\).

A tempting wrong answer is \(\frac15\), but that comes from averaging the denominators, which is not how fractions are averaged.

ANSWER 2: C

Problem 3:

The numbers are

\[
-2,\ 4,\ 6,\ 9,\ 12
\]

We are rearranging them using these rules:

- The largest number, \(12\), is not first, but is in one of the first three places.
- The smallest number, \(-2\), is not last, but is in one of the last three places.
- The median number, \(6\), is not first or last.

The largest number \(12\) must be in position 2 or 3.

The smallest number \(-2\) must be in position 3 or 4.

The median \(6\) must be in position 2, 3, or 4.

So the numbers \(12\), \(-2\), and \(6\) occupy the middle three positions in some order.

That means the first and last positions must be filled by the remaining two numbers:

\[
4 \quad \text{and} \quad 9
\]

Their average is

\[
\frac{4+9}{2}=\frac{13}{2}=6.5
\]

So the average of the first and last numbers is \(6.5\).

ANSWER 3: C

Problem 4:

We are given

\[
991+993+995+997+999=5000-N
\]

Add the five numbers. Since they are evenly spaced, their average is the middle number:

\[
995
\]

There are 5 numbers, so the sum is

\[
5 \cdot 995=4975
\]

Thus

\[
4975=5000-N
\]

Solve for \(N\):

\[
N=5000-4975=25
\]

So \(N=25\).

ANSWER 4: E

Problem 5:

We need to count 3-digit positive integers whose digits multiply to 24.

Since the product is 24, no digit can be 0.

Factor:

\[
24=2^3\cdot 3
\]

Now list digit triples whose product is 24:

\[
1,3,8
\]

\[
1,4,6
\]

\[
2,2,6
\]

\[
2,3,4
\]

Now count rearrangements.

For \(1,3,8\), all digits are different:

\[
3!=6
\]

For \(1,4,6\), all digits are different:

\[
3!=6
\]

For \(2,2,6\), two digits are the same, so there are

\[
\frac{3!}{2!}=3
\]

For \(2,3,4\), all digits are different:

\[
3!=6
\]

Total:

\[
6+6+3+6=21
\]

So there are 21 such integers.

ANSWER 5: D

Problem 6:

Brent starts with 4 goldfish and quadruples every month.

After \(m\) months, Brent has

\[
4\cdot 4^m
\]

Gretel starts with 128 goldfish and doubles every month.

After \(m\) months, Gretel has

\[
128\cdot 2^m
\]

We want them equal:

\[
4\cdot 4^m=128\cdot 2^m
\]

Rewrite everything as powers of 2:

\[
4=2^2,\quad 4^m=2^{2m},\quad 128=2^7
\]

So

\[
2^2\cdot 2^{2m}=2^7\cdot 2^m
\]

\[
2^{2+2m}=2^{7+m}
\]

Therefore,

\[
2+2m=7+m
\]

\[
m=5
\]

So in 5 months, they will have the same number of goldfish.

ANSWER 6: B

Problem 7:

We have an isosceles triangle with

\[
AB=BC=29
\]

and base

\[
AC=42
\]

The altitude from \(B\) to \(AC\) bisects the base, so each half is

\[
\frac{42}{2}=21
\]

Now use the Pythagorean Theorem to find the height \(h\):

\[
h^2+21^2=29^2
\]

\[
h^2+441=841
\]

\[
h^2=400
\]

\[
h=20
\]

Now find the area:

\[
\text{Area}=\frac12 \cdot \text{base} \cdot \text{height}
\]

\[
\text{Area}=\frac12 \cdot 42 \cdot 20=420
\]

So the area is 420.

ANSWER 7: B

Problem 8:

We need isosceles triangles with integer side lengths and perimeter 23.

Let the equal sides each have length \(a\), and let the base have length \(b\).

Then

\[
2a+b=23
\]

So

\[
b=23-2a
\]

For a triangle, the sum of the two equal sides must be greater than the base:

\[
2a>b
\]

Substitute \(b=23-2a\):

\[
2a>23-2a
\]

\[
4a>23
\]

So

\[
a>5.75
\]

Thus

\[
a\ge 6
\]

Also, \(b\) must be positive:

\[
23-2a>0
\]

\[
2a<23
\]

\[
a<11.5
\]

Thus

\[
a\le 11
\]

So possible values of \(a\) are

\[
6,7,8,9,10,11
\]

That gives 6 different triangles.

ANSWER 8: C

Problem 9:

We draw three digits from

\[
1,2,3,4
\]

and form a three-digit number.

A number is divisible by 3 if the sum of its digits is divisible by 3.

The sum of all four digits is

\[
1+2+3+4=10
\]

We choose three digits, which is the same as leaving out one digit.

If we leave out digit \(d\), then the chosen digit sum is

\[
10-d
\]

We need

\[
10-d
\]

to be divisible by 3.

Check the possible omitted digits:

- Omit 1: sum is \(9\), divisible by 3.
- Omit 2: sum is \(8\), not divisible by 3.
- Omit 3: sum is \(7\), not divisible by 3.
- Omit 4: sum is \(6\), divisible by 3.

So 2 of the 4 possible sets of digits work.

For each chosen set, all rearrangements are also multiples of 3, since digit sum does not change.

Thus the probability is

\[
\frac{2}{4}=\frac12
\]

ANSWER 9: C

Problem 10:

If the phone is on but not being used, the battery lasts 24 hours, so it uses

\[
\frac{1}{24}
\]

of the battery per hour.

If the phone is being used constantly, the battery lasts 3 hours, so it uses

\[
\frac{1}{3}
\]

of the battery per hour.

Since the last recharge, the phone has been on for 9 hours, and Niki used it for 60 minutes, which is 1 hour.

So the phone was:

- used for 1 hour
- on but not used for 8 hours

Battery used while talking:

\[
1\cdot \frac13=\frac13
\]

Battery used while on but idle:

\[
8\cdot \frac{1}{24}=\frac{8}{24}=\frac13
\]

Total battery used:

\[
\frac13+\frac13=\frac23
\]

Battery remaining:

\[
1-\frac23=\frac13
\]

Now she will leave the phone on but not talk, so it uses \(\frac{1}{24}\)

Problem 1:
**Restatement:** We are asked to find how many times greater the value of the place occupied by the digit 9 is compared to the value of the place occupied by the digit 3 in the number 74982.1035.

**Step-by-step Working:**
1. Identify the place value of the digit 9: In 74982.1035, the 9 is in the hundreds place. Its value is $9 \times 100$, so the place value is $100$.
2. Identify the place value of the digit 3: The 3 is in the thousandths place (three places to the right of the decimal point). Its value is $3 \times 0.001$, so the place value is $0.001$ (or $\frac{1}{1000}$).
3. Calculate the ratio: To find out how many times as great the hundreds place is compared to the thousandths place, we divide the two place values:
$$\frac{100}{0.001} = 100 \times 1000 = 100,000$$

**Ruling out wrong choices:** 
- A (1,000) and B (10,000) are tempting if you miscount the decimal places between the two digits.
- D (1,000,000) would be the answer if the 3 were in the ten-thousandths place.

ANSWER 1: C

Problem 2:
**Restatement:** We need to find the number that is exactly halfway between the fractions $\frac{1}{6}$ and $\frac{1}{4}$.

**Step-by-step Working:**
1. The number halfway between two numbers is their average. We need to calculate $\frac{\frac{1}{6} + \frac{1}{4}}{2}$.
2. Find a common denominator to add the fractions. The least common multiple of 6 and 4 is 12.
$$\frac{1}{6} = \frac{2}{12}$$
$$\frac{1}{4} = \frac{3}{12}$$
3. Add the fractions:
$$\frac{2}{12} + \frac{3}{12} = \frac{5}{12}$$
4. Divide the sum by 2 to find the average:
$$\frac{\frac{5}{12}}{2} = \frac{5}{12} \times \frac{1}{2} = \frac{5}{24}$$

**Ruling out wrong choices:**
- B ($\frac{1}{5}$) is a common mistake made by incorrectly "averaging" the denominators (halfway between 4 and 6).
- E ($\frac{5}{12}$) is the sum of the two fractions, forgetting to divide by 2.

ANSWER 2: C

Problem 3:
**Restatement:** We are given the numbers $-2, 4, 6, 9,$ and $12$. They are placed in five positions based on specific rules. We need to find the average of the numbers in the first and last positions.

**Step-by-step Working:**
1. Identify the key numbers:
   - Largest: $12$
   - Smallest: $-2$
   - Median (middle value when sorted): $6$
   - Remaining numbers: $4$ and $9$
2. Analyze the rules for the 5 positions (1, 2, 3, 4, 5):
   - The largest ($12$) isn't first, but is in the first three places. So, $12$ must be in position 2 or 3. This means $12$ is NOT in position 1 or 5.
   - The smallest ($-2$) isn't last, but is in the last three places. So, $-2$ must be in position 3 or 4. This means $-2$ is NOT in position 1 or 5.
   - The median ($6$) isn't first or last. So, $6$ is NOT in position 1 or 5.
3. Determine the first and last numbers:
   Since $12$, $-2$, and $6$ are all forbidden from being in position 1 and position 5, the only numbers left to fill the first and last spots are $4$ and $9$.
4. Calculate the average of the first and last numbers:
   $$\frac{4 + 9}{2} = \frac{13}{2} = 6.5$$

**Ruling out wrong choices:**
- If you mistakenly thought the median was part of the ends, you might get 5 (average of 4 and 6) or 7.5 (average of 6 and 9). 

ANSWER 3: C

Problem 4:
**Restatement:** We are given the equation $991 + 993 + 995 + 997 + 999 = 5000 - N$ and need to solve for $N$.

**Step-by-step Working:**
1. Notice that each number on the left side is slightly less than 1000. We can rewrite each term:
   $991 = 1000 - 9$
   $993 = 1000 - 7$
   $995 = 1000 - 5$
   $997 = 1000 - 3$
   $999 = 1000 - 1$
2. Substitute these into the sum:
   $(1000 - 9) + (1000 - 7) + (1000 - 5) + (1000 - 3) + (1000 - 1)$
3. Group the 1000s and the subtracted amounts:
   $= 5000 - (9 + 7 + 5 + 3 + 1)$
4. Add the subtracted amounts:
   $9 + 7 + 5 + 3 + 1 = 25$
5. The sum is $5000 - 25$. Therefore, $N = 25$.

**Ruling out wrong choices:**
- A (5) is tempting if you just count the number of terms being added.
- C (15) or D (20) might result from an arithmetic error when adding the odd numbers.

ANSWER 4: E

Problem 5:
**Restatement:** We need to find how many 3-digit positive integers exist such that the product of their three digits is exactly 24.

**Step-by-step Working:**
1. Find all combinations of 3 single-digit numbers (from 1 to 9) that multiply to 24.
   - The prime factorization of 24 is $2^3 \times 3$.
   - Possible sets of digits:
     - $\{1, 3, 8\}$
     - $\{1, 4, 6\}$
     - $\{2, 2, 6\}$
     - $\{2, 3, 4\}$
   (No other combinations work. For example, we cannot use 9, 7, or 5 because 24 is not divisible by them.)
2. Count the number of permutations (arrangements) for each set to form 3-digit numbers:
   - $\{1, 3, 8\}$: 3 distinct digits $\rightarrow 3! = 6$ numbers.
   - $\{1, 4, 6\}$: 3 distinct digits $\rightarrow 3! = 6$ numbers.
   - $\{2, 3, 4\}$: 3 distinct digits $\rightarrow 3! = 6$ numbers.
   - $\{2, 2, 6\}$: 2 identical digits $\rightarrow \frac{3!}{2!} = 3$ numbers.
3. Add them up:
   $6 + 6 + 6 + 3 = 21$

**Ruling out wrong choices:**
- E (24) is tempting if one assumes all 4 sets have 6 permutations ($4 \times 6 = 24$).
- C (18) is tempting if one misses the $\{2, 2, 6\}$ set entirely.

ANSWER 5: D

Problem 6:
**Restatement:** Brent starts with 4 goldfish that quadruple every month. Gretel starts with 128 goldfish that double every month. We need to find the number of months until they have the same amount.

**Step-by-step Working:**
1. Let $m$ be the number of months.
2. Write an expression for Brent's goldfish:
   $B(m) = 4 \times 4^m$
3. Write an expression for Gretel's goldfish:
   $G(m) = 128 \times 2^m$
4. Set them equal to each other:
   $4 \times 4^m = 128 \times 2^m$
5. Express all bases as powers of 2 to easily solve for $m$:
   $4 = 2^2$
   $128 = 2^7$
   $2^2 \times (2^2)^m = 2^7 \times 2^m$
   $2^2 \times 2^{2m} = 2^{m+7}$
   $2^{2m+2} = 2^{m+7}$
6. Equate the exponents:
   $2m + 2 = m + 7$
   $m = 5$

**Ruling out wrong choices:**
- A (4) or C (6) could be obtained if one makes an arithmetic error 

Problem 1:
The problem asks us to compare the place value of the digits 9 and 3 in the number 74982.1035. 
First, we identify the positions of 9 and 3. The digit 9 is in the thousands place, and the digit 3 is in the thousandths place. 
The value of the place occupied by 9 is 1000, and the value of the place occupied by 3 is 0.001. 
To find how many times as great the value of the place occupied by 9 is compared to the value of the place occupied by 3, we calculate 1000 / 0.001. 
This is equal to 1000 * 1000 = 1,000,000. 
Let's examine the choices: A. 1,000, B. 10,000, C. 100,000, D. 1,000,000, E. 10,000,000. 
Our calculation matches choice D.

ANSWER 1: D

Problem 2:
The problem asks for the number halfway between 1/6 and 1/4. 
To find the number halfway between two numbers, we need to calculate their average. 
First, we need to find a common denominator for 1/6 and 1/4. The least common denominator is 12. 
1/6 = 2/12 and 1/4 = 3/12. 
The average is (2/12 + 3/12) / 2 = (5/12) / 2 = 5/24. 
Let's examine the choices: A. 1/10, B. 1/5, C. 5/24, D. 7/24, E. 5/12. 
Our calculation matches choice C.

ANSWER 2: C

Problem 3:
The numbers given are −2, 4, 6, 9, and 12. We need to rearrange them according to the given rules and find the average of the first and last numbers. 
First, let's arrange the numbers in ascending order: −2, 4, 6, 9, 12. 
The largest number is 12, the smallest is −2, and the median is 6. 
According to the rules: the largest isn't first but is in one of the first three places, the smallest isn't last but is in one of the last three places, and the median isn't first or last. 
Let's try to satisfy these conditions. 
Since the median (6) can't be first or last, it must be in the second, third, or fourth position. 
If 6 is in the third position, the arrangement could be: x, y, 6, z, w. 
Given that 12 is in one of the first three places and −2 is in one of the last three places, a possible arrangement is: 9, 12, 6, 4, −2 or 4, 12, 6, 9, −2. 
However, we need to check if these satisfy all conditions. 
For 9, 12, 6, 4, −2: 12 is in the second position (satisfies the first condition), and −2 is in the last position (does not satisfy the second condition because −2 is last). 
For 4, 12, 6, 9, −2: again, −2 is last, so it doesn't satisfy the condition. 
Let's consider 6 in the second position: x, 6, y, z, w. 
A possible arrangement could be: 9, 6, 12, −2, 4. 
Here, 12 is in the third position, and −2 is in the fourth, satisfying the conditions that 12 is in one of the first three places and −2 is in one of the last three. 
The first number is 9, and the last is 4. The average is (9 + 4) / 2 = 6.5. 
Let's examine the choices: A. 3.5, B. 5, C. 6.5, D. 7.5, E. 8. 
Our calculation matches choice C.

ANSWER 3: C

Problem 4:
The problem gives the equation 991 + 993 + 995 + 997 + 999 = 5000 − N and asks to find N. 
First, let's simplify the left-hand side of the equation. 
991 + 993 + 995 + 997 + 999 can be rewritten as (1000 - 9) + (1000 - 7) + (1000 - 5) + (1000 - 3) + (1000 - 1). 
This equals 5*1000 - (9 + 7 + 5 + 3 + 1) = 5000 - 25. 
So, 5000 - 25 = 5000 - N. 
This implies that N = 25. 
Let's examine the choices: A. 5, B. 10, C. 15, D. 20, E. 25. 
Our calculation matches choice E.

ANSWER 4: E

Problem 5:
The problem asks us to find how many 3-digit positive integers have digits whose product equals 24. 
To solve this, we need to list the factor combinations of 24 and see which ones can form 3-digit numbers. 
The prime factorization of 24 is 2^3 * 3. 
Possible combinations for the digits are derived from the factors: (1, 1, 24) - not valid since 24 is not a single digit; (1, 2, 12) - not valid; (1, 3, 8); (1, 4, 6); (2, 2, 6); (2, 3, 4); and (2, 2, 6). 
Let's count the permutations for valid combinations: 
- For (1, 3, 8), the permutations are 138, 183, 318, 381, 813, 831, which gives 6 numbers. 
- For (1, 4, 6), the permutations are 146, 164, 416, 461, 614, 641, which gives 6 numbers. 
- For (2, 2, 6), the permutations are 226, 262, 622, which gives 3 numbers. 
- For (2, 3, 4), the permutations are 234, 243, 324, 342, 423, 432, which gives 6 numbers. 
Adding these up: 6 + 6 + 3 + 6 = 21. 
Let's examine the choices: A. 12, B. 15, C. 18, D. 21, E. 24. 
Our calculation matches choice D.

ANSWER 5: D

Problem 6:
Brent has goldfish that quadruple every month, and Gretel has goldfish that double every month. 
Initially, Brent has 4 goldfish, and Gretel has 128 goldfish. 
Let's denote the number of months from the initial time as 'n'. 
After 'n' months, Brent will have 4 * (4^n) goldfish, and Gretel will have 128 * (2^n) goldfish. 
We need to find 'n' when 4 * (4^n) = 128 * (2^n). 
Simplifying, we get 4^(n+1) = 128 * 2^n. 
Since 4 = 2^2, we can rewrite this as (2^2)^(n+1) = 2^7 * 2^n. 
This simplifies to 2^(2n+2) = 2^(n+7). 
Equating the exponents, we get 2n + 2 = n + 7. 
Solving for 'n', we get n = 5. 
Let's examine the choices: A. 4, B. 5, C. 6, D. 7, E. 8. 
Our calculation matches choice B.

ANSWER 6: B

Problem 7:
In ▵ABC, AB = BC = 29, and AC = 42. 
Since AB = BC, ▵ABC is isosceles. 
To find the area, we need the height. 
Let's draw the altitude from B to AC, meeting AC at point D. 
Since ▵ABC is isosceles, D is the midpoint of AC. 
So, AD = DC = 42 / 2 = 21. 
Using the Pythagorean theorem in ▵ADB, we have BD^2 = AB^2 - AD^2 = 29^2 - 21^2. 
Calculating, BD^2 = 841 - 441 = 400, so BD = 20. 
The area of ▵ABC = 0.5 * base * height = 0.5 * 42 * 20 = 420. 
Let's examine the choices: A. 100, B. 420, C. 500, D. 609, E. 701. 
Our calculation matches choice B.

ANSWER 7: B

Problem 8:
The problem asks for the number of different isosceles triangles with integer side lengths and a perimeter of 23. 
For an isosceles triangle, two sides are equal. Let's denote the equal sides as 'a' and the base as 'b'. 
So, 2a + b = 23, and a + a > b (triangle inequality). 
This implies that 2a > b, or 2a > 23 - 2a, which simplifies to 4a > 23, or a > 5.75. 
Since 'a' is an integer, a ≥ 6. 
Also, a + b > a, which i

Problem 1:

We are asked about the **place** occupied by the digit 9 and the digit 3 in

\[
74982.1035
\]

The digit 9 is in the **hundreds** place, so the value of its place is

\[
100
\]

The digit 3 is in the **thousandths** place, so the value of its place is

\[
0.001=\frac{1}{1000}
\]

Now compare:

\[
100 \div 0.001 = 100 \div \frac{1}{1000}=100 \times 1000=100000
\]

So the place of the 9 is \(100{,}000\) times as great as the place of the 3.

Choices A and B are too small, and D and E are too large.

ANSWER 1: C

Problem 2:

We need the number halfway between

\[
\frac16 \quad \text{and} \quad \frac14
\]

The halfway point is the average:

\[
\frac{\frac16+\frac14}{2}
\]

Find a common denominator:

\[
\frac16=\frac{2}{12}, \qquad \frac14=\frac{3}{12}
\]

So

\[
\frac16+\frac14=\frac{2}{12}+\frac{3}{12}=\frac{5}{12}
\]

Now divide by 2:

\[
\frac{5}{12}\div 2=\frac{5}{12}\cdot \frac12=\frac{5}{24}
\]

So the number halfway between them is \(\frac{5}{24}\).

A tempting wrong answer is \(\frac15\), but that comes from averaging the denominators, which is not how fractions are averaged.

ANSWER 2: C

Problem 3:

The numbers are

\[
-2,\ 4,\ 6,\ 9,\ 12
\]

We are rearranging them using these rules:

- The largest number, \(12\), is not first, but is in one of the first three places.
- The smallest number, \(-2\), is not last, but is in one of the last three places.
- The median number, \(6\), is not first or last.

The largest number \(12\) must be in position 2 or 3.

The smallest number \(-2\) must be in position 3 or 4.

The median \(6\) must be in position 2, 3, or 4.

So the numbers \(12\), \(-2\), and \(6\) occupy the middle three positions in some order.

That means the first and last positions must be filled by the remaining two numbers:

\[
4 \quad \text{and} \quad 9
\]

Their average is

\[
\frac{4+9}{2}=\frac{13}{2}=6.5
\]

So the average of the first and last numbers is \(6.5\).

ANSWER 3: C

Problem 4:

We are given

\[
991+993+995+997+999=5000-N
\]

Add the five numbers. Since they are evenly spaced, their average is the middle number:

\[
995
\]

There are 5 numbers, so the sum is

\[
5 \cdot 995=4975
\]

Thus

\[
4975=5000-N
\]

Solve for \(N\):

\[
N=5000-4975=25
\]

So \(N=25\).

ANSWER 4: E

Problem 5:

We need to count 3-digit positive integers whose digits multiply to 24.

Since the product is 24, no digit can be 0.

Factor:

\[
24=2^3\cdot 3
\]

Now list digit triples whose product is 24:

\[
1,3,8
\]

\[
1,4,6
\]

\[
2,2,6
\]

\[
2,3,4
\]

Now count rearrangements.

For \(1,3,8\), all digits are different:

\[
3!=6
\]

For \(1,4,6\), all digits are different:

\[
3!=6
\]

For \(2,2,6\), two digits are the same, so there are

\[
\frac{3!}{2!}=3
\]

For \(2,3,4\), all digits are different:

\[
3!=6
\]

Total:

\[
6+6+3+6=21
\]

So there are 21 such integers.

ANSWER 5: D

Problem 6:

Brent starts with 4 goldfish and quadruples every month.

After \(m\) months, Brent has

\[
4\cdot 4^m
\]

Gretel starts with 128 goldfish and doubles every month.

After \(m\) months, Gretel has

\[
128\cdot 2^m
\]

We want them equal:

\[
4\cdot 4^m=128\cdot 2^m
\]

Rewrite everything as powers of 2:

\[
4=2^2,\quad 4^m=2^{2m},\quad 128=2^7
\]

So

\[
2^2\cdot 2^{2m}=2^7\cdot 2^m
\]

\[
2^{2+2m}=2^{7+m}
\]

Therefore,

\[
2+2m=7+m
\]

\[
m=5
\]

So in 5 months, they will have the same number of goldfish.

ANSWER 6: B

Problem 7:

We have an isosceles triangle with

\[
AB=BC=29
\]

and base

\[
AC=42
\]

The altitude from \(B\) to \(AC\) bisects the base, so each half is

\[
\frac{42}{2}=21
\]

Now use the Pythagorean Theorem to find the height \(h\):

\[
h^2+21^2=29^2
\]

\[
h^2+441=841
\]

\[
h^2=400
\]

\[
h=20
\]

Now find the area:

\[
\text{Area}=\frac12 \cdot \text{base} \cdot \text{height}
\]

\[
\text{Area}=\frac12 \cdot 42 \cdot 20=420
\]

So the area is 420.

ANSWER 7: B

Problem 8:

We need isosceles triangles with integer side lengths and perimeter 23.

Let the equal sides each have length \(a\), and let the base have length \(b\).

Then

\[
2a+b=23
\]

So

\[
b=23-2a
\]

For a triangle, the sum of the two equal sides must be greater than the base:

\[
2a>b
\]

Substitute \(b=23-2a\):

\[
2a>23-2a
\]

\[
4a>23
\]

So

\[
a>5.75
\]

Thus

\[
a\ge 6
\]

Also, \(b\) must be positive:

\[
23-2a>0
\]

\[
2a<23
\]

\[
a<11.5
\]

Thus

\[
a\le 11
\]

So possible values of \(a\) are

\[
6,7,8,9,10,11
\]

That gives 6 different triangles.

ANSWER 8: C

Problem 9:

We draw three digits from

\[
1,2,3,4
\]

and form a three-digit number.

A number is divisible by 3 if the sum of its digits is divisible by 3.

The sum of all four digits is

\[
1+2+3+4=10
\]

We choose three digits, which is the same as leaving out one digit.

If we leave out digit \(d\), then the chosen digit sum is

\[
10-d
\]

We need

\[
10-d
\]

to be divisible by 3.

Check the possible omitted digits:

- Omit 1: sum is \(9\), divisible by 3.
- Omit 2: sum is \(8\), not divisible by 3.
- Omit 3: sum is \(7\), not divisible by 3.
- Omit 4: sum is \(6\), divisible by 3.

So 2 of the 4 possible sets of digits work.

For each chosen set, all rearrangements are also multiples of 3, since digit sum does not change.

Thus the probability is

\[
\frac{2}{4}=\frac12
\]

ANSWER 9: C

Problem 10:

If the phone is on but not being used, the battery lasts 24 hours, so it uses

\[
\frac{1}{24}
\]

of the battery per hour.

If the phone is being used constantly, the battery lasts 3 hours, so it uses

\[
\frac{1}{3}
\]

of the battery per hour.

Since the last recharge, the phone has been on for 9 hours, and Niki used it for 60 minutes, which is 1 hour.

So the phone was:

- used for 1 hour
- on but not used for 8 hours

Battery used while talking:

\[
1\cdot \frac13=\frac13
\]

Battery used while on but idle:

\[
8\cdot \frac{1}{24}=\frac{8}{24}=\frac13
\]

Total battery used:

\[
\frac13+\frac13=\frac23
\]

Battery remaining:

\[
1-\frac23=\frac13
\]

Now she will leave the phone on but not talk, so it uses \(\frac{1}{24}\)

Problem 1:
**Restatement:** We are asked to find how many times greater the value of the place occupied by the digit 9 is compared to the value of the place occupied by the digit 3 in the number 74982.1035.

**Step-by-step Working:**
1. Identify the place value of the digit 9: In 74982.1035, the 9 is in the hundreds place. Its value is $9 \times 100$, so the place value is $100$.
2. Identify the place value of the digit 3: The 3 is in the thousandths place (three places to the right of the decimal point). Its value is $3 \times 0.001$, so the place value is $0.001$ (or $\frac{1}{1000}$).
3. Calculate the ratio: To find out how many times as great the hundreds place is compared to the thousandths place, we divide the two place values:
$$\frac{100}{0.001} = 100 \times 1000 = 100,000$$

**Ruling out wrong choices:** 
- A (1,000) and B (10,000) are tempting if you miscount the decimal places between the two digits.
- D (1,000,000) would be the answer if the 3 were in the ten-thousandths place.

ANSWER 1: C

Problem 2:
**Restatement:** We need to find the number that is exactly halfway between the fractions $\frac{1}{6}$ and $\frac{1}{4}$.

**Step-by-step Working:**
1. The number halfway between two numbers is their average. We need to calculate $\frac{\frac{1}{6} + \frac{1}{4}}{2}$.
2. Find a common denominator to add the fractions. The least common multiple of 6 and 4 is 12.
$$\frac{1}{6} = \frac{2}{12}$$
$$\frac{1}{4} = \frac{3}{12}$$
3. Add the fractions:
$$\frac{2}{12} + \frac{3}{12} = \frac{5}{12}$$
4. Divide the sum by 2 to find the average:
$$\frac{\frac{5}{12}}{2} = \frac{5}{12} \times \frac{1}{2} = \frac{5}{24}$$

**Ruling out wrong choices:**
- B ($\frac{1}{5}$) is a common mistake made by incorrectly "averaging" the denominators (halfway between 4 and 6).
- E ($\frac{5}{12}$) is the sum of the two fractions, forgetting to divide by 2.

ANSWER 2: C

Problem 3:
**Restatement:** We are given the numbers $-2, 4, 6, 9,$ and $12$. They are placed in five positions based on specific rules. We need to find the average of the numbers in the first and last positions.

**Step-by-step Working:**
1. Identify the key numbers:
   - Largest: $12$
   - Smallest: $-2$
   - Median (middle value when sorted): $6$
   - Remaining numbers: $4$ and $9$
2. Analyze the rules for the 5 positions (1, 2, 3, 4, 5):
   - The largest ($12$) isn't first, but is in the first three places. So, $12$ must be in position 2 or 3. This means $12$ is NOT in position 1 or 5.
   - The smallest ($-2$) isn't last, but is in the last three places. So, $-2$ must be in position 3 or 4. This means $-2$ is NOT in position 1 or 5.
   - The median ($6$) isn't first or last. So, $6$ is NOT in position 1 or 5.
3. Determine the first and last numbers:
   Since $12$, $-2$, and $6$ are all forbidden from being in position 1 and position 5, the only numbers left to fill the first and last spots are $4$ and $9$.
4. Calculate the average of the first and last numbers:
   $$\frac{4 + 9}{2} = \frac{13}{2} = 6.5$$

**Ruling out wrong choices:**
- If you mistakenly thought the median was part of the ends, you might get 5 (average of 4 and 6) or 7.5 (average of 6 and 9). 

ANSWER 3: C

Problem 4:
**Restatement:** We are given the equation $991 + 993 + 995 + 997 + 999 = 5000 - N$ and need to solve for $N$.

**Step-by-step Working:**
1. Notice that each number on the left side is slightly less than 1000. We can rewrite each term:
   $991 = 1000 - 9$
   $993 = 1000 - 7$
   $995 = 1000 - 5$
   $997 = 1000 - 3$
   $999 = 1000 - 1$
2. Substitute these into the sum:
   $(1000 - 9) + (1000 - 7) + (1000 - 5) + (1000 - 3) + (1000 - 1)$
3. Group the 1000s and the subtracted amounts:
   $= 5000 - (9 + 7 + 5 + 3 + 1)$
4. Add the subtracted amounts:
   $9 + 7 + 5 + 3 + 1 = 25$
5. The sum is $5000 - 25$. Therefore, $N = 25$.

**Ruling out wrong choices:**
- A (5) is tempting if you just count the number of terms being added.
- C (15) or D (20) might result from an arithmetic error when adding the odd numbers.

ANSWER 4: E

Problem 5:
**Restatement:** We need to find how many 3-digit positive integers exist such that the product of their three digits is exactly 24.

**Step-by-step Working:**
1. Find all combinations of 3 single-digit numbers (from 1 to 9) that multiply to 24.
   - The prime factorization of 24 is $2^3 \times 3$.
   - Possible sets of digits:
     - $\{1, 3, 8\}$
     - $\{1, 4, 6\}$
     - $\{2, 2, 6\}$
     - $\{2, 3, 4\}$
   (No other combinations work. For example, we cannot use 9, 7, or 5 because 24 is not divisible by them.)
2. Count the number of permutations (arrangements) for each set to form 3-digit numbers:
   - $\{1, 3, 8\}$: 3 distinct digits $\rightarrow 3! = 6$ numbers.
   - $\{1, 4, 6\}$: 3 distinct digits $\rightarrow 3! = 6$ numbers.
   - $\{2, 3, 4\}$: 3 distinct digits $\rightarrow 3! = 6$ numbers.
   - $\{2, 2, 6\}$: 2 identical digits $\rightarrow \frac{3!}{2!} = 3$ numbers.
3. Add them up:
   $6 + 6 + 6 + 3 = 21$

**Ruling out wrong choices:**
- E (24) is tempting if one assumes all 4 sets have 6 permutations ($4 \times 6 = 24$).
- C (18) is tempting if one misses the $\{2, 2, 6\}$ set entirely.

ANSWER 5: D

Problem 6:
**Restatement:** Brent starts with 4 goldfish that quadruple every month. Gretel starts with 128 goldfish that double every month. We need to find the number of months until they have the same amount.

**Step-by-step Working:**
1. Let $m$ be the number of months.
2. Write an expression for Brent's goldfish:
   $B(m) = 4 \times 4^m$
3. Write an expression for Gretel's goldfish:
   $G(m) = 128 \times 2^m$
4. Set them equal to each other:
   $4 \times 4^m = 128 \times 2^m$
5. Express all bases as powers of 2 to easily solve for $m$:
   $4 = 2^2$
   $128 = 2^7$
   $2^2 \times (2^2)^m = 2^7 \times 2^m$
   $2^2 \times 2^{2m} = 2^{m+7}$
   $2^{2m+2} = 2^{m+7}$
6. Equate the exponents:
   $2m + 2 = m + 7$
   $m = 5$

**Ruling out wrong choices:**
- A (4) or C (6) could be obtained if one makes an arithmetic error 

Problem 1:
The problem asks us to compare the place value of the digits 9 and 3 in the number 74982.1035. 
First, we identify the positions of 9 and 3. The digit 9 is in the thousands place, and the digit 3 is in the thousandths place. 
The value of the place occupied by 9 is 1000, and the value of the place occupied by 3 is 0.001. 
To find how many times as great the value of the place occupied by 9 is compared to the value of the place occupied by 3, we calculate 1000 / 0.001. 
This is equal to 1000 * 1000 = 1,000,000. 
Let's examine the choices: A. 1,000, B. 10,000, C. 100,000, D. 1,000,000, E. 10,000,000. 
Our calculation matches choice D.

ANSWER 1: D

Problem 2:
The problem asks for the number halfway between 1/6 and 1/4. 
To find the number halfway between two numbers, we need to calculate their average. 
First, we need to find a common denominator for 1/6 and 1/4. The least common denominator is 12. 
1/6 = 2/12 and 1/4 = 3/12. 
The average is (2/12 + 3/12) / 2 = (5/12) / 2 = 5/24. 
Let's examine the choices: A. 1/10, B. 1/5, C. 5/24, D. 7/24, E. 5/12. 
Our calculation matches choice C.

ANSWER 2: C

Problem 3:
The numbers given are −2, 4, 6, 9, and 12. We need to rearrange them according to the given rules and find the average of the first and last numbers. 
First, let's arrange the numbers in ascending order: −2, 4, 6, 9, 12. 
The largest number is 12, the smallest is −2, and the median is 6. 
According to the rules: the largest isn't first but is in one of the first three places, the smallest isn't last but is in one of the last three places, and the median isn't first or last. 
Let's try to satisfy these conditions. 
Since the median (6) can't be first or last, it must be in the second, third, or fourth position. 
If 6 is in the third position, the arrangement could be: x, y, 6, z, w. 
Given that 12 is in one of the first three places and −2 is in one of the last three places, a possible arrangement is: 9, 12, 6, 4, −2 or 4, 12, 6, 9, −2. 
However, we need to check if these satisfy all conditions. 
For 9, 12, 6, 4, −2: 12 is in the second position (satisfies the first condition), and −2 is in the last position (does not satisfy the second condition because −2 is last). 
For 4, 12, 6, 9, −2: again, −2 is last, so it doesn't satisfy the condition. 
Let's consider 6 in the second position: x, 6, y, z, w. 
A possible arrangement could be: 9, 6, 12, −2, 4. 
Here, 12 is in the third position, and −2 is in the fourth, satisfying the conditions that 12 is in one of the first three places and −2 is in one of the last three. 
The first number is 9, and the last is 4. The average is (9 + 4) / 2 = 6.5. 
Let's examine the choices: A. 3.5, B. 5, C. 6.5, D. 7.5, E. 8. 
Our calculation matches choice C.

ANSWER 3: C

Problem 4:
The problem gives the equation 991 + 993 + 995 + 997 + 999 = 5000 − N and asks to find N. 
First, let's simplify the left-hand side of the equation. 
991 + 993 + 995 + 997 + 999 can be rewritten as (1000 - 9) + (1000 - 7) + (1000 - 5) + (1000 - 3) + (1000 - 1). 
This equals 5*1000 - (9 + 7 + 5 + 3 + 1) = 5000 - 25. 
So, 5000 - 25 = 5000 - N. 
This implies that N = 25. 
Let's examine the choices: A. 5, B. 10, C. 15, D. 20, E. 25. 
Our calculation matches choice E.

ANSWER 4: E

Problem 5:
The problem asks us to find how many 3-digit positive integers have digits whose product equals 24. 
To solve this, we need to list the factor combinations of 24 and see which ones can form 3-digit numbers. 
The prime factorization of 24 is 2^3 * 3. 
Possible combinations for the digits are derived from the factors: (1, 1, 24) - not valid since 24 is not a single digit; (1, 2, 12) - not valid; (1, 3, 8); (1, 4, 6); (2, 2, 6); (2, 3, 4); and (2, 2, 6). 
Let's count the permutations for valid combinations: 
- For (1, 3, 8), the permutations are 138, 183, 318, 381, 813, 831, which gives 6 numbers. 
- For (1, 4, 6), the permutations are 146, 164, 416, 461, 614, 641, which gives 6 numbers. 
- For (2, 2, 6), the permutations are 226, 262, 622, which gives 3 numbers. 
- For (2, 3, 4), the permutations are 234, 243, 324, 342, 423, 432, which gives 6 numbers. 
Adding these up: 6 + 6 + 3 + 6 = 21. 
Let's examine the choices: A. 12, B. 15, C. 18, D. 21, E. 24. 
Our calculation matches choice D.

ANSWER 5: D

Problem 6:
Brent has goldfish that quadruple every month, and Gretel has goldfish that double every month. 
Initially, Brent has 4 goldfish, and Gretel has 128 goldfish. 
Let's denote the number of months from the initial time as 'n'. 
After 'n' months, Brent will have 4 * (4^n) goldfish, and Gretel will have 128 * (2^n) goldfish. 
We need to find 'n' when 4 * (4^n) = 128 * (2^n). 
Simplifying, we get 4^(n+1) = 128 * 2^n. 
Since 4 = 2^2, we can rewrite this as (2^2)^(n+1) = 2^7 * 2^n. 
This simplifies to 2^(2n+2) = 2^(n+7). 
Equating the exponents, we get 2n + 2 = n + 7. 
Solving for 'n', we get n = 5. 
Let's examine the choices: A. 4, B. 5, C. 6, D. 7, E. 8. 
Our calculation matches choice B.

ANSWER 6: B

Problem 7:
In ▵ABC, AB = BC = 29, and AC = 42. 
Since AB = BC, ▵ABC is isosceles. 
To find the area, we need the height. 
Let's draw the altitude from B to AC, meeting AC at point D. 
Since ▵ABC is isosceles, D is the midpoint of AC. 
So, AD = DC = 42 / 2 = 21. 
Using the Pythagorean theorem in ▵ADB, we have BD^2 = AB^2 - AD^2 = 29^2 - 21^2. 
Calculating, BD^2 = 841 - 441 = 400, so BD = 20. 
The area of ▵ABC = 0.5 * base * height = 0.5 * 42 * 20 = 420. 
Let's examine the choices: A. 100, B. 420, C. 500, D. 609, E. 701. 
Our calculation matches choice B.

ANSWER 7: B

Problem 8:
The problem asks for the number of different isosceles triangles with integer side lengths and a perimeter of 23. 
For an isosceles triangle, two sides are equal. Let's denote the equal sides as 'a' and the base as 'b'. 
So, 2a + b = 23, and a + a > b (triangle inequality). 
This implies that 2a > b, or 2a > 23 - 2a, which simplifies to 4a > 23, or a > 5.75. 
Since 'a' is an integer, a ≥ 6. 
Also, a + b > a, which i

Problem 1:

We are asked about the **place** occupied by the digit 9 and the digit 3 in

\[
74982.1035
\]

The digit 9 is in the **hundreds** place, so the value of its place is

\[
100
\]

The digit 3 is in the **thousandths** place, so the value of its place is

\[
0.001=\frac{1}{1000}
\]

Now compare:

\[
100 \div 0.001 = 100 \div \frac{1}{1000}=100 \times 1000=100000
\]

So the place of the 9 is \(100{,}000\) times as great as the place of the 3.

Choices A and B are too small, and D and E are too large.

ANSWER 1: C

Problem 2:

We need the number halfway between

\[
\frac16 \quad \text{and} \quad \frac14
\]

The halfway point is the average:

\[
\frac{\frac16+\frac14}{2}
\]

Find a common denominator:

\[
\frac16=\frac{2}{12}, \qquad \frac14=\frac{3}{12}
\]

So

\[
\frac16+\frac14=\frac{2}{12}+\frac{3}{12}=\frac{5}{12}
\]

Now divide by 2:

\[
\frac{5}{12}\div 2=\frac{5}{12}\cdot \frac12=\frac{5}{24}
\]

So the number halfway between them is \(\frac{5}{24}\).

A tempting wrong answer is \(\frac15\), but that comes from averaging the denominators, which is not how fractions are averaged.

ANSWER 2: C

Problem 3:

The numbers are

\[
-2,\ 4,\ 6,\ 9,\ 12
\]

We are rearranging them using these rules:

- The largest number, \(12\), is not first, but is in one of the first three places.
- The smallest number, \(-2\), is not last, but is in one of the last three places.
- The median number, \(6\), is not first or last.

The largest number \(12\) must be in position 2 or 3.

The smallest number \(-2\) must be in position 3 or 4.

The median \(6\) must be in position 2, 3, or 4.

So the numbers \(12\), \(-2\), and \(6\) occupy the middle three positions in some order.

That means the first and last positions must be filled by the remaining two numbers:

\[
4 \quad \text{and} \quad 9
\]

Their average is

\[
\frac{4+9}{2}=\frac{13}{2}=6.5
\]

So the average of the first and last numbers is \(6.5\).

ANSWER 3: C

Problem 4:

We are given

\[
991+993+995+997+999=5000-N
\]

Add the five numbers. Since they are evenly spaced, their average is the middle number:

\[
995
\]

There are 5 numbers, so the sum is

\[
5 \cdot 995=4975
\]

Thus

\[
4975=5000-N
\]

Solve for \(N\):

\[
N=5000-4975=25
\]

So \(N=25\).

ANSWER 4: E

Problem 5:

We need to count 3-digit positive integers whose digits multiply to 24.

Since the product is 24, no digit can be 0.

Factor:

\[
24=2^3\cdot 3
\]

Now list digit triples whose product is 24:

\[
1,3,8
\]

\[
1,4,6
\]

\[
2,2,6
\]

\[
2,3,4
\]

Now count rearrangements.

For \(1,3,8\), all digits are different:

\[
3!=6
\]

For \(1,4,6\), all digits are different:

\[
3!=6
\]

For \(2,2,6\), two digits are the same, so there are

\[
\frac{3!}{2!}=3
\]

For \(2,3,4\), all digits are different:

\[
3!=6
\]

Total:

\[
6+6+3+6=21
\]

So there are 21 such integers.

ANSWER 5: D

Problem 6:

Brent starts with 4 goldfish and quadruples every month.

After \(m\) months, Brent has

\[
4\cdot 4^m
\]

Gretel starts with 128 goldfish and doubles every month.

After \(m\) months, Gretel has

\[
128\cdot 2^m
\]

We want them equal:

\[
4\cdot 4^m=128\cdot 2^m
\]

Rewrite everything as powers of 2:

\[
4=2^2,\quad 4^m=2^{2m},\quad 128=2^7
\]

So

\[
2^2\cdot 2^{2m}=2^7\cdot 2^m
\]

\[
2^{2+2m}=2^{7+m}
\]

Therefore,

\[
2+2m=7+m
\]

\[
m=5
\]

So in 5 months, they will have the same number of goldfish.

ANSWER 6: B

Problem 7:

We have an isosceles triangle with

\[
AB=BC=29
\]

and base

\[
AC=42
\]

The altitude from \(B\) to \(AC\) bisects the base, so each half is

\[
\frac{42}{2}=21
\]

Now use the Pythagorean Theorem to find the height \(h\):

\[
h^2+21^2=29^2
\]

\[
h^2+441=841
\]

\[
h^2=400
\]

\[
h=20
\]

Now find the area:

\[
\text{Area}=\frac12 \cdot \text{base} \cdot \text{height}
\]

\[
\text{Area}=\frac12 \cdot 42 \cdot 20=420
\]

So the area is 420.

ANSWER 7: B

Problem 8:

We need isosceles triangles with integer side lengths and perimeter 23.

Let the equal sides each have length \(a\), and let the base have length \(b\).

Then

\[
2a+b=23
\]

So

\[
b=23-2a
\]

For a triangle, the sum of the two equal sides must be greater than the base:

\[
2a>b
\]

Substitute \(b=23-2a\):

\[
2a>23-2a
\]

\[
4a>23
\]

So

\[
a>5.75
\]

Thus

\[
a\ge 6
\]

Also, \(b\) must be positive:

\[
23-2a>0
\]

\[
2a<23
\]

\[
a<11.5
\]

Thus

\[
a\le 11
\]

So possible values of \(a\) are

\[
6,7,8,9,10,11
\]

That gives 6 different triangles.

ANSWER 8: C

Problem 9:

We draw three digits from

\[
1,2,3,4
\]

and form a three-digit number.

A number is divisible by 3 if the sum of its digits is divisible by 3.

The sum of all four digits is

\[
1+2+3+4=10
\]

We choose three digits, which is the same as leaving out one digit.

If we leave out digit \(d\), then the chosen digit sum is

\[
10-d
\]

We need

\[
10-d
\]

to be divisible by 3.

Check the possible omitted digits:

- Omit 1: sum is \(9\), divisible by 3.
- Omit 2: sum is \(8\), not divisible by 3.
- Omit 3: sum is \(7\), not divisible by 3.
- Omit 4: sum is \(6\), divisible by 3.

So 2 of the 4 possible sets of digits work.

For each chosen set, all rearrangements are also multiples of 3, since digit sum does not change.

Thus the probability is

\[
\frac{2}{4}=\frac12
\]

ANSWER 9: C

Problem 10:

If the phone is on but not being used, the battery lasts 24 hours, so it uses

\[
\frac{1}{24}
\]

of the battery per hour.

If the phone is being used constantly, the battery lasts 3 hours, so it uses

\[
\frac{1}{3}
\]

of the battery per hour.

Since the last recharge, the phone has been on for 9 hours, and Niki used it for 60 minutes, which is 1 hour.

So the phone was:

- used for 1 hour
- on but not used for 8 hours

Battery used while talking:

\[
1\cdot \frac13=\frac13
\]

Battery used while on but idle:

\[
8\cdot \frac{1}{24}=\frac{8}{24}=\frac13
\]

Total battery used:

\[
\frac13+\frac13=\frac23
\]

Battery remaining:

\[
1-\frac23=\frac13
\]

Now she will leave the phone on but not talk, so it uses \(\frac{1}{24}\)

Problem 1:
**Restatement:** We are asked to find how many times greater the value of the place occupied by the digit 9 is compared to the value of the place occupied by the digit 3 in the number 74982.1035.

**Step-by-step Working:**
1. Identify the place value of the digit 9: In 74982.1035, the 9 is in the hundreds place. Its value is $9 \times 100$, so the place value is $100$.
2. Identify the place value of the digit 3: The 3 is in the thousandths place (three places to the right of the decimal point). Its value is $3 \times 0.001$, so the place value is $0.001$ (or $\frac{1}{1000}$).
3. Calculate the ratio: To find out how many times as great the hundreds place is compared to the thousandths place, we divide the two place values:
$$\frac{100}{0.001} = 100 \times 1000 = 100,000$$

**Ruling out wrong choices:** 
- A (1,000) and B (10,000) are tempting if you miscount the decimal places between the two digits.
- D (1,000,000) would be the answer if the 3 were in the ten-thousandths place.

ANSWER 1: C

Problem 2:
**Restatement:** We need to find the number that is exactly halfway between the fractions $\frac{1}{6}$ and $\frac{1}{4}$.

**Step-by-step Working:**
1. The number halfway between two numbers is their average. We need to calculate $\frac{\frac{1}{6} + \frac{1}{4}}{2}$.
2. Find a common denominator to add the fractions. The least common multiple of 6 and 4 is 12.
$$\frac{1}{6} = \frac{2}{12}$$
$$\frac{1}{4} = \frac{3}{12}$$
3. Add the fractions:
$$\frac{2}{12} + \frac{3}{12} = \frac{5}{12}$$
4. Divide the sum by 2 to find the average:
$$\frac{\frac{5}{12}}{2} = \frac{5}{12} \times \frac{1}{2} = \frac{5}{24}$$

**Ruling out wrong choices:**
- B ($\frac{1}{5}$) is a common mistake made by incorrectly "averaging" the denominators (halfway between 4 and 6).
- E ($\frac{5}{12}$) is the sum of the two fractions, forgetting to divide by 2.

ANSWER 2: C

Problem 3:
**Restatement:** We are given the numbers $-2, 4, 6, 9,$ and $12$. They are placed in five positions based on specific rules. We need to find the average of the numbers in the first and last positions.

**Step-by-step Working:**
1. Identify the key numbers:
   - Largest: $12$
   - Smallest: $-2$
   - Median (middle value when sorted): $6$
   - Remaining numbers: $4$ and $9$
2. Analyze the rules for the 5 positions (1, 2, 3, 4, 5):
   - The largest ($12$) isn't first, but is in the first three places. So, $12$ must be in position 2 or 3. This means $12$ is NOT in position 1 or 5.
   - The smallest ($-2$) isn't last, but is in the last three places. So, $-2$ must be in position 3 or 4. This means $-2$ is NOT in position 1 or 5.
   - The median ($6$) isn't first or last. So, $6$ is NOT in position 1 or 5.
3. Determine the first and last numbers:
   Since $12$, $-2$, and $6$ are all forbidden from being in position 1 and position 5, the only numbers left to fill the first and last spots are $4$ and $9$.
4. Calculate the average of the first and last numbers:
   $$\frac{4 + 9}{2} = \frac{13}{2} = 6.5$$

**Ruling out wrong choices:**
- If you mistakenly thought the median was part of the ends, you might get 5 (average of 4 and 6) or 7.5 (average of 6 and 9). 

ANSWER 3: C

Problem 4:
**Restatement:** We are given the equation $991 + 993 + 995 + 997 + 999 = 5000 - N$ and need to solve for $N$.

**Step-by-step Working:**
1. Notice that each number on the left side is slightly less than 1000. We can rewrite each term:
   $991 = 1000 - 9$
   $993 = 1000 - 7$
   $995 = 1000 - 5$
   $997 = 1000 - 3$
   $999 = 1000 - 1$
2. Substitute these into the sum:
   $(1000 - 9) + (1000 - 7) + (1000 - 5) + (1000 - 3) + (1000 - 1)$
3. Group the 1000s and the subtracted amounts:
   $= 5000 - (9 + 7 + 5 + 3 + 1)$
4. Add the subtracted amounts:
   $9 + 7 + 5 + 3 + 1 = 25$
5. The sum is $5000 - 25$. Therefore, $N = 25$.

**Ruling out wrong choices:**
- A (5) is tempting if you just count the number of terms being added.
- C (15) or D (20) might result from an arithmetic error when adding the odd numbers.

ANSWER 4: E

Problem 5:
**Restatement:** We need to find how many 3-digit positive integers exist such that the product of their three digits is exactly 24.

**Step-by-step Working:**
1. Find all combinations of 3 single-digit numbers (from 1 to 9) that multiply to 24.
   - The prime factorization of 24 is $2^3 \times 3$.
   - Possible sets of digits:
     - $\{1, 3, 8\}$
     - $\{1, 4, 6\}$
     - $\{2, 2, 6\}$
     - $\{2, 3, 4\}$
   (No other combinations work. For example, we cannot use 9, 7, or 5 because 24 is not divisible by them.)
2. Count the number of permutations (arrangements) for each set to form 3-digit numbers:
   - $\{1, 3, 8\}$: 3 distinct digits $\rightarrow 3! = 6$ numbers.
   - $\{1, 4, 6\}$: 3 distinct digits $\rightarrow 3! = 6$ numbers.
   - $\{2, 3, 4\}$: 3 distinct digits $\rightarrow 3! = 6$ numbers.
   - $\{2, 2, 6\}$: 2 identical digits $\rightarrow \frac{3!}{2!} = 3$ numbers.
3. Add them up:
   $6 + 6 + 6 + 3 = 21$

**Ruling out wrong choices:**
- E (24) is tempting if one assumes all 4 sets have 6 permutations ($4 \times 6 = 24$).
- C (18) is tempting if one misses the $\{2, 2, 6\}$ set entirely.

ANSWER 5: D

Problem 6:
**Restatement:** Brent starts with 4 goldfish that quadruple every month. Gretel starts with 128 goldfish that double every month. We need to find the number of months until they have the same amount.

**Step-by-step Working:**
1. Let $m$ be the number of months.
2. Write an expression for Brent's goldfish:
   $B(m) = 4 \times 4^m$
3. Write an expression for Gretel's goldfish:
   $G(m) = 128 \times 2^m$
4. Set them equal to each other:
   $4 \times 4^m = 128 \times 2^m$
5. Express all bases as powers of 2 to easily solve for $m$:
   $4 = 2^2$
   $128 = 2^7$
   $2^2 \times (2^2)^m = 2^7 \times 2^m$
   $2^2 \times 2^{2m} = 2^{m+7}$
   $2^{2m+2} = 2^{m+7}$
6. Equate the exponents:
   $2m + 2 = m + 7$
   $m = 5$

**Ruling out wrong choices:**
- A (4) or C (6) could be obtained if one makes an arithmetic error 

Problem 1:
The problem asks us to compare the place value of the digits 9 and 3 in the number 74982.1035. 
First, we identify the positions of 9 and 3. The digit 9 is in the thousands place, and the digit 3 is in the thousandths place. 
The value of the place occupied by 9 is 1000, and the value of the place occupied by 3 is 0.001. 
To find how many times as great the value of the place occupied by 9 is compared to the value of the place occupied by 3, we calculate 1000 / 0.001. 
This is equal to 1000 * 1000 = 1,000,000. 
Let's examine the choices: A. 1,000, B. 10,000, C. 100,000, D. 1,000,000, E. 10,000,000. 
Our calculation matches choice D.

ANSWER 1: D

Problem 2:
The problem asks for the number halfway between 1/6 and 1/4. 
To find the number halfway between two numbers, we need to calculate their average. 
First, we need to find a common denominator for 1/6 and 1/4. The least common denominator is 12. 
1/6 = 2/12 and 1/4 = 3/12. 
The average is (2/12 + 3/12) / 2 = (5/12) / 2 = 5/24. 
Let's examine the choices: A. 1/10, B. 1/5, C. 5/24, D. 7/24, E. 5/12. 
Our calculation matches choice C.

ANSWER 2: C

Problem 3:
The numbers given are −2, 4, 6, 9, and 12. We need to rearrange them according to the given rules and find the average of the first and last numbers. 
First, let's arrange the numbers in ascending order: −2, 4, 6, 9, 12. 
The largest number is 12, the smallest is −2, and the median is 6. 
According to the rules: the largest isn't first but is in one of the first three places, the smallest isn't last but is in one of the last three places, and the median isn't first or last. 
Let's try to satisfy these conditions. 
Since the median (6) can't be first or last, it must be in the second, third, or fourth position. 
If 6 is in the third position, the arrangement could be: x, y, 6, z, w. 
Given that 12 is in one of the first three places and −2 is in one of the last three places, a possible arrangement is: 9, 12, 6, 4, −2 or 4, 12, 6, 9, −2. 
However, we need to check if these satisfy all conditions. 
For 9, 12, 6, 4, −2: 12 is in the second position (satisfies the first condition), and −2 is in the last position (does not satisfy the second condition because −2 is last). 
For 4, 12, 6, 9, −2: again, −2 is last, so it doesn't satisfy the condition. 
Let's consider 6 in the second position: x, 6, y, z, w. 
A possible arrangement could be: 9, 6, 12, −2, 4. 
Here, 12 is in the third position, and −2 is in the fourth, satisfying the conditions that 12 is in one of the first three places and −2 is in one of the last three. 
The first number is 9, and the last is 4. The average is (9 + 4) / 2 = 6.5. 
Let's examine the choices: A. 3.5, B. 5, C. 6.5, D. 7.5, E. 8. 
Our calculation matches choice C.

ANSWER 3: C

Problem 4:
The problem gives the equation 991 + 993 + 995 + 997 + 999 = 5000 − N and asks to find N. 
First, let's simplify the left-hand side of the equation. 
991 + 993 + 995 + 997 + 999 can be rewritten as (1000 - 9) + (1000 - 7) + (1000 - 5) + (1000 - 3) + (1000 - 1). 
This equals 5*1000 - (9 + 7 + 5 + 3 + 1) = 5000 - 25. 
So, 5000 - 25 = 5000 - N. 
This implies that N = 25. 
Let's examine the choices: A. 5, B. 10, C. 15, D. 20, E. 25. 
Our calculation matches choice E.

ANSWER 4: E

Problem 5:
The problem asks us to find how many 3-digit positive integers have digits whose product equals 24. 
To solve this, we need to list the factor combinations of 24 and see which ones can form 3-digit numbers. 
The prime factorization of 24 is 2^3 * 3. 
Possible combinations for the digits are derived from the factors: (1, 1, 24) - not valid since 24 is not a single digit; (1, 2, 12) - not valid; (1, 3, 8); (1, 4, 6); (2, 2, 6); (2, 3, 4); and (2, 2, 6). 
Let's count the permutations for valid combinations: 
- For (1, 3, 8), the permutations are 138, 183, 318, 381, 813, 831, which gives 6 numbers. 
- For (1, 4, 6), the permutations are 146, 164, 416, 461, 614, 641, which gives 6 numbers. 
- For (2, 2, 6), the permutations are 226, 262, 622, which gives 3 numbers. 
- For (2, 3, 4), the permutations are 234, 243, 324, 342, 423, 432, which gives 6 numbers. 
Adding these up: 6 + 6 + 3 + 6 = 21. 
Let's examine the choices: A. 12, B. 15, C. 18, D. 21, E. 24. 
Our calculation matches choice D.

ANSWER 5: D

Problem 6:
Brent has goldfish that quadruple every month, and Gretel has goldfish that double every month. 
Initially, Brent has 4 goldfish, and Gretel has 128 goldfish. 
Let's denote the number of months from the initial time as 'n'. 
After 'n' months, Brent will have 4 * (4^n) goldfish, and Gretel will have 128 * (2^n) goldfish. 
We need to find 'n' when 4 * (4^n) = 128 * (2^n). 
Simplifying, we get 4^(n+1) = 128 * 2^n. 
Since 4 = 2^2, we can rewrite this as (2^2)^(n+1) = 2^7 * 2^n. 
This simplifies to 2^(2n+2) = 2^(n+7). 
Equating the exponents, we get 2n + 2 = n + 7. 
Solving for 'n', we get n = 5. 
Let's examine the choices: A. 4, B. 5, C. 6, D. 7, E. 8. 
Our calculation matches choice B.

ANSWER 6: B

Problem 7:
In ▵ABC, AB = BC = 29, and AC = 42. 
Since AB = BC, ▵ABC is isosceles. 
To find the area, we need the height. 
Let's draw the altitude from B to AC, meeting AC at point D. 
Since ▵ABC is isosceles, D is the midpoint of AC. 
So, AD = DC = 42 / 2 = 21. 
Using the Pythagorean theorem in ▵ADB, we have BD^2 = AB^2 - AD^2 = 29^2 - 21^2. 
Calculating, BD^2 = 841 - 441 = 400, so BD = 20. 
The area of ▵ABC = 0.5 * base * height = 0.5 * 42 * 20 = 420. 
Let's examine the choices: A. 100, B. 420, C. 500, D. 609, E. 701. 
Our calculation matches choice B.

ANSWER 7: B

Problem 8:
The problem asks for the number of different isosceles triangles with integer side lengths and a perimeter of 23. 
For an isosceles triangle, two sides are equal. Let's denote the equal sides as 'a' and the base as 'b'. 
So, 2a + b = 23, and a + a > b (triangle inequality). 
This implies that 2a > b, or 2a > 23 - 2a, which simplifies to 4a > 23, or a > 5.75. 
Since 'a' is an integer, a ≥ 6. 
Also, a + b > a, which i

Problem 1:

We are asked about the **place** occupied by the digit 9 and the digit 3 in

\[
74982.1035
\]

The digit 9 is in the **hundreds** place, so the value of its place is

\[
100
\]

The digit 3 is in the **thousandths** place, so the value of its place is

\[
0.001=\frac{1}{1000}
\]

Now compare:

\[
100 \div 0.001 = 100 \div \frac{1}{1000}=100 \times 1000=100000
\]

So the place of the 9 is \(100{,}000\) times as great as the place of the 3.

Choices A and B are too small, and D and E are too large.

ANSWER 1: C

Problem 2:

We need the number halfway between

\[
\frac16 \quad \text{and} \quad \frac14
\]

The halfway point is the average:

\[
\frac{\frac16+\frac14}{2}
\]

Find a common denominator:

\[
\frac16=\frac{2}{12}, \qquad \frac14=\frac{3}{12}
\]

So

\[
\frac16+\frac14=\frac{2}{12}+\frac{3}{12}=\frac{5}{12}
\]

Now divide by 2:

\[
\frac{5}{12}\div 2=\frac{5}{12}\cdot \frac12=\frac{5}{24}
\]

So the number halfway between them is \(\frac{5}{24}\).

A tempting wrong answer is \(\frac15\), but that comes from averaging the denominators, which is not how fractions are averaged.

ANSWER 2: C

Problem 3:

The numbers are

\[
-2,\ 4,\ 6,\ 9,\ 12
\]

We are rearranging them using these rules:

- The largest number, \(12\), is not first, but is in one of the first three places.
- The smallest number, \(-2\), is not last, but is in one of the last three places.
- The median number, \(6\), is not first or last.

The largest number \(12\) must be in position 2 or 3.

The smallest number \(-2\) must be in position 3 or 4.

The median \(6\) must be in position 2, 3, or 4.

So the numbers \(12\), \(-2\), and \(6\) occupy the middle three positions in some order.

That means the first and last positions must be filled by the remaining two numbers:

\[
4 \quad \text{and} \quad 9
\]

Their average is

\[
\frac{4+9}{2}=\frac{13}{2}=6.5
\]

So the average of the first and last numbers is \(6.5\).

ANSWER 3: C

Problem 4:

We are given

\[
991+993+995+997+999=5000-N
\]

Add the five numbers. Since they are evenly spaced, their average is the middle number:

\[
995
\]

There are 5 numbers, so the sum is

\[
5 \cdot 995=4975
\]

Thus

\[
4975=5000-N
\]

Solve for \(N\):

\[
N=5000-4975=25
\]

So \(N=25\).

ANSWER 4: E

Problem 5:

We need to count 3-digit positive integers whose digits multiply to 24.

Since the product is 24, no digit can be 0.

Factor:

\[
24=2^3\cdot 3
\]

Now list digit triples whose product is 24:

\[
1,3,8
\]

\[
1,4,6
\]

\[
2,2,6
\]

\[
2,3,4
\]

Now count rearrangements.

For \(1,3,8\), all digits are different:

\[
3!=6
\]

For \(1,4,6\), all digits are different:

\[
3!=6
\]

For \(2,2,6\), two digits are the same, so there are

\[
\frac{3!}{2!}=3
\]

For \(2,3,4\), all digits are different:

\[
3!=6
\]

Total:

\[
6+6+3+6=21
\]

So there are 21 such integers.

ANSWER 5: D

Problem 6:

Brent starts with 4 goldfish and quadruples every month.

After \(m\) months, Brent has

\[
4\cdot 4^m
\]

Gretel starts with 128 goldfish and doubles every month.

After \(m\) months, Gretel has

\[
128\cdot 2^m
\]

We want them equal:

\[
4\cdot 4^m=128\cdot 2^m
\]

Rewrite everything as powers of 2:

\[
4=2^2,\quad 4^m=2^{2m},\quad 128=2^7
\]

So

\[
2^2\cdot 2^{2m}=2^7\cdot 2^m
\]

\[
2^{2+2m}=2^{7+m}
\]

Therefore,

\[
2+2m=7+m
\]

\[
m=5
\]

So in 5 months, they will have the same number of goldfish.

ANSWER 6: B

Problem 7:

We have an isosceles triangle with

\[
AB=BC=29
\]

and base

\[
AC=42
\]

The altitude from \(B\) to \(AC\) bisects the base, so each half is

\[
\frac{42}{2}=21
\]

Now use the Pythagorean Theorem to find the height \(h\):

\[
h^2+21^2=29^2
\]

\[
h^2+441=841
\]

\[
h^2=400
\]

\[
h=20
\]

Now find the area:

\[
\text{Area}=\frac12 \cdot \text{base} \cdot \text{height}
\]

\[
\text{Area}=\frac12 \cdot 42 \cdot 20=420
\]

So the area is 420.

ANSWER 7: B

Problem 8:

We need isosceles triangles with integer side lengths and perimeter 23.

Let the equal sides each have length \(a\), and let the base have length \(b\).

Then

\[
2a+b=23
\]

So

\[
b=23-2a
\]

For a triangle, the sum of the two equal sides must be greater than the base:

\[
2a>b
\]

Substitute \(b=23-2a\):

\[
2a>23-2a
\]

\[
4a>23
\]

So

\[
a>5.75
\]

Thus

\[
a\ge 6
\]

Also, \(b\) must be positive:

\[
23-2a>0
\]

\[
2a<23
\]

\[
a<11.5
\]

Thus

\[
a\le 11
\]

So possible values of \(a\) are

\[
6,7,8,9,10,11
\]

That gives 6 different triangles.

ANSWER 8: C

Problem 9:

We draw three digits from

\[
1,2,3,4
\]

and form a three-digit number.

A number is divisible by 3 if the sum of its digits is divisible by 3.

The sum of all four digits is

\[
1+2+3+4=10
\]

We choose three digits, which is the same as leaving out one digit.

If we leave out digit \(d\), then the chosen digit sum is

\[
10-d
\]

We need

\[
10-d
\]

to be divisible by 3.

Check the possible omitted digits:

- Omit 1: sum is \(9\), divisible by 3.
- Omit 2: sum is \(8\), not divisible by 3.
- Omit 3: sum is \(7\), not divisible by 3.
- Omit 4: sum is \(6\), divisible by 3.

So 2 of the 4 possible sets of digits work.

For each chosen set, all rearrangements are also multiples of 3, since digit sum does not change.

Thus the probability is

\[
\frac{2}{4}=\frac12
\]

ANSWER 9: C

Problem 10:

If the phone is on but not being used, the battery lasts 24 hours, so it uses

\[
\frac{1}{24}
\]

of the battery per hour.

If the phone is being used constantly, the battery lasts 3 hours, so it uses

\[
\frac{1}{3}
\]

of the battery per hour.

Since the last recharge, the phone has been on for 9 hours, and Niki used it for 60 minutes, which is 1 hour.

So the phone was:

- used for 1 hour
- on but not used for 8 hours

Battery used while talking:

\[
1\cdot \frac13=\frac13
\]

Battery used while on but idle:

\[
8\cdot \frac{1}{24}=\frac{8}{24}=\frac13
\]

Total battery used:

\[
\frac13+\frac13=\frac23
\]

Battery remaining:

\[
1-\frac23=\frac13
\]

Now she will leave the phone on but not talk, so it uses \(\frac{1}{24}\)

Problem 1:
**Restatement:** We are asked to find how many times greater the value of the place occupied by the digit 9 is compared to the value of the place occupied by the digit 3 in the number 74982.1035.

**Step-by-step Working:**
1. Identify the place value of the digit 9: In 74982.1035, the 9 is in the hundreds place. Its value is $9 \times 100$, so the place value is $100$.
2. Identify the place value of the digit 3: The 3 is in the thousandths place (three places to the right of the decimal point). Its value is $3 \times 0.001$, so the place value is $0.001$ (or $\frac{1}{1000}$).
3. Calculate the ratio: To find out how many times as great the hundreds place is compared to the thousandths place, we divide the two place values:
$$\frac{100}{0.001} = 100 \times 1000 = 100,000$$

**Ruling out wrong choices:** 
- A (1,000) and B (10,000) are tempting if you miscount the decimal places between the two digits.
- D (1,000,000) would be the answer if the 3 were in the ten-thousandths place.

ANSWER 1: C

Problem 2:
**Restatement:** We need to find the number that is exactly halfway between the fractions $\frac{1}{6}$ and $\frac{1}{4}$.

**Step-by-step Working:**
1. The number halfway between two numbers is their average. We need to calculate $\frac{\frac{1}{6} + \frac{1}{4}}{2}$.
2. Find a common denominator to add the fractions. The least common multiple of 6 and 4 is 12.
$$\frac{1}{6} = \frac{2}{12}$$
$$\frac{1}{4} = \frac{3}{12}$$
3. Add the fractions:
$$\frac{2}{12} + \frac{3}{12} = \frac{5}{12}$$
4. Divide the sum by 2 to find the average:
$$\frac{\frac{5}{12}}{2} = \frac{5}{12} \times \frac{1}{2} = \frac{5}{24}$$

**Ruling out wrong choices:**
- B ($\frac{1}{5}$) is a common mistake made by incorrectly "averaging" the denominators (halfway between 4 and 6).
- E ($\frac{5}{12}$) is the sum of the two fractions, forgetting to divide by 2.

ANSWER 2: C

Problem 3:
**Restatement:** We are given the numbers $-2, 4, 6, 9,$ and $12$. They are placed in five positions based on specific rules. We need to find the average of the numbers in the first and last positions.

**Step-by-step Working:**
1. Identify the key numbers:
   - Largest: $12$
   - Smallest: $-2$
   - Median (middle value when sorted): $6$
   - Remaining numbers: $4$ and $9$
2. Analyze the rules for the 5 positions (1, 2, 3, 4, 5):
   - The largest ($12$) isn't first, but is in the first three places. So, $12$ must be in position 2 or 3. This means $12$ is NOT in position 1 or 5.
   - The smallest ($-2$) isn't last, but is in the last three places. So, $-2$ must be in position 3 or 4. This means $-2$ is NOT in position 1 or 5.
   - The median ($6$) isn't first or last. So, $6$ is NOT in position 1 or 5.
3. Determine the first and last numbers:
   Since $12$, $-2$, and $6$ are all forbidden from being in position 1 and position 5, the only numbers left to fill the first and last spots are $4$ and $9$.
4. Calculate the average of the first and last numbers:
   $$\frac{4 + 9}{2} = \frac{13}{2} = 6.5$$

**Ruling out wrong choices:**
- If you mistakenly thought the median was part of the ends, you might get 5 (average of 4 and 6) or 7.5 (average of 6 and 9). 

ANSWER 3: C

Problem 4:
**Restatement:** We are given the equation $991 + 993 + 995 + 997 + 999 = 5000 - N$ and need to solve for $N$.

**Step-by-step Working:**
1. Notice that each number on the left side is slightly less than 1000. We can rewrite each term:
   $991 = 1000 - 9$
   $993 = 1000 - 7$
   $995 = 1000 - 5$
   $997 = 1000 - 3$
   $999 = 1000 - 1$
2. Substitute these into the sum:
   $(1000 - 9) + (1000 - 7) + (1000 - 5) + (1000 - 3) + (1000 - 1)$
3. Group the 1000s and the subtracted amounts:
   $= 5000 - (9 + 7 + 5 + 3 + 1)$
4. Add the subtracted amounts:
   $9 + 7 + 5 + 3 + 1 = 25$
5. The sum is $5000 - 25$. Therefore, $N = 25$.

**Ruling out wrong choices:**
- A (5) is tempting if you just count the number of terms being added.
- C (15) or D (20) might result from an arithmetic error when adding the odd numbers.

ANSWER 4: E

Problem 5:
**Restatement:** We need to find how many 3-digit positive integers exist such that the product of their three digits is exactly 24.

**Step-by-step Working:**
1. Find all combinations of 3 single-digit numbers (from 1 to 9) that multiply to 24.
   - The prime factorization of 24 is $2^3 \times 3$.
   - Possible sets of digits:
     - $\{1, 3, 8\}$
     - $\{1, 4, 6\}$
     - $\{2, 2, 6\}$
     - $\{2, 3, 4\}$
   (No other combinations work. For example, we cannot use 9, 7, or 5 because 24 is not divisible by them.)
2. Count the number of permutations (arrangements) for each set to form 3-digit numbers:
   - $\{1, 3, 8\}$: 3 distinct digits $\rightarrow 3! = 6$ numbers.
   - $\{1, 4, 6\}$: 3 distinct digits $\rightarrow 3! = 6$ numbers.
   - $\{2, 3, 4\}$: 3 distinct digits $\rightarrow 3! = 6$ numbers.
   - $\{2, 2, 6\}$: 2 identical digits $\rightarrow \frac{3!}{2!} = 3$ numbers.
3. Add them up:
   $6 + 6 + 6 + 3 = 21$

**Ruling out wrong choices:**
- E (24) is tempting if one assumes all 4 sets have 6 permutations ($4 \times 6 = 24$).
- C (18) is tempting if one misses the $\{2, 2, 6\}$ set entirely.

ANSWER 5: D

Problem 6:
**Restatement:** Brent starts with 4 goldfish that quadruple every month. Gretel starts with 128 goldfish that double every month. We need to find the number of months until they have the same amount.

**Step-by-step Working:**
1. Let $m$ be the number of months.
2. Write an expression for Brent's goldfish:
   $B(m) = 4 \times 4^m$
3. Write an expression for Gretel's goldfish:
   $G(m) = 128 \times 2^m$
4. Set them equal to each other:
   $4 \times 4^m = 128 \times 2^m$
5. Express all bases as powers of 2 to easily solve for $m$:
   $4 = 2^2$
   $128 = 2^7$
   $2^2 \times (2^2)^m = 2^7 \times 2^m$
   $2^2 \times 2^{2m} = 2^{m+7}$
   $2^{2m+2} = 2^{m+7}$
6. Equate the exponents:
   $2m + 2 = m + 7$
   $m = 5$

**Ruling out wrong choices:**
- A (4) or C (6) could be obtained if one makes an arithmetic error 

Problem 1:
The problem asks us to compare the place value of the digits 9 and 3 in the number 74982.1035. 
First, we identify the positions of 9 and 3. The digit 9 is in the thousands place, and the digit 3 is in the thousandths place. 
The value of the place occupied by 9 is 1000, and the value of the place occupied by 3 is 0.001. 
To find how many times as great the value of the place occupied by 9 is compared to the value of the place occupied by 3, we calculate 1000 / 0.001. 
This is equal to 1000 * 1000 = 1,000,000. 
Let's examine the choices: A. 1,000, B. 10,000, C. 100,000, D. 1,000,000, E. 10,000,000. 
Our calculation matches choice D.

ANSWER 1: D

Problem 2:
The problem asks for the number halfway between 1/6 and 1/4. 
To find the number halfway between two numbers, we need to calculate their average. 
First, we need to find a common denominator for 1/6 and 1/4. The least common denominator is 12. 
1/6 = 2/12 and 1/4 = 3/12. 
The average is (2/12 + 3/12) / 2 = (5/12) / 2 = 5/24. 
Let's examine the choices: A. 1/10, B. 1/5, C. 5/24, D. 7/24, E. 5/12. 
Our calculation matches choice C.

ANSWER 2: C

Problem 3:
The numbers given are −2, 4, 6, 9, and 12. We need to rearrange them according to the given rules and find the average of the first and last numbers. 
First, let's arrange the numbers in ascending order: −2, 4, 6, 9, 12. 
The largest number is 12, the smallest is −2, and the median is 6. 
According to the rules: the largest isn't first but is in one of the first three places, the smallest isn't last but is in one of the last three places, and the median isn't first or last. 
Let's try to satisfy these conditions. 
Since the median (6) can't be first or last, it must be in the second, third, or fourth position. 
If 6 is in the third position, the arrangement could be: x, y, 6, z, w. 
Given that 12 is in one of the first three places and −2 is in one of the last three places, a possible arrangement is: 9, 12, 6, 4, −2 or 4, 12, 6, 9, −2. 
However, we need to check if these satisfy all conditions. 
For 9, 12, 6, 4, −2: 12 is in the second position (satisfies the first condition), and −2 is in the last position (does not satisfy the second condition because −2 is last). 
For 4, 12, 6, 9, −2: again, −2 is last, so it doesn't satisfy the condition. 
Let's consider 6 in the second position: x, 6, y, z, w. 
A possible arrangement could be: 9, 6, 12, −2, 4. 
Here, 12 is in the third position, and −2 is in the fourth, satisfying the conditions that 12 is in one of the first three places and −2 is in one of the last three. 
The first number is 9, and the last is 4. The average is (9 + 4) / 2 = 6.5. 
Let's examine the choices: A. 3.5, B. 5, C. 6.5, D. 7.5, E. 8. 
Our calculation matches choice C.

ANSWER 3: C

Problem 4:
The problem gives the equation 991 + 993 + 995 + 997 + 999 = 5000 − N and asks to find N. 
First, let's simplify the left-hand side of the equation. 
991 + 993 + 995 + 997 + 999 can be rewritten as (1000 - 9) + (1000 - 7) + (1000 - 5) + (1000 - 3) + (1000 - 1). 
This equals 5*1000 - (9 + 7 + 5 + 3 + 1) = 5000 - 25. 
So, 5000 - 25 = 5000 - N. 
This implies that N = 25. 
Let's examine the choices: A. 5, B. 10, C. 15, D. 20, E. 25. 
Our calculation matches choice E.

ANSWER 4: E

Problem 5:
The problem asks us to find how many 3-digit positive integers have digits whose product equals 24. 
To solve this, we need to list the factor combinations of 24 and see which ones can form 3-digit numbers. 
The prime factorization of 24 is 2^3 * 3. 
Possible combinations for the digits are derived from the factors: (1, 1, 24) - not valid since 24 is not a single digit; (1, 2, 12) - not valid; (1, 3, 8); (1, 4, 6); (2, 2, 6); (2, 3, 4); and (2, 2, 6). 
Let's count the permutations for valid combinations: 
- For (1, 3, 8), the permutations are 138, 183, 318, 381, 813, 831, which gives 6 numbers. 
- For (1, 4, 6), the permutations are 146, 164, 416, 461, 614, 641, which gives 6 numbers. 
- For (2, 2, 6), the permutations are 226, 262, 622, which gives 3 numbers. 
- For (2, 3, 4), the permutations are 234, 243, 324, 342, 423, 432, which gives 6 numbers. 
Adding these up: 6 + 6 + 3 + 6 = 21. 
Let's examine the choices: A. 12, B. 15, C. 18, D. 21, E. 24. 
Our calculation matches choice D.

ANSWER 5: D

Problem 6:
Brent has goldfish that quadruple every month, and Gretel has goldfish that double every month. 
Initially, Brent has 4 goldfish, and Gretel has 128 goldfish. 
Let's denote the number of months from the initial time as 'n'. 
After 'n' months, Brent will have 4 * (4^n) goldfish, and Gretel will have 128 * (2^n) goldfish. 
We need to find 'n' when 4 * (4^n) = 128 * (2^n). 
Simplifying, we get 4^(n+1) = 128 * 2^n. 
Since 4 = 2^2, we can rewrite this as (2^2)^(n+1) = 2^7 * 2^n. 
This simplifies to 2^(2n+2) = 2^(n+7). 
Equating the exponents, we get 2n + 2 = n + 7. 
Solving for 'n', we get n = 5. 
Let's examine the choices: A. 4, B. 5, C. 6, D. 7, E. 8. 
Our calculation matches choice B.

ANSWER 6: B

Problem 7:
In ▵ABC, AB = BC = 29, and AC = 42. 
Since AB = BC, ▵ABC is isosceles. 
To find the area, we need the height. 
Let's draw the altitude from B to AC, meeting AC at point D. 
Since ▵ABC is isosceles, D is the midpoint of AC. 
So, AD = DC = 42 / 2 = 21. 
Using the Pythagorean theorem in ▵ADB, we have BD^2 = AB^2 - AD^2 = 29^2 - 21^2. 
Calculating, BD^2 = 841 - 441 = 400, so BD = 20. 
The area of ▵ABC = 0.5 * base * height = 0.5 * 42 * 20 = 420. 
Let's examine the choices: A. 100, B. 420, C. 500, D. 609, E. 701. 
Our calculation matches choice B.

ANSWER 7: B

Problem 8:
The problem asks for the number of different isosceles triangles with integer side lengths and a perimeter of 23. 
For an isosceles triangle, two sides are equal. Let's denote the equal sides as 'a' and the base as 'b'. 
So, 2a + b = 23, and a + a > b (triangle inequality). 
This implies that 2a > b, or 2a > 23 - 2a, which simplifies to 4a > 23, or a > 5.75. 
Since 'a' is an integer, a ≥ 6. 
Also, a + b > a, which i

Problem 1:

We are asked about the **place** occupied by the digit 9 and the digit 3 in

\[
74982.1035
\]

The digit 9 is in the **hundreds** place, so the value of its place is

\[
100
\]

The digit 3 is in the **thousandths** place, so the value of its place is

\[
0.001=\frac{1}{1000}
\]

Now compare:

\[
100 \div 0.001 = 100 \div \frac{1}{1000}=100 \times 1000=100000
\]

So the place of the 9 is \(100{,}000\) times as great as the place of the 3.

Choices A and B are too small, and D and E are too large.

ANSWER 1: C

Problem 2:

We need the number halfway between

\[
\frac16 \quad \text{and} \quad \frac14
\]

The halfway point is the average:

\[
\frac{\frac16+\frac14}{2}
\]

Find a common denominator:

\[
\frac16=\frac{2}{12}, \qquad \frac14=\frac{3}{12}
\]

So

\[
\frac16+\frac14=\frac{2}{12}+\frac{3}{12}=\frac{5}{12}
\]

Now divide by 2:

\[
\frac{5}{12}\div 2=\frac{5}{12}\cdot \frac12=\frac{5}{24}
\]

So the number halfway between them is \(\frac{5}{24}\).

A tempting wrong answer is \(\frac15\), but that comes from averaging the denominators, which is not how fractions are averaged.

ANSWER 2: C

Problem 3:

The numbers are

\[
-2,\ 4,\ 6,\ 9,\ 12
\]

We are rearranging them using these rules:

- The largest number, \(12\), is not first, but is in one of the first three places.
- The smallest number, \(-2\), is not last, but is in one of the last three places.
- The median number, \(6\), is not first or last.

The largest number \(12\) must be in position 2 or 3.

The smallest number \(-2\) must be in position 3 or 4.

The median \(6\) must be in position 2, 3, or 4.

So the numbers \(12\), \(-2\), and \(6\) occupy the middle three positions in some order.

That means the first and last positions must be filled by the remaining two numbers:

\[
4 \quad \text{and} \quad 9
\]

Their average is

\[
\frac{4+9}{2}=\frac{13}{2}=6.5
\]

So the average of the first and last numbers is \(6.5\).

ANSWER 3: C

Problem 4:

We are given

\[
991+993+995+997+999=5000-N
\]

Add the five numbers. Since they are evenly spaced, their average is the middle number:

\[
995
\]

There are 5 numbers, so the sum is

\[
5 \cdot 995=4975
\]

Thus

\[
4975=5000-N
\]

Solve for \(N\):

\[
N=5000-4975=25
\]

So \(N=25\).

ANSWER 4: E

Problem 5:

We need to count 3-digit positive integers whose digits multiply to 24.

Since the product is 24, no digit can be 0.

Factor:

\[
24=2^3\cdot 3
\]

Now list digit triples whose product is 24:

\[
1,3,8
\]

\[
1,4,6
\]

\[
2,2,6
\]

\[
2,3,4
\]

Now count rearrangements.

For \(1,3,8\), all digits are different:

\[
3!=6
\]

For \(1,4,6\), all digits are different:

\[
3!=6
\]

For \(2,2,6\), two digits are the same, so there are

\[
\frac{3!}{2!}=3
\]

For \(2,3,4\), all digits are different:

\[
3!=6
\]

Total:

\[
6+6+3+6=21
\]

So there are 21 such integers.

ANSWER 5: D

Problem 6:

Brent starts with 4 goldfish and quadruples every month.

After \(m\) months, Brent has

\[
4\cdot 4^m
\]

Gretel starts with 128 goldfish and doubles every month.

After \(m\) months, Gretel has

\[
128\cdot 2^m
\]

We want them equal:

\[
4\cdot 4^m=128\cdot 2^m
\]

Rewrite everything as powers of 2:

\[
4=2^2,\quad 4^m=2^{2m},\quad 128=2^7
\]

So

\[
2^2\cdot 2^{2m}=2^7\cdot 2^m
\]

\[
2^{2+2m}=2^{7+m}
\]

Therefore,

\[
2+2m=7+m
\]

\[
m=5
\]

So in 5 months, they will have the same number of goldfish.

ANSWER 6: B

Problem 7:

We have an isosceles triangle with

\[
AB=BC=29
\]

and base

\[
AC=42
\]

The altitude from \(B\) to \(AC\) bisects the base, so each half is

\[
\frac{42}{2}=21
\]

Now use the Pythagorean Theorem to find the height \(h\):

\[
h^2+21^2=29^2
\]

\[
h^2+441=841
\]

\[
h^2=400
\]

\[
h=20
\]

Now find the area:

\[
\text{Area}=\frac12 \cdot \text{base} \cdot \text{height}
\]

\[
\text{Area}=\frac12 \cdot 42 \cdot 20=420
\]

So the area is 420.

ANSWER 7: B

Problem 8:

We need isosceles triangles with integer side lengths and perimeter 23.

Let the equal sides each have length \(a\), and let the base have length \(b\).

Then

\[
2a+b=23
\]

So

\[
b=23-2a
\]

For a triangle, the sum of the two equal sides must be greater than the base:

\[
2a>b
\]

Substitute \(b=23-2a\):

\[
2a>23-2a
\]

\[
4a>23
\]

So

\[
a>5.75
\]

Thus

\[
a\ge 6
\]

Also, \(b\) must be positive:

\[
23-2a>0
\]

\[
2a<23
\]

\[
a<11.5
\]

Thus

\[
a\le 11
\]

So possible values of \(a\) are

\[
6,7,8,9,10,11
\]

That gives 6 different triangles.

ANSWER 8: C

Problem 9:

We draw three digits from

\[
1,2,3,4
\]

and form a three-digit number.

A number is divisible by 3 if the sum of its digits is divisible by 3.

The sum of all four digits is

\[
1+2+3+4=10
\]

We choose three digits, which is the same as leaving out one digit.

If we leave out digit \(d\), then the chosen digit sum is

\[
10-d
\]

We need

\[
10-d
\]

to be divisible by 3.

Check the possible omitted digits:

- Omit 1: sum is \(9\), divisible by 3.
- Omit 2: sum is \(8\), not divisible by 3.
- Omit 3: sum is \(7\), not divisible by 3.
- Omit 4: sum is \(6\), divisible by 3.

So 2 of the 4 possible sets of digits work.

For each chosen set, all rearrangements are also multiples of 3, since digit sum does not change.

Thus the probability is

\[
\frac{2}{4}=\frac12
\]

ANSWER 9: C

Problem 10:

If the phone is on but not being used, the battery lasts 24 hours, so it uses

\[
\frac{1}{24}
\]

of the battery per hour.

If the phone is being used constantly, the battery lasts 3 hours, so it uses

\[
\frac{1}{3}
\]

of the battery per hour.

Since the last recharge, the phone has been on for 9 hours, and Niki used it for 60 minutes, which is 1 hour.

So the phone was:

- used for 1 hour
- on but not used for 8 hours

Battery used while talking:

\[
1\cdot \frac13=\frac13
\]

Battery used while on but idle:

\[
8\cdot \frac{1}{24}=\frac{8}{24}=\frac13
\]

Total battery used:

\[
\frac13+\frac13=\frac23
\]

Battery remaining:

\[
1-\frac23=\frac13
\]

Now she will leave the phone on but not talk, so it uses \(\frac{1}{24}\)

Problem 1:
**Restatement:** We are asked to find how many times greater the value of the place occupied by the digit 9 is compared to the value of the place occupied by the digit 3 in the number 74982.1035.

**Step-by-step Working:**
1. Identify the place value of the digit 9: In 74982.1035, the 9 is in the hundreds place. Its value is $9 \times 100$, so the place value is $100$.
2. Identify the place value of the digit 3: The 3 is in the thousandths place (three places to the right of the decimal point). Its value is $3 \times 0.001$, so the place value is $0.001$ (or $\frac{1}{1000}$).
3. Calculate the ratio: To find out how many times as great the hundreds place is compared to the thousandths place, we divide the two place values:
$$\frac{100}{0.001} = 100 \times 1000 = 100,000$$

**Ruling out wrong choices:** 
- A (1,000) and B (10,000) are tempting if you miscount the decimal places between the two digits.
- D (1,000,000) would be the answer if the 3 were in the ten-thousandths place.

ANSWER 1: C

Problem 2:
**Restatement:** We need to find the number that is exactly halfway between the fractions $\frac{1}{6}$ and $\frac{1}{4}$.

**Step-by-step Working:**
1. The number halfway between two numbers is their average. We need to calculate $\frac{\frac{1}{6} + \frac{1}{4}}{2}$.
2. Find a common denominator to add the fractions. The least common multiple of 6 and 4 is 12.
$$\frac{1}{6} = \frac{2}{12}$$
$$\frac{1}{4} = \frac{3}{12}$$
3. Add the fractions:
$$\frac{2}{12} + \frac{3}{12} = \frac{5}{12}$$
4. Divide the sum by 2 to find the average:
$$\frac{\frac{5}{12}}{2} = \frac{5}{12} \times \frac{1}{2} = \frac{5}{24}$$

**Ruling out wrong choices:**
- B ($\frac{1}{5}$) is a common mistake made by incorrectly "averaging" the denominators (halfway between 4 and 6).
- E ($\frac{5}{12}$) is the sum of the two fractions, forgetting to divide by 2.

ANSWER 2: C

Problem 3:
**Restatement:** We are given the numbers $-2, 4, 6, 9,$ and $12$. They are placed in five positions based on specific rules. We need to find the average of the numbers in the first and last positions.

**Step-by-step Working:**
1. Identify the key numbers:
   - Largest: $12$
   - Smallest: $-2$
   - Median (middle value when sorted): $6$
   - Remaining numbers: $4$ and $9$
2. Analyze the rules for the 5 positions (1, 2, 3, 4, 5):
   - The largest ($12$) isn't first, but is in the first three places. So, $12$ must be in position 2 or 3. This means $12$ is NOT in position 1 or 5.
   - The smallest ($-2$) isn't last, but is in the last three places. So, $-2$ must be in position 3 or 4. This means $-2$ is NOT in position 1 or 5.
   - The median ($6$) isn't first or last. So, $6$ is NOT in position 1 or 5.
3. Determine the first and last numbers:
   Since $12$, $-2$, and $6$ are all forbidden from being in position 1 and position 5, the only numbers left to fill the first and last spots are $4$ and $9$.
4. Calculate the average of the first and last numbers:
   $$\frac{4 + 9}{2} = \frac{13}{2} = 6.5$$

**Ruling out wrong choices:**
- If you mistakenly thought the median was part of the ends, you might get 5 (average of 4 and 6) or 7.5 (average of 6 and 9). 

ANSWER 3: C

Problem 4:
**Restatement:** We are given the equation $991 + 993 + 995 + 997 + 999 = 5000 - N$ and need to solve for $N$.

**Step-by-step Working:**
1. Notice that each number on the left side is slightly less than 1000. We can rewrite each term:
   $991 = 1000 - 9$
   $993 = 1000 - 7$
   $995 = 1000 - 5$
   $997 = 1000 - 3$
   $999 = 1000 - 1$
2. Substitute these into the sum:
   $(1000 - 9) + (1000 - 7) + (1000 - 5) + (1000 - 3) + (1000 - 1)$
3. Group the 1000s and the subtracted amounts:
   $= 5000 - (9 + 7 + 5 + 3 + 1)$
4. Add the subtracted amounts:
   $9 + 7 + 5 + 3 + 1 = 25$
5. The sum is $5000 - 25$. Therefore, $N = 25$.

**Ruling out wrong choices:**
- A (5) is tempting if you just count the number of terms being added.
- C (15) or D (20) might result from an arithmetic error when adding the odd numbers.

ANSWER 4: E

Problem 5:
**Restatement:** We need to find how many 3-digit positive integers exist such that the product of their three digits is exactly 24.

**Step-by-step Working:**
1. Find all combinations of 3 single-digit numbers (from 1 to 9) that multiply to 24.
   - The prime factorization of 24 is $2^3 \times 3$.
   - Possible sets of digits:
     - $\{1, 3, 8\}$
     - $\{1, 4, 6\}$
     - $\{2, 2, 6\}$
     - $\{2, 3, 4\}$
   (No other combinations work. For example, we cannot use 9, 7, or 5 because 24 is not divisible by them.)
2. Count the number of permutations (arrangements) for each set to form 3-digit numbers:
   - $\{1, 3, 8\}$: 3 distinct digits $\rightarrow 3! = 6$ numbers.
   - $\{1, 4, 6\}$: 3 distinct digits $\rightarrow 3! = 6$ numbers.
   - $\{2, 3, 4\}$: 3 distinct digits $\rightarrow 3! = 6$ numbers.
   - $\{2, 2, 6\}$: 2 identical digits $\rightarrow \frac{3!}{2!} = 3$ numbers.
3. Add them up:
   $6 + 6 + 6 + 3 = 21$

**Ruling out wrong choices:**
- E (24) is tempting if one assumes all 4 sets have 6 permutations ($4 \times 6 = 24$).
- C (18) is tempting if one misses the $\{2, 2, 6\}$ set entirely.

ANSWER 5: D

Problem 6:
**Restatement:** Brent starts with 4 goldfish that quadruple every month. Gretel starts with 128 goldfish that double every month. We need to find the number of months until they have the same amount.

**Step-by-step Working:**
1. Let $m$ be the number of months.
2. Write an expression for Brent's goldfish:
   $B(m) = 4 \times 4^m$
3. Write an expression for Gretel's goldfish:
   $G(m) = 128 \times 2^m$
4. Set them equal to each other:
   $4 \times 4^m = 128 \times 2^m$
5. Express all bases as powers of 2 to easily solve for $m$:
   $4 = 2^2$
   $128 = 2^7$
   $2^2 \times (2^2)^m = 2^7 \times 2^m$
   $2^2 \times 2^{2m} = 2^{m+7}$
   $2^{2m+2} = 2^{m+7}$
6. Equate the exponents:
   $2m + 2 = m + 7$
   $m = 5$

**Ruling out wrong choices:**
- A (4) or C (6) could be obtained if one makes an arithmetic error 

Problem 1:
The problem asks us to compare the place value of the digits 9 and 3 in the number 74982.1035. 
First, we identify the positions of 9 and 3. The digit 9 is in the thousands place, and the digit 3 is in the thousandths place. 
The value of the place occupied by 9 is 1000, and the value of the place occupied by 3 is 0.001. 
To find how many times as great the value of the place occupied by 9 is compared to the value of the place occupied by 3, we calculate 1000 / 0.001. 
This is equal to 1000 * 1000 = 1,000,000. 
Let's examine the choices: A. 1,000, B. 10,000, C. 100,000, D. 1,000,000, E. 10,000,000. 
Our calculation matches choice D.

ANSWER 1: D

Problem 2:
The problem asks for the number halfway between 1/6 and 1/4. 
To find the number halfway between two numbers, we need to calculate their average. 
First, we need to find a common denominator for 1/6 and 1/4. The least common denominator is 12. 
1/6 = 2/12 and 1/4 = 3/12. 
The average is (2/12 + 3/12) / 2 = (5/12) / 2 = 5/24. 
Let's examine the choices: A. 1/10, B. 1/5, C. 5/24, D. 7/24, E. 5/12. 
Our calculation matches choice C.

ANSWER 2: C

Problem 3:
The numbers given are −2, 4, 6, 9, and 12. We need to rearrange them according to the given rules and find the average of the first and last numbers. 
First, let's arrange the numbers in ascending order: −2, 4, 6, 9, 12. 
The largest number is 12, the smallest is −2, and the median is 6. 
According to the rules: the largest isn't first but is in one of the first three places, the smallest isn't last but is in one of the last three places, and the median isn't first or last. 
Let's try to satisfy these conditions. 
Since the median (6) can't be first or last, it must be in the second, third, or fourth position. 
If 6 is in the third position, the arrangement could be: x, y, 6, z, w. 
Given that 12 is in one of the first three places and −2 is in one of the last three places, a possible arrangement is: 9, 12, 6, 4, −2 or 4, 12, 6, 9, −2. 
However, we need to check if these satisfy all conditions. 
For 9, 12, 6, 4, −2: 12 is in the second position (satisfies the first condition), and −2 is in the last position (does not satisfy the second condition because −2 is last). 
For 4, 12, 6, 9, −2: again, −2 is last, so it doesn't satisfy the condition. 
Let's consider 6 in the second position: x, 6, y, z, w. 
A possible arrangement could be: 9, 6, 12, −2, 4. 
Here, 12 is in the third position, and −2 is in the fourth, satisfying the conditions that 12 is in one of the first three places and −2 is in one of the last three. 
The first number is 9, and the last is 4. The average is (9 + 4) / 2 = 6.5. 
Let's examine the choices: A. 3.5, B. 5, C. 6.5, D. 7.5, E. 8. 
Our calculation matches choice C.

ANSWER 3: C

Problem 4:
The problem gives the equation 991 + 993 + 995 + 997 + 999 = 5000 − N and asks to find N. 
First, let's simplify the left-hand side of the equation. 
991 + 993 + 995 + 997 + 999 can be rewritten as (1000 - 9) + (1000 - 7) + (1000 - 5) + (1000 - 3) + (1000 - 1). 
This equals 5*1000 - (9 + 7 + 5 + 3 + 1) = 5000 - 25. 
So, 5000 - 25 = 5000 - N. 
This implies that N = 25. 
Let's examine the choices: A. 5, B. 10, C. 15, D. 20, E. 25. 
Our calculation matches choice E.

ANSWER 4: E

Problem 5:
The problem asks us to find how many 3-digit positive integers have digits whose product equals 24. 
To solve this, we need to list the factor combinations of 24 and see which ones can form 3-digit numbers. 
The prime factorization of 24 is 2^3 * 3. 
Possible combinations for the digits are derived from the factors: (1, 1, 24) - not valid since 24 is not a single digit; (1, 2, 12) - not valid; (1, 3, 8); (1, 4, 6); (2, 2, 6); (2, 3, 4); and (2, 2, 6). 
Let's count the permutations for valid combinations: 
- For (1, 3, 8), the permutations are 138, 183, 318, 381, 813, 831, which gives 6 numbers. 
- For (1, 4, 6), the permutations are 146, 164, 416, 461, 614, 641, which gives 6 numbers. 
- For (2, 2, 6), the permutations are 226, 262, 622, which gives 3 numbers. 
- For (2, 3, 4), the permutations are 234, 243, 324, 342, 423, 432, which gives 6 numbers. 
Adding these up: 6 + 6 + 3 + 6 = 21. 
Let's examine the choices: A. 12, B. 15, C. 18, D. 21, E. 24. 
Our calculation matches choice D.

ANSWER 5: D

Problem 6:
Brent has goldfish that quadruple every month, and Gretel has goldfish that double every month. 
Initially, Brent has 4 goldfish, and Gretel has 128 goldfish. 
Let's denote the number of months from the initial time as 'n'. 
After 'n' months, Brent will have 4 * (4^n) goldfish, and Gretel will have 128 * (2^n) goldfish. 
We need to find 'n' when 4 * (4^n) = 128 * (2^n). 
Simplifying, we get 4^(n+1) = 128 * 2^n. 
Since 4 = 2^2, we can rewrite this as (2^2)^(n+1) = 2^7 * 2^n. 
This simplifies to 2^(2n+2) = 2^(n+7). 
Equating the exponents, we get 2n + 2 = n + 7. 
Solving for 'n', we get n = 5. 
Let's examine the choices: A. 4, B. 5, C. 6, D. 7, E. 8. 
Our calculation matches choice B.

ANSWER 6: B

Problem 7:
In ▵ABC, AB = BC = 29, and AC = 42. 
Since AB = BC, ▵ABC is isosceles. 
To find the area, we need the height. 
Let's draw the altitude from B to AC, meeting AC at point D. 
Since ▵ABC is isosceles, D is the midpoint of AC. 
So, AD = DC = 42 / 2 = 21. 
Using the Pythagorean theorem in ▵ADB, we have BD^2 = AB^2 - AD^2 = 29^2 - 21^2. 
Calculating, BD^2 = 841 - 441 = 400, so BD = 20. 
The area of ▵ABC = 0.5 * base * height = 0.5 * 42 * 20 = 420. 
Let's examine the choices: A. 100, B. 420, C. 500, D. 609, E. 701. 
Our calculation matches choice B.

ANSWER 7: B

Problem 8:
The problem asks for the number of different isosceles triangles with integer side lengths and a perimeter of 23. 
For an isosceles triangle, two sides are equal. Let's denote the equal sides as 'a' and the base as 'b'. 
So, 2a + b = 23, and a + a > b (triangle inequality). 
This implies that 2a > b, or 2a > 23 - 2a, which simplifies to 4a > 23, or a > 5.75. 
Since 'a' is an integer, a ≥ 6. 
Also, a + b > a, which i

Problem 1:

We are asked about the **place** occupied by the digit 9 and the digit 3 in

\[
74982.1035
\]

The digit 9 is in the **hundreds** place, so the value of its place is

\[
100
\]

The digit 3 is in the **thousandths** place, so the value of its place is

\[
0.001=\frac{1}{1000}
\]

Now compare:

\[
100 \div 0.001 = 100 \div \frac{1}{1000}=100 \times 1000=100000
\]

So the place of the 9 is \(100{,}000\) times as great as the place of the 3.

Choices A and B are too small, and D and E are too large.

ANSWER 1: C

Problem 2:

We need the number halfway between

\[
\frac16 \quad \text{and} \quad \frac14
\]

The halfway point is the average:

\[
\frac{\frac16+\frac14}{2}
\]

Find a common denominator:

\[
\frac16=\frac{2}{12}, \qquad \frac14=\frac{3}{12}
\]

So

\[
\frac16+\frac14=\frac{2}{12}+\frac{3}{12}=\frac{5}{12}
\]

Now divide by 2:

\[
\frac{5}{12}\div 2=\frac{5}{12}\cdot \frac12=\frac{5}{24}
\]

So the number halfway between them is \(\frac{5}{24}\).

A tempting wrong answer is \(\frac15\), but that comes from averaging the denominators, which is not how fractions are averaged.

ANSWER 2: C

Problem 3:

The numbers are

\[
-2,\ 4,\ 6,\ 9,\ 12
\]

We are rearranging them using these rules:

- The largest number, \(12\), is not first, but is in one of the first three places.
- The smallest number, \(-2\), is not last, but is in one of the last three places.
- The median number, \(6\), is not first or last.

The largest number \(12\) must be in position 2 or 3.

The smallest number \(-2\) must be in position 3 or 4.

The median \(6\) must be in position 2, 3, or 4.

So the numbers \(12\), \(-2\), and \(6\) occupy the middle three positions in some order.

That means the first and last positions must be filled by the remaining two numbers:

\[
4 \quad \text{and} \quad 9
\]

Their average is

\[
\frac{4+9}{2}=\frac{13}{2}=6.5
\]

So the average of the first and last numbers is \(6.5\).

ANSWER 3: C

Problem 4:

We are given

\[
991+993+995+997+999=5000-N
\]

Add the five numbers. Since they are evenly spaced, their average is the middle number:

\[
995
\]

There are 5 numbers, so the sum is

\[
5 \cdot 995=4975
\]

Thus

\[
4975=5000-N
\]

Solve for \(N\):

\[
N=5000-4975=25
\]

So \(N=25\).

ANSWER 4: E

Problem 5:

We need to count 3-digit positive integers whose digits multiply to 24.

Since the product is 24, no digit can be 0.

Factor:

\[
24=2^3\cdot 3
\]

Now list digit triples whose product is 24:

\[
1,3,8
\]

\[
1,4,6
\]

\[
2,2,6
\]

\[
2,3,4
\]

Now count rearrangements.

For \(1,3,8\), all digits are different:

\[
3!=6
\]

For \(1,4,6\), all digits are different:

\[
3!=6
\]

For \(2,2,6\), two digits are the same, so there are

\[
\frac{3!}{2!}=3
\]

For \(2,3,4\), all digits are different:

\[
3!=6
\]

Total:

\[
6+6+3+6=21
\]

So there are 21 such integers.

ANSWER 5: D

Problem 6:

Brent starts with 4 goldfish and quadruples every month.

After \(m\) months, Brent has

\[
4\cdot 4^m
\]

Gretel starts with 128 goldfish and doubles every month.

After \(m\) months, Gretel has

\[
128\cdot 2^m
\]

We want them equal:

\[
4\cdot 4^m=128\cdot 2^m
\]

Rewrite everything as powers of 2:

\[
4=2^2,\quad 4^m=2^{2m},\quad 128=2^7
\]

So

\[
2^2\cdot 2^{2m}=2^7\cdot 2^m
\]

\[
2^{2+2m}=2^{7+m}
\]

Therefore,

\[
2+2m=7+m
\]

\[
m=5
\]

So in 5 months, they will have the same number of goldfish.

ANSWER 6: B

Problem 7:

We have an isosceles triangle with

\[
AB=BC=29
\]

and base

\[
AC=42
\]

The altitude from \(B\) to \(AC\) bisects the base, so each half is

\[
\frac{42}{2}=21
\]

Now use the Pythagorean Theorem to find the height \(h\):

\[
h^2+21^2=29^2
\]

\[
h^2+441=841
\]

\[
h^2=400
\]

\[
h=20
\]

Now find the area:

\[
\text{Area}=\frac12 \cdot \text{base} \cdot \text{height}
\]

\[
\text{Area}=\frac12 \cdot 42 \cdot 20=420
\]

So the area is 420.

ANSWER 7: B

Problem 8:

We need isosceles triangles with integer side lengths and perimeter 23.

Let the equal sides each have length \(a\), and let the base have length \(b\).

Then

\[
2a+b=23
\]

So

\[
b=23-2a
\]

For a triangle, the sum of the two equal sides must be greater than the base:

\[
2a>b
\]

Substitute \(b=23-2a\):

\[
2a>23-2a
\]

\[
4a>23
\]

So

\[
a>5.75
\]

Thus

\[
a\ge 6
\]

Also, \(b\) must be positive:

\[
23-2a>0
\]

\[
2a<23
\]

\[
a<11.5
\]

Thus

\[
a\le 11
\]

So possible values of \(a\) are

\[
6,7,8,9,10,11
\]

That gives 6 different triangles.

ANSWER 8: C

Problem 9:

We draw three digits from

\[
1,2,3,4
\]

and form a three-digit number.

A number is divisible by 3 if the sum of its digits is divisible by 3.

The sum of all four digits is

\[
1+2+3+4=10
\]

We choose three digits, which is the same as leaving out one digit.

If we leave out digit \(d\), then the chosen digit sum is

\[
10-d
\]

We need

\[
10-d
\]

to be divisible by 3.

Check the possible omitted digits:

- Omit 1: sum is \(9\), divisible by 3.
- Omit 2: sum is \(8\), not divisible by 3.
- Omit 3: sum is \(7\), not divisible by 3.
- Omit 4: sum is \(6\), divisible by 3.

So 2 of the 4 possible sets of digits work.

For each chosen set, all rearrangements are also multiples of 3, since digit sum does not change.

Thus the probability is

\[
\frac{2}{4}=\frac12
\]

ANSWER 9: C

Problem 10:

If the phone is on but not being used, the battery lasts 24 hours, so it uses

\[
\frac{1}{24}
\]

of the battery per hour.

If the phone is being used constantly, the battery lasts 3 hours, so it uses

\[
\frac{1}{3}
\]

of the battery per hour.

Since the last recharge, the phone has been on for 9 hours, and Niki used it for 60 minutes, which is 1 hour.

So the phone was:

- used for 1 hour
- on but not used for 8 hours

Battery used while talking:

\[
1\cdot \frac13=\frac13
\]

Battery used while on but idle:

\[
8\cdot \frac{1}{24}=\frac{8}{24}=\frac13
\]

Total battery used:

\[
\frac13+\frac13=\frac23
\]

Battery remaining:

\[
1-\frac23=\frac13
\]

Now she will leave the phone on but not talk, so it uses \(\frac{1}{24}\)

Problem 1:
**Restatement:** We are asked to find how many times greater the value of the place occupied by the digit 9 is compared to the value of the place occupied by the digit 3 in the number 74982.1035.

**Step-by-step Working:**
1. Identify the place value of the digit 9: In 74982.1035, the 9 is in the hundreds place. Its value is $9 \times 100$, so the place value is $100$.
2. Identify the place value of the digit 3: The 3 is in the thousandths place (three places to the right of the decimal point). Its value is $3 \times 0.001$, so the place value is $0.001$ (or $\frac{1}{1000}$).
3. Calculate the ratio: To find out how many times as great the hundreds place is compared to the thousandths place, we divide the two place values:
$$\frac{100}{0.001} = 100 \times 1000 = 100,000$$

**Ruling out wrong choices:** 
- A (1,000) and B (10,000) are tempting if you miscount the decimal places between the two digits.
- D (1,000,000) would be the answer if the 3 were in the ten-thousandths place.

ANSWER 1: C

Problem 2:
**Restatement:** We need to find the number that is exactly halfway between the fractions $\frac{1}{6}$ and $\frac{1}{4}$.

**Step-by-step Working:**
1. The number halfway between two numbers is their average. We need to calculate $\frac{\frac{1}{6} + \frac{1}{4}}{2}$.
2. Find a common denominator to add the fractions. The least common multiple of 6 and 4 is 12.
$$\frac{1}{6} = \frac{2}{12}$$
$$\frac{1}{4} = \frac{3}{12}$$
3. Add the fractions:
$$\frac{2}{12} + \frac{3}{12} = \frac{5}{12}$$
4. Divide the sum by 2 to find the average:
$$\frac{\frac{5}{12}}{2} = \frac{5}{12} \times \frac{1}{2} = \frac{5}{24}$$

**Ruling out wrong choices:**
- B ($\frac{1}{5}$) is a common mistake made by incorrectly "averaging" the denominators (halfway between 4 and 6).
- E ($\frac{5}{12}$) is the sum of the two fractions, forgetting to divide by 2.

ANSWER 2: C

Problem 3:
**Restatement:** We are given the numbers $-2, 4, 6, 9,$ and $12$. They are placed in five positions based on specific rules. We need to find the average of the numbers in the first and last positions.

**Step-by-step Working:**
1. Identify the key numbers:
   - Largest: $12$
   - Smallest: $-2$
   - Median (middle value when sorted): $6$
   - Remaining numbers: $4$ and $9$
2. Analyze the rules for the 5 positions (1, 2, 3, 4, 5):
   - The largest ($12$) isn't first, but is in the first three places. So, $12$ must be in position 2 or 3. This means $12$ is NOT in position 1 or 5.
   - The smallest ($-2$) isn't last, but is in the last three places. So, $-2$ must be in position 3 or 4. This means $-2$ is NOT in position 1 or 5.
   - The median ($6$) isn't first or last. So, $6$ is NOT in position 1 or 5.
3. Determine the first and last numbers:
   Since $12$, $-2$, and $6$ are all forbidden from being in position 1 and position 5, the only numbers left to fill the first and last spots are $4$ and $9$.
4. Calculate the average of the first and last numbers:
   $$\frac{4 + 9}{2} = \frac{13}{2} = 6.5$$

**Ruling out wrong choices:**
- If you mistakenly thought the median was part of the ends, you might get 5 (average of 4 and 6) or 7.5 (average of 6 and 9). 

ANSWER 3: C

Problem 4:
**Restatement:** We are given the equation $991 + 993 + 995 + 997 + 999 = 5000 - N$ and need to solve for $N$.

**Step-by-step Working:**
1. Notice that each number on the left side is slightly less than 1000. We can rewrite each term:
   $991 = 1000 - 9$
   $993 = 1000 - 7$
   $995 = 1000 - 5$
   $997 = 1000 - 3$
   $999 = 1000 - 1$
2. Substitute these into the sum:
   $(1000 - 9) + (1000 - 7) + (1000 - 5) + (1000 - 3) + (1000 - 1)$
3. Group the 1000s and the subtracted amounts:
   $= 5000 - (9 + 7 + 5 + 3 + 1)$
4. Add the subtracted amounts:
   $9 + 7 + 5 + 3 + 1 = 25$
5. The sum is $5000 - 25$. Therefore, $N = 25$.

**Ruling out wrong choices:**
- A (5) is tempting if you just count the number of terms being added.
- C (15) or D (20) might result from an arithmetic error when adding the odd numbers.

ANSWER 4: E

Problem 5:
**Restatement:** We need to find how many 3-digit positive integers exist such that the product of their three digits is exactly 24.

**Step-by-step Working:**
1. Find all combinations of 3 single-digit numbers (from 1 to 9) that multiply to 24.
   - The prime factorization of 24 is $2^3 \times 3$.
   - Possible sets of digits:
     - $\{1, 3, 8\}$
     - $\{1, 4, 6\}$
     - $\{2, 2, 6\}$
     - $\{2, 3, 4\}$
   (No other combinations work. For example, we cannot use 9, 7, or 5 because 24 is not divisible by them.)
2. Count the number of permutations (arrangements) for each set to form 3-digit numbers:
   - $\{1, 3, 8\}$: 3 distinct digits $\rightarrow 3! = 6$ numbers.
   - $\{1, 4, 6\}$: 3 distinct digits $\rightarrow 3! = 6$ numbers.
   - $\{2, 3, 4\}$: 3 distinct digits $\rightarrow 3! = 6$ numbers.
   - $\{2, 2, 6\}$: 2 identical digits $\rightarrow \frac{3!}{2!} = 3$ numbers.
3. Add them up:
   $6 + 6 + 6 + 3 = 21$

**Ruling out wrong choices:**
- E (24) is tempting if one assumes all 4 sets have 6 permutations ($4 \times 6 = 24$).
- C (18) is tempting if one misses the $\{2, 2, 6\}$ set entirely.

ANSWER 5: D

Problem 6:
**Restatement:** Brent starts with 4 goldfish that quadruple every month. Gretel starts with 128 goldfish that double every month. We need to find the number of months until they have the same amount.

**Step-by-step Working:**
1. Let $m$ be the number of months.
2. Write an expression for Brent's goldfish:
   $B(m) = 4 \times 4^m$
3. Write an expression for Gretel's goldfish:
   $G(m) = 128 \times 2^m$
4. Set them equal to each other:
   $4 \times 4^m = 128 \times 2^m$
5. Express all bases as powers of 2 to easily solve for $m$:
   $4 = 2^2$
   $128 = 2^7$
   $2^2 \times (2^2)^m = 2^7 \times 2^m$
   $2^2 \times 2^{2m} = 2^{m+7}$
   $2^{2m+2} = 2^{m+7}$
6. Equate the exponents:
   $2m + 2 = m + 7$
   $m = 5$

**Ruling out wrong choices:**
- A (4) or C (6) could be obtained if one makes an arithmetic error