🔬 Model Lab

## Step 1: Problem 1 - Understanding the problem
We need to find the fraction of 7-digit telephone numbers that begin with 9 and end with 0, given that the numbers cannot start with 0 or 1.

## Step 2: Problem 1 - Calculating total possible telephone numbers
The first digit can be any number from 2 to 9, so there are 8 possibilities. The remaining 6 digits can be any number from 0 to 9, so there are 10 possibilities for each. Thus, the total number of telephone numbers is $8 \times 10^6$.

## Step 3: Problem 1 - Calculating the number of telephone numbers that begin with 9 and end with 0
For a number to begin with 9 and end with 0, the first digit must be 9 and the last digit must be 0. The first digit has 1 possibility (9), the last digit has 1 possibility (0), and the remaining 5 digits have 10 possibilities each. So, the total number of such telephone numbers is $1 \times 10^5 \times 1$.

## Step 4: Problem 1 - Calculating the fraction
The fraction of telephone numbers that begin with 9 and end with 0 is $\frac{1 \times 10^5 \times 1}{8 \times 10^6} = \frac{10^5}{8 \times 10^6} = \frac{1}{80}$.

## Step 5: Problem 1 - Answer
ANSWER 1: B

## Step 6: Problem 2 - Understanding the problem
We are given 25 integers divided into 5 groups of 5 numbers each. We need to find the least possible value of M, the median of the medians of these groups.

## Step 7: Problem 2 - Analyzing the medians
To minimize M, we should minimize the medians of the 5 groups. The integers are from 1 to 25. The median of each group will be the third number when the numbers are arranged in ascending order.

## Step 8: Problem 2 - Distributing numbers to minimize M
To minimize M, we can distribute the numbers such that the medians of the groups are as small as possible. One way to do this is by grouping the smallest numbers together and ensuring that the medians are as low as possible.

## Step 9: Problem 2 - Calculating the medians of the groups
Let's distribute the numbers: Group 1 = {1, 2, 3, 24, 25}, Group 2 = {4, 5, 6, 22, 23}, Group 3 = {7, 8, 9, 20, 21}, Group 4 = {10, 11, 12, 18, 19}, Group 5 = {13, 14, 15, 16, 17}. The medians are 3, 6, 9, 12, 15.

## Step 10: Problem 2 - Finding M
The medians of the groups are 3, 6, 9, 12, 15. The median of these medians (M) is 9.

## Step 11: Problem 2 - Answer
ANSWER 2: A

## Step 12: Problem 3 - Understanding the problem
A cricket hops between 4 leaves with equal probability. We need to find the probability that it returns to the starting leaf after 4 hops.

## Step 13: Problem 3 - Analyzing the hops
After the first hop, the cricket is on a different leaf. For it to return to the starting leaf after 4 hops, it must be on a different leaf after the third hop as well.

## Step 14: Problem 3 - Calculating the probability
Let's denote the probability of being on the starting leaf after n hops as $P_n$. $P_0 = 1$, $P_1 = 0$. For $n \geq 1$, $P_{n+1} = \frac{1}{3}(1 - P_n)$ because the probability of not being on the starting leaf after n hops is $1 - P_n$, and from any other leaf, there's a $\frac{1}{3}$ chance of hopping to the starting leaf.

## Step 15: Problem 3 - Computing $P_4$
$P_2 = \frac{1}{3}$, $P_3 = \frac{1}{3}(1 - \frac{1}{3}) = \frac{2}{9}$, $P_4 = \frac{1}{3}(1 - \frac{2}{9}) = \frac{1}{3} \times \frac{7}{9} = \frac{7}{27}$.

## Step 16: Problem 3 - Answer
ANSWER 3: E

## Step 17: Problem 4 - Understanding the problem
The letters and digits are cycled separately and combined. We need to find when "AJHSME 1989" appears again.

## Step 18: Problem 4 - Analyzing the cycles
The letters cycle every 6 steps, and the digits cycle every 4 steps (since there are 4 digits and they are distinct when considering their positions). The combined cycle will be the least common multiple (LCM) of 6 and 4.

## Step 19: Problem 4 - Calculating the LCM of 6 and 4
The LCM of 6 and 4 is 12.

## Step 20: Problem 4 - Answer
ANSWER 4: C

## Step 21: Problem 5 - Understanding the problem
A price is increased by a certain percent and then decreased by the same percent, resulting in 84% of the original price.

## Step 22: Problem 5 - Setting up the equation
Let $x$ be the original price and $p$ be the percentage change. The new price after increase is $x(1 + \frac{p}{100})$, and after decrease is $x(1 + \frac{p}{100})(1 - \frac{p}{100}) = 0.84x$.

## Step 23: Problem 5 - Solving for $p$
$x(1 - \frac{p^2}{10000}) = 0.84x$. Simplifying gives $1 - \frac{p^2}{10000} = 0.84$. Thus, $\frac{p^2}{10000} = 0.16$, so $p^2 = 1600$, and $p = 40$.

## Step 24: Problem 5 - Answer
ANSWER 5: E

## Step 25: Problem 6 - Understanding the problem
Harold tosses a nickel four times. We need to find the probability of getting at least as many heads as tails.

## Step 26: Problem 6 - Enumerating the outcomes
The possible outcomes where heads are at least as many as tails are: 4 heads, 3 heads and 1 tail, and 2 heads and 2 tails.

## Step 27: Problem 6 - Calculating the probabilities
The total number of outcomes is $2^4 = 16$. The number of ways to get 4 heads is 1, 3 heads and 1 tail is $\binom{4}{3} = 4$, and 2 heads and 2 tails is $\binom{4}{2} = 6$. So, the total favorable outcomes are $1 + 4 + 6 = 11$.

## Step 28: Problem 6 - Calculating the probability
The probability is $\frac{11}{16}$.

## Step 29: Problem 6 - Answer
ANSWER 6: E

## Step 30: Problem 7 - Understanding the problem
A machine applies a rule to a positive integer N: if N is even, output N/2; if N is odd, output 3N + 1. We need to find the sum of all N that result in 1 after 6 steps.

## Step 31: Problem 7 - Working backwards from 1
After 6 steps, the output is 1. So, before that, it was 2 (since 2 is even and 2/2 = 1). Before that, it could be 4. Continuing this process backwards: 4 -> 2 -> 1 (3 steps), 8 -> 4 -> 2 -> 1 (4 steps), 16 -> 8 -> 4 -> 2 -> 1 (5 steps), and so on, until we reach 6 steps.

## Step 32: Problem 7 - Enumerating the possible N
For 6 steps: $2^6 = 64$ is a possible N (since 64 -> 32 -> 16 -> 8 -> 4 -> 2 -> 1). Working bac

Problem 1:

We are asked what fraction of possible telephone numbers begin with 9 and end with 0.

A 7-digit telephone number cannot begin with 0 or 1.

Total possible telephone numbers:

- First digit: 2 through 9, so 8 choices.
- Remaining 6 digits: each has 10 choices.

So total numbers:

\[
8 \cdot 10^6
\]

Now count numbers that begin with 9 and end with 0:

- First digit fixed as 9: 1 choice.
- Last digit fixed as 0: 1 choice.
- Middle 5 digits: each has 10 choices.

So favorable numbers:

\[
10^5
\]

Thus the fraction is:

\[
\frac{10^5}{8\cdot 10^6}=\frac{1}{80}
\]

The tempting answer \(\frac{1}{100}\) would be correct only if all first digits were equally allowed, including 0 and 1, but they are not.

ANSWER 1: B

Problem 2:

We have the integers 1 through 25 separated into five groups of 5 numbers each. We find the median of each group, then take the median of those five medians. We want the least possible value of \(M\).

For a group of 5 numbers to have median at most \(m\), at least 3 numbers in that group must be at most \(m\).

Since \(M\) is the median of the five medians, at least 3 of the group medians must be at most \(M\).

Therefore, to make \(M\) as small as possible, we need at least:

\[
3 \times 3 = 9
\]

numbers that are at most \(M\).

The 9 smallest numbers are:

\[
1,2,3,4,5,6,7,8,9
\]

So \(M\) cannot be less than 9.

Now we show that \(M=9\) is possible. For example:

\[
\{1,2,3,22,23\} \quad \text{median } 3
\]

\[
\{4,5,6,24,25\} \quad \text{median } 6
\]

\[
\{7,8,9,20,21\} \quad \text{median } 9
\]

The remaining numbers can form two groups with medians larger than 9:

\[
\{10,11,12,13,14\} \quad \text{median } 12
\]

\[
\{15,16,17,18,19\} \quad \text{median } 17
\]

The five medians are:

\[
3,6,9,12,17
\]

Their median is 9.

So the least possible value is 9.

ANSWER 2: A

Problem 3:

The cricket starts on one leaf and hops to one of the other 3 leaves each time. We want the probability that after 4 hops it is back where it started.

Let \(p_n\) be the probability that the cricket is on its starting leaf after \(n\) hops.

Initially:

\[
p_0=1
\]

After one hop, it cannot be on the starting leaf:

\[
p_1=0
\]

If the cricket is not on the starting leaf, then on the next hop it has probability \(\frac13\) of hopping back to the starting leaf. Therefore:

\[
p_{n+1}=\frac{1-p_n}{3}
\]

Now compute:

\[
p_2=\frac{1-0}{3}=\frac13
\]

\[
p_3=\frac{1-\frac13}{3}=\frac{\frac23}{3}=\frac29
\]

\[
p_4=\frac{1-\frac29}{3}=\frac{\frac79}{3}=\frac7{27}
\]

So the probability is:

\[
\frac7{27}
\]

Choice A, \(\frac29\), is the probability after 3 hops, not 4 hops.

ANSWER 3: E

Problem 4:

The letters and digits are cycled separately.

The letters are:

\[
AJHSME
\]

There are 6 letters, so the letters return to their original order after 6 cycles.

The digits are:

\[
1989
\]

There are 4 digits, so the digits return to their original order after 4 cycles.

We need both the letters and the digits to return to their original arrangements at the same time.

So we need the least common multiple of 6 and 4:

\[
\operatorname{lcm}(6,4)=12
\]

Thus \(AJHSME\ 1989\) appears again for the first time on line 12.

Choice A, 6, would return only the letters, not necessarily the digits. Choice B, 10, returns neither. Choice E, 24, is a common multiple but not the first one.

ANSWER 4: C

Problem 5:

Let the original price be \(P\).

Suppose the price is increased by \(r\), where \(r\) is written as a decimal. Then the new price is:

\[
P(1+r)
\]

Then it is decreased by the same percent, so the final price is:

\[
P(1+r)(1-r)
\]

We are told this equals 84% of the original price:

\[
P(1+r)(1-r)=0.84P
\]

Cancel \(P\):

\[
(1+r)(1-r)=0.84
\]

\[
1-r^2=0.84
\]

\[
r^2=0.16
\]

\[
r=0.4
\]

So the percent is:

\[
40\%
\]

A common mistake is to think an increase and decrease by the same percent cancel each other, but they do not because the second percent is taken from a different price.

ANSWER 5: E

Problem 6:

Harold tosses a nickel 4 times. We want the probability that he gets at least as many heads as tails.

There are \(2^4=16\) equally likely outcomes.

“At least as many heads as tails” means:

\[
H \ge T
\]

Since there are 4 tosses total, this means Harold must get 2, 3, or 4 heads.

Count the outcomes:

\[
\binom42=6
\]

\[
\binom43=4
\]

\[
\binom44=1
\]

Total favorable outcomes:

\[
6+4+1=11
\]

So the probability is:

\[
\frac{11}{16}
\]

The tempting answer \(\frac12\) would count only strictly more heads than tails and would forget the tie case.

ANSWER 6: E

Problem 7:

The machine rule is:

- If \(N\) is even, output \(\frac N2\).
- If \(N\) is odd, output \(3N+1\).

We need all starting positive integers \(N\) that become 1 after exactly 6 steps.

It is easiest to work backward from 1.

If a number outputs \(y\), then possible previous numbers are:

1. \(2y\), because if the previous number was even, halving it gives \(y\).
2. \(\frac{y-1}{3}\), but only if this is a positive odd integer.

Now work backward 6 steps.

Step 0:

\[
\{1\}
\]

Step 1 predecessors of 1:

\[
\{2\}
\]

Step 2 predecessors of 2:

\[
\{4\}
\]

Step 3 predecessors of 4:

\[
2\cdot 4=8
\]

and

\[
\frac{4-1}{3}=1
\]

So:

\[
\{8,1\}
\]

Step 4:

Predecessors of 8: \(16\)

Predecessors of 1: \(2\)

So:

\[
\{16,2\}
\]

Step 5:

Predecessors of 16:

\[
32,\quad \frac{16-1}{3}=5
\]

Predecessors of 2:

\[
4
\]

So:

\[
\{32,5,4\}
\]

Step 6:

Predecessors of 32:

\[
64
\]

Predecessors of 5:

\[
10
\]

Predecessors of 4:

\[
8,\quad 1
\]

So the possible starting values are:

\[
64,10,8,1
\]

Their sum is:

\[
64+10+8+1=83
\]

ANSWER 7: E

Problem 8:

Let the list be:

\[
a_1,a_2,a_3,a_4,a_5,a_6,a_7,a_8
\]

Each new number is the product of the two previous numbers:

\[
a_3=a_1a_2
\]

\[
a_4=a_2a_3
\]

and so on.

We are told the last three numbers are:

\[
a_6=16,\quad a_7=64,\quad a_8=1024
\]

Check:

\[
16\cdot 64=1024
\]

Now work backward.

Since:

\[
a_7=a_5a_6
\]

we have:

\[
64=

Problem 1:
We are asked to find the fraction of possible 7-digit telephone numbers that begin with 9 and end with 0, given that a telephone number cannot begin with 0 or 1.

First, let's find the total number of possible telephone numbers:
- The first digit can be any number from 2 to 9, which gives 8 possibilities.
- The remaining 6 digits can be any number from 0 to 9, giving 10 possibilities each.
Total possible telephone numbers = 8 × 10 × 10 × 10 × 10 × 10 × 10 = 8,000,000.

Next, let's find the number of telephone numbers that begin with 9 and end with 0:
- The first digit must be 9 (1 possibility).
- The last digit must be 0 (1 possibility).
- The middle 5 digits can be any number from 0 to 9 (10 possibilities each).
Total such numbers = 1 × 10 × 10 × 10 × 10 × 10 × 1 = 100,000.

The fraction of telephone numbers that meet this condition is:
100,000 / 8,000,000 = 1 / 80.

ANSWER 1: B

Problem 2:
We are asked to find the least possible value of M, which is the median of the medians of five groups of 5 numbers each, chosen from the integers 1 through 25.

Let the 5 groups be G1, G2, G3, G4, and G5, and let their respective medians be m1, m2, m3, m4, and m5. 
Assume the medians are ordered such that m1 < m2 < m3 < m4 < m5. The median of these medians is M = m3.
To minimize m3, we must make the numbers in the groups as small as possible. 

For m3 to be a certain value, there must be at least 3 groups (G1, G2, G3) whose medians are less than or equal to m3. 
In each of these 3 groups, there are 3 numbers that are less than or equal to the group's median. 
Therefore, there are at least 3 groups × 3 numbers = 9 numbers that are less than or equal to m3.

Since the numbers are distinct integers from 1 to 25, the 9th smallest possible integer is 9. Thus, m3 must be at least 9. 
We can verify that 9 is possible with the following grouping:
G1: 1, 2, 7, 24, 25 (median 7)
G2: 3, 4, 8, 22, 23 (median 8)
G3: 5, 6, 9, 20, 21 (median 9)
G4: 10, 11, 12, 18, 19 (median 12)
G5: 13, 14, 15, 16, 17 (median 15)
The medians are 7, 8, 9, 12, and 15. Their median is 9.

ANSWER 2: A

Problem 3:
We want to find the probability that a cricket, which hops to one of 3 other leaves with equal probability, returns to its starting leaf after exactly 4 hops.

Let the leaves be A, B, C, and D, and assume the cricket starts at leaf A.
Let P(n) be the probability that the cricket is at leaf A after n hops.
- P(0) = 1 (it starts at A).
- For any hop n > 0, the cricket can only land on A if it was NOT on A in the previous hop. If it was on B, C, or D, it has a 1/3 chance of hopping to A.
Thus, the recursive formula is P(n) = (1 - P(n-1)) × (1/3).

Let's calculate step-by-step:
- P(1) = (1 - 1) × 1/3 = 0
- P(2) = (1 - 0) × 1/3 = 1/3
- P(3) = (1 - 1/3) × 1/3 = (2/3) × (1/3) = 2/9
- P(4) = (1 - 2/9) × 1/3 = (7/9) × (1/3) = 7/27

ANSWER 3: E

Problem 4:
We are given a sequence of letters "AJHSME" (6 characters) and a sequence of digits "1989" (4 characters). They are cycled separately line by line. We want to find the line number where the original combination "AJHSME 1989" appears again for the first time.

The letters cycle back to their original arrangement every 6 lines.
The digits cycle back to their original arrangement every 4 lines.
For both to return to their original arrangement simultaneously, the line number must be a common multiple of both 6 and 4. 

The least common multiple (LCM) of 6 and 4 is 12.
Therefore, on line 12, the letters will have completed exactly 2 full cycles and the digits will have completed exactly 3 full cycles, returning to "AJHSME 1989".

ANSWER 4: C

Problem 5:
Let the original price of the shirt be P, and let the percentage of increase and decrease be x.
When the price is increased by x percent, it becomes P(1 + x/100).
When this new price is decreased by x percent, it becomes P(1 + x/100)(1 - x/100).

We are given that the resulting price is 84% of the original price, so:
P(1 + x/100)(1 - x/100) = 0.84P

Divide both sides by P:
(1 + x/100)(1 - x/100) = 0.84
1 - (x/100)² = 0.84
(x/100)² = 1 - 0.84
(x/100)² = 0.16

Taking the square root of both sides:
x/100 = 0.4
x = 40

The price was increased and decreased by 40%.

ANSWER 5: E

Problem 6:
Harold tosses a nickel four times. We want to find the probability that he gets at least as many heads as tails.
"At least as many heads as tails" in 4 tosses means getting 2, 3, or 4 heads.

The total number of possible outcomes when tossing a coin 4 times is 2⁴ = 16.
Using combinations, we find the number of ways to get 2, 3, or 4 heads:
- 2 heads: 4C2 = 6 ways
- 3 heads: 4C3 = 4 ways
- 4 heads: 4C4 = 1 way

Total favorable outcomes = 6 + 4 + 1 = 11.
The probability is 11 / 16.

ANSWER 6: E

Problem 7:
The machine follows the Collatz-like rule:
- If N is even, output N/2
- If N is odd, output 3N + 1

We are told a 6-step process ends in 1. Let's work backward from Step 6 to find all possible starting positive integers N (Step 0).
- Step 6: 1
- Step 5: Must be 2 (since 3x + 1 = 1 gives x = 0, which is not a positive integer).
- Step 4: Must be 4 (since 3x + 1 = 2 gives x = 1/3, not an integer).
- Step 3: Can be 8 (since 8/2 = 4) or 1 (since 3(1) + 1 = 4).

Branch 1 (Step 3 is 8):
- Step 2: Must be 16 (since 3x + 1 = 8 gives x = 7/3).
- Step 1: Can be 32 (32/2 = 16) or 5 (3(5) + 1 = 16).
- Step 0 (from 32): Must be 64 (since 3x + 1 = 32 gives x = 31/3).
- Step 0 (from 5): Must be 10 (since 3x + 1 = 5 gives x = 4/3).

Branch 2 (Step 3 is 1):
- Step 2: Must be 2.
- Step 1: Must be 4.
- Step 0: Can be 8 (8/2 = 4) or 1 (3(1) + 1 = 4).

The possible starting integers N are 64, 10, 8, and 1.
The sum of all such integers is 64 + 10 + 8 + 1 = 83.

ANSWER 7: E

Problem 8:
We are given a sequence of 8 numbers where each number is the product of the two preceding numbers: a_n = a_{n-2} × a_{n-1}.
We know the last three numbers: a_6 = 16, a_7 = 64, and a_8 = 1024.
Let's work backward using the rule a_{n-2} = a_n / a_{n-1}:
- a_5 = a_7 / a_6 = 64 / 16 = 4
- a_4 

## Step 1: Problem 1 - Understanding the problem
We need to find the fraction of 7-digit telephone numbers that begin with 9 and end with 0, given that the numbers cannot start with 0 or 1.

## Step 2: Problem 1 - Calculating total possible telephone numbers
The first digit can be any number from 2 to 9, so there are 8 possibilities. The remaining 6 digits can be any number from 0 to 9, so there are 10 possibilities for each. Thus, the total number of telephone numbers is $8 \times 10^6$.

## Step 3: Problem 1 - Calculating the number of telephone numbers that begin with 9 and end with 0
For a number to begin with 9 and end with 0, the first digit must be 9 and the last digit must be 0. The first digit has 1 possibility (9), the last digit has 1 possibility (0), and the remaining 5 digits have 10 possibilities each. So, the total number of such telephone numbers is $1 \times 10^5 \times 1$.

## Step 4: Problem 1 - Calculating the fraction
The fraction of telephone numbers that begin with 9 and end with 0 is $\frac{1 \times 10^5 \times 1}{8 \times 10^6} = \frac{10^5}{8 \times 10^6} = \frac{1}{80}$.

## Step 5: Problem 1 - Answer
ANSWER 1: B

## Step 6: Problem 2 - Understanding the problem
We are given 25 integers divided into 5 groups of 5 numbers each. We need to find the least possible value of M, the median of the medians of these groups.

## Step 7: Problem 2 - Analyzing the medians
To minimize M, we should minimize the medians of the 5 groups. The integers are from 1 to 25. The median of each group will be the third number when the numbers are arranged in ascending order.

## Step 8: Problem 2 - Distributing numbers to minimize M
To minimize M, we can distribute the numbers such that the medians of the groups are as small as possible. One way to do this is by grouping the smallest numbers together and ensuring that the medians are as low as possible.

## Step 9: Problem 2 - Calculating the medians of the groups
Let's distribute the numbers: Group 1 = {1, 2, 3, 24, 25}, Group 2 = {4, 5, 6, 22, 23}, Group 3 = {7, 8, 9, 20, 21}, Group 4 = {10, 11, 12, 18, 19}, Group 5 = {13, 14, 15, 16, 17}. The medians are 3, 6, 9, 12, 15.

## Step 10: Problem 2 - Finding M
The medians of the groups are 3, 6, 9, 12, 15. The median of these medians (M) is 9.

## Step 11: Problem 2 - Answer
ANSWER 2: A

## Step 12: Problem 3 - Understanding the problem
A cricket hops between 4 leaves with equal probability. We need to find the probability that it returns to the starting leaf after 4 hops.

## Step 13: Problem 3 - Analyzing the hops
After the first hop, the cricket is on a different leaf. For it to return to the starting leaf after 4 hops, it must be on a different leaf after the third hop as well.

## Step 14: Problem 3 - Calculating the probability
Let's denote the probability of being on the starting leaf after n hops as $P_n$. $P_0 = 1$, $P_1 = 0$. For $n \geq 1$, $P_{n+1} = \frac{1}{3}(1 - P_n)$ because the probability of not being on the starting leaf after n hops is $1 - P_n$, and from any other leaf, there's a $\frac{1}{3}$ chance of hopping to the starting leaf.

## Step 15: Problem 3 - Computing $P_4$
$P_2 = \frac{1}{3}$, $P_3 = \frac{1}{3}(1 - \frac{1}{3}) = \frac{2}{9}$, $P_4 = \frac{1}{3}(1 - \frac{2}{9}) = \frac{1}{3} \times \frac{7}{9} = \frac{7}{27}$.

## Step 16: Problem 3 - Answer
ANSWER 3: E

## Step 17: Problem 4 - Understanding the problem
The letters and digits are cycled separately and combined. We need to find when "AJHSME 1989" appears again.

## Step 18: Problem 4 - Analyzing the cycles
The letters cycle every 6 steps, and the digits cycle every 4 steps (since there are 4 digits and they are distinct when considering their positions). The combined cycle will be the least common multiple (LCM) of 6 and 4.

## Step 19: Problem 4 - Calculating the LCM of 6 and 4
The LCM of 6 and 4 is 12.

## Step 20: Problem 4 - Answer
ANSWER 4: C

## Step 21: Problem 5 - Understanding the problem
A price is increased by a certain percent and then decreased by the same percent, resulting in 84% of the original price.

## Step 22: Problem 5 - Setting up the equation
Let $x$ be the original price and $p$ be the percentage change. The new price after increase is $x(1 + \frac{p}{100})$, and after decrease is $x(1 + \frac{p}{100})(1 - \frac{p}{100}) = 0.84x$.

## Step 23: Problem 5 - Solving for $p$
$x(1 - \frac{p^2}{10000}) = 0.84x$. Simplifying gives $1 - \frac{p^2}{10000} = 0.84$. Thus, $\frac{p^2}{10000} = 0.16$, so $p^2 = 1600$, and $p = 40$.

## Step 24: Problem 5 - Answer
ANSWER 5: E

## Step 25: Problem 6 - Understanding the problem
Harold tosses a nickel four times. We need to find the probability of getting at least as many heads as tails.

## Step 26: Problem 6 - Enumerating the outcomes
The possible outcomes where heads are at least as many as tails are: 4 heads, 3 heads and 1 tail, and 2 heads and 2 tails.

## Step 27: Problem 6 - Calculating the probabilities
The total number of outcomes is $2^4 = 16$. The number of ways to get 4 heads is 1, 3 heads and 1 tail is $\binom{4}{3} = 4$, and 2 heads and 2 tails is $\binom{4}{2} = 6$. So, the total favorable outcomes are $1 + 4 + 6 = 11$.

## Step 28: Problem 6 - Calculating the probability
The probability is $\frac{11}{16}$.

## Step 29: Problem 6 - Answer
ANSWER 6: E

## Step 30: Problem 7 - Understanding the problem
A machine applies a rule to a positive integer N: if N is even, output N/2; if N is odd, output 3N + 1. We need to find the sum of all N that result in 1 after 6 steps.

## Step 31: Problem 7 - Working backwards from 1
After 6 steps, the output is 1. So, before that, it was 2 (since 2 is even and 2/2 = 1). Before that, it could be 4. Continuing this process backwards: 4 -> 2 -> 1 (3 steps), 8 -> 4 -> 2 -> 1 (4 steps), 16 -> 8 -> 4 -> 2 -> 1 (5 steps), and so on, until we reach 6 steps.

## Step 32: Problem 7 - Enumerating the possible N
For 6 steps: $2^6 = 64$ is a possible N (since 64 -> 32 -> 16 -> 8 -> 4 -> 2 -> 1). Working bac

Problem 1:

We are asked what fraction of possible telephone numbers begin with 9 and end with 0.

A 7-digit telephone number cannot begin with 0 or 1.

Total possible telephone numbers:

- First digit: 2 through 9, so 8 choices.
- Remaining 6 digits: each has 10 choices.

So total numbers:

\[
8 \cdot 10^6
\]

Now count numbers that begin with 9 and end with 0:

- First digit fixed as 9: 1 choice.
- Last digit fixed as 0: 1 choice.
- Middle 5 digits: each has 10 choices.

So favorable numbers:

\[
10^5
\]

Thus the fraction is:

\[
\frac{10^5}{8\cdot 10^6}=\frac{1}{80}
\]

The tempting answer \(\frac{1}{100}\) would be correct only if all first digits were equally allowed, including 0 and 1, but they are not.

ANSWER 1: B

Problem 2:

We have the integers 1 through 25 separated into five groups of 5 numbers each. We find the median of each group, then take the median of those five medians. We want the least possible value of \(M\).

For a group of 5 numbers to have median at most \(m\), at least 3 numbers in that group must be at most \(m\).

Since \(M\) is the median of the five medians, at least 3 of the group medians must be at most \(M\).

Therefore, to make \(M\) as small as possible, we need at least:

\[
3 \times 3 = 9
\]

numbers that are at most \(M\).

The 9 smallest numbers are:

\[
1,2,3,4,5,6,7,8,9
\]

So \(M\) cannot be less than 9.

Now we show that \(M=9\) is possible. For example:

\[
\{1,2,3,22,23\} \quad \text{median } 3
\]

\[
\{4,5,6,24,25\} \quad \text{median } 6
\]

\[
\{7,8,9,20,21\} \quad \text{median } 9
\]

The remaining numbers can form two groups with medians larger than 9:

\[
\{10,11,12,13,14\} \quad \text{median } 12
\]

\[
\{15,16,17,18,19\} \quad \text{median } 17
\]

The five medians are:

\[
3,6,9,12,17
\]

Their median is 9.

So the least possible value is 9.

ANSWER 2: A

Problem 3:

The cricket starts on one leaf and hops to one of the other 3 leaves each time. We want the probability that after 4 hops it is back where it started.

Let \(p_n\) be the probability that the cricket is on its starting leaf after \(n\) hops.

Initially:

\[
p_0=1
\]

After one hop, it cannot be on the starting leaf:

\[
p_1=0
\]

If the cricket is not on the starting leaf, then on the next hop it has probability \(\frac13\) of hopping back to the starting leaf. Therefore:

\[
p_{n+1}=\frac{1-p_n}{3}
\]

Now compute:

\[
p_2=\frac{1-0}{3}=\frac13
\]

\[
p_3=\frac{1-\frac13}{3}=\frac{\frac23}{3}=\frac29
\]

\[
p_4=\frac{1-\frac29}{3}=\frac{\frac79}{3}=\frac7{27}
\]

So the probability is:

\[
\frac7{27}
\]

Choice A, \(\frac29\), is the probability after 3 hops, not 4 hops.

ANSWER 3: E

Problem 4:

The letters and digits are cycled separately.

The letters are:

\[
AJHSME
\]

There are 6 letters, so the letters return to their original order after 6 cycles.

The digits are:

\[
1989
\]

There are 4 digits, so the digits return to their original order after 4 cycles.

We need both the letters and the digits to return to their original arrangements at the same time.

So we need the least common multiple of 6 and 4:

\[
\operatorname{lcm}(6,4)=12
\]

Thus \(AJHSME\ 1989\) appears again for the first time on line 12.

Choice A, 6, would return only the letters, not necessarily the digits. Choice B, 10, returns neither. Choice E, 24, is a common multiple but not the first one.

ANSWER 4: C

Problem 5:

Let the original price be \(P\).

Suppose the price is increased by \(r\), where \(r\) is written as a decimal. Then the new price is:

\[
P(1+r)
\]

Then it is decreased by the same percent, so the final price is:

\[
P(1+r)(1-r)
\]

We are told this equals 84% of the original price:

\[
P(1+r)(1-r)=0.84P
\]

Cancel \(P\):

\[
(1+r)(1-r)=0.84
\]

\[
1-r^2=0.84
\]

\[
r^2=0.16
\]

\[
r=0.4
\]

So the percent is:

\[
40\%
\]

A common mistake is to think an increase and decrease by the same percent cancel each other, but they do not because the second percent is taken from a different price.

ANSWER 5: E

Problem 6:

Harold tosses a nickel 4 times. We want the probability that he gets at least as many heads as tails.

There are \(2^4=16\) equally likely outcomes.

“At least as many heads as tails” means:

\[
H \ge T
\]

Since there are 4 tosses total, this means Harold must get 2, 3, or 4 heads.

Count the outcomes:

\[
\binom42=6
\]

\[
\binom43=4
\]

\[
\binom44=1
\]

Total favorable outcomes:

\[
6+4+1=11
\]

So the probability is:

\[
\frac{11}{16}
\]

The tempting answer \(\frac12\) would count only strictly more heads than tails and would forget the tie case.

ANSWER 6: E

Problem 7:

The machine rule is:

- If \(N\) is even, output \(\frac N2\).
- If \(N\) is odd, output \(3N+1\).

We need all starting positive integers \(N\) that become 1 after exactly 6 steps.

It is easiest to work backward from 1.

If a number outputs \(y\), then possible previous numbers are:

1. \(2y\), because if the previous number was even, halving it gives \(y\).
2. \(\frac{y-1}{3}\), but only if this is a positive odd integer.

Now work backward 6 steps.

Step 0:

\[
\{1\}
\]

Step 1 predecessors of 1:

\[
\{2\}
\]

Step 2 predecessors of 2:

\[
\{4\}
\]

Step 3 predecessors of 4:

\[
2\cdot 4=8
\]

and

\[
\frac{4-1}{3}=1
\]

So:

\[
\{8,1\}
\]

Step 4:

Predecessors of 8: \(16\)

Predecessors of 1: \(2\)

So:

\[
\{16,2\}
\]

Step 5:

Predecessors of 16:

\[
32,\quad \frac{16-1}{3}=5
\]

Predecessors of 2:

\[
4
\]

So:

\[
\{32,5,4\}
\]

Step 6:

Predecessors of 32:

\[
64
\]

Predecessors of 5:

\[
10
\]

Predecessors of 4:

\[
8,\quad 1
\]

So the possible starting values are:

\[
64,10,8,1
\]

Their sum is:

\[
64+10+8+1=83
\]

ANSWER 7: E

Problem 8:

Let the list be:

\[
a_1,a_2,a_3,a_4,a_5,a_6,a_7,a_8
\]

Each new number is the product of the two previous numbers:

\[
a_3=a_1a_2
\]

\[
a_4=a_2a_3
\]

and so on.

We are told the last three numbers are:

\[
a_6=16,\quad a_7=64,\quad a_8=1024
\]

Check:

\[
16\cdot 64=1024
\]

Now work backward.

Since:

\[
a_7=a_5a_6
\]

we have:

\[
64=

Problem 1:
We are asked to find the fraction of possible 7-digit telephone numbers that begin with 9 and end with 0, given that a telephone number cannot begin with 0 or 1.

First, let's find the total number of possible telephone numbers:
- The first digit can be any number from 2 to 9, which gives 8 possibilities.
- The remaining 6 digits can be any number from 0 to 9, giving 10 possibilities each.
Total possible telephone numbers = 8 × 10 × 10 × 10 × 10 × 10 × 10 = 8,000,000.

Next, let's find the number of telephone numbers that begin with 9 and end with 0:
- The first digit must be 9 (1 possibility).
- The last digit must be 0 (1 possibility).
- The middle 5 digits can be any number from 0 to 9 (10 possibilities each).
Total such numbers = 1 × 10 × 10 × 10 × 10 × 10 × 1 = 100,000.

The fraction of telephone numbers that meet this condition is:
100,000 / 8,000,000 = 1 / 80.

ANSWER 1: B

Problem 2:
We are asked to find the least possible value of M, which is the median of the medians of five groups of 5 numbers each, chosen from the integers 1 through 25.

Let the 5 groups be G1, G2, G3, G4, and G5, and let their respective medians be m1, m2, m3, m4, and m5. 
Assume the medians are ordered such that m1 < m2 < m3 < m4 < m5. The median of these medians is M = m3.
To minimize m3, we must make the numbers in the groups as small as possible. 

For m3 to be a certain value, there must be at least 3 groups (G1, G2, G3) whose medians are less than or equal to m3. 
In each of these 3 groups, there are 3 numbers that are less than or equal to the group's median. 
Therefore, there are at least 3 groups × 3 numbers = 9 numbers that are less than or equal to m3.

Since the numbers are distinct integers from 1 to 25, the 9th smallest possible integer is 9. Thus, m3 must be at least 9. 
We can verify that 9 is possible with the following grouping:
G1: 1, 2, 7, 24, 25 (median 7)
G2: 3, 4, 8, 22, 23 (median 8)
G3: 5, 6, 9, 20, 21 (median 9)
G4: 10, 11, 12, 18, 19 (median 12)
G5: 13, 14, 15, 16, 17 (median 15)
The medians are 7, 8, 9, 12, and 15. Their median is 9.

ANSWER 2: A

Problem 3:
We want to find the probability that a cricket, which hops to one of 3 other leaves with equal probability, returns to its starting leaf after exactly 4 hops.

Let the leaves be A, B, C, and D, and assume the cricket starts at leaf A.
Let P(n) be the probability that the cricket is at leaf A after n hops.
- P(0) = 1 (it starts at A).
- For any hop n > 0, the cricket can only land on A if it was NOT on A in the previous hop. If it was on B, C, or D, it has a 1/3 chance of hopping to A.
Thus, the recursive formula is P(n) = (1 - P(n-1)) × (1/3).

Let's calculate step-by-step:
- P(1) = (1 - 1) × 1/3 = 0
- P(2) = (1 - 0) × 1/3 = 1/3
- P(3) = (1 - 1/3) × 1/3 = (2/3) × (1/3) = 2/9
- P(4) = (1 - 2/9) × 1/3 = (7/9) × (1/3) = 7/27

ANSWER 3: E

Problem 4:
We are given a sequence of letters "AJHSME" (6 characters) and a sequence of digits "1989" (4 characters). They are cycled separately line by line. We want to find the line number where the original combination "AJHSME 1989" appears again for the first time.

The letters cycle back to their original arrangement every 6 lines.
The digits cycle back to their original arrangement every 4 lines.
For both to return to their original arrangement simultaneously, the line number must be a common multiple of both 6 and 4. 

The least common multiple (LCM) of 6 and 4 is 12.
Therefore, on line 12, the letters will have completed exactly 2 full cycles and the digits will have completed exactly 3 full cycles, returning to "AJHSME 1989".

ANSWER 4: C

Problem 5:
Let the original price of the shirt be P, and let the percentage of increase and decrease be x.
When the price is increased by x percent, it becomes P(1 + x/100).
When this new price is decreased by x percent, it becomes P(1 + x/100)(1 - x/100).

We are given that the resulting price is 84% of the original price, so:
P(1 + x/100)(1 - x/100) = 0.84P

Divide both sides by P:
(1 + x/100)(1 - x/100) = 0.84
1 - (x/100)² = 0.84
(x/100)² = 1 - 0.84
(x/100)² = 0.16

Taking the square root of both sides:
x/100 = 0.4
x = 40

The price was increased and decreased by 40%.

ANSWER 5: E

Problem 6:
Harold tosses a nickel four times. We want to find the probability that he gets at least as many heads as tails.
"At least as many heads as tails" in 4 tosses means getting 2, 3, or 4 heads.

The total number of possible outcomes when tossing a coin 4 times is 2⁴ = 16.
Using combinations, we find the number of ways to get 2, 3, or 4 heads:
- 2 heads: 4C2 = 6 ways
- 3 heads: 4C3 = 4 ways
- 4 heads: 4C4 = 1 way

Total favorable outcomes = 6 + 4 + 1 = 11.
The probability is 11 / 16.

ANSWER 6: E

Problem 7:
The machine follows the Collatz-like rule:
- If N is even, output N/2
- If N is odd, output 3N + 1

We are told a 6-step process ends in 1. Let's work backward from Step 6 to find all possible starting positive integers N (Step 0).
- Step 6: 1
- Step 5: Must be 2 (since 3x + 1 = 1 gives x = 0, which is not a positive integer).
- Step 4: Must be 4 (since 3x + 1 = 2 gives x = 1/3, not an integer).
- Step 3: Can be 8 (since 8/2 = 4) or 1 (since 3(1) + 1 = 4).

Branch 1 (Step 3 is 8):
- Step 2: Must be 16 (since 3x + 1 = 8 gives x = 7/3).
- Step 1: Can be 32 (32/2 = 16) or 5 (3(5) + 1 = 16).
- Step 0 (from 32): Must be 64 (since 3x + 1 = 32 gives x = 31/3).
- Step 0 (from 5): Must be 10 (since 3x + 1 = 5 gives x = 4/3).

Branch 2 (Step 3 is 1):
- Step 2: Must be 2.
- Step 1: Must be 4.
- Step 0: Can be 8 (8/2 = 4) or 1 (3(1) + 1 = 4).

The possible starting integers N are 64, 10, 8, and 1.
The sum of all such integers is 64 + 10 + 8 + 1 = 83.

ANSWER 7: E

Problem 8:
We are given a sequence of 8 numbers where each number is the product of the two preceding numbers: a_n = a_{n-2} × a_{n-1}.
We know the last three numbers: a_6 = 16, a_7 = 64, and a_8 = 1024.
Let's work backward using the rule a_{n-2} = a_n / a_{n-1}:
- a_5 = a_7 / a_6 = 64 / 16 = 4
- a_4 

## Step 1: Problem 1 - Understanding the problem
We need to find the fraction of 7-digit telephone numbers that begin with 9 and end with 0, given that the numbers cannot start with 0 or 1.

## Step 2: Problem 1 - Calculating total possible telephone numbers
The first digit can be any number from 2 to 9, so there are 8 possibilities. The remaining 6 digits can be any number from 0 to 9, so there are 10 possibilities for each. Thus, the total number of telephone numbers is $8 \times 10^6$.

## Step 3: Problem 1 - Calculating the number of telephone numbers that begin with 9 and end with 0
For a number to begin with 9 and end with 0, the first digit must be 9 and the last digit must be 0. The first digit has 1 possibility (9), the last digit has 1 possibility (0), and the remaining 5 digits have 10 possibilities each. So, the total number of such telephone numbers is $1 \times 10^5 \times 1$.

## Step 4: Problem 1 - Calculating the fraction
The fraction of telephone numbers that begin with 9 and end with 0 is $\frac{1 \times 10^5 \times 1}{8 \times 10^6} = \frac{10^5}{8 \times 10^6} = \frac{1}{80}$.

## Step 5: Problem 1 - Answer
ANSWER 1: B

## Step 6: Problem 2 - Understanding the problem
We are given 25 integers divided into 5 groups of 5 numbers each. We need to find the least possible value of M, the median of the medians of these groups.

## Step 7: Problem 2 - Analyzing the medians
To minimize M, we should minimize the medians of the 5 groups. The integers are from 1 to 25. The median of each group will be the third number when the numbers are arranged in ascending order.

## Step 8: Problem 2 - Distributing numbers to minimize M
To minimize M, we can distribute the numbers such that the medians of the groups are as small as possible. One way to do this is by grouping the smallest numbers together and ensuring that the medians are as low as possible.

## Step 9: Problem 2 - Calculating the medians of the groups
Let's distribute the numbers: Group 1 = {1, 2, 3, 24, 25}, Group 2 = {4, 5, 6, 22, 23}, Group 3 = {7, 8, 9, 20, 21}, Group 4 = {10, 11, 12, 18, 19}, Group 5 = {13, 14, 15, 16, 17}. The medians are 3, 6, 9, 12, 15.

## Step 10: Problem 2 - Finding M
The medians of the groups are 3, 6, 9, 12, 15. The median of these medians (M) is 9.

## Step 11: Problem 2 - Answer
ANSWER 2: A

## Step 12: Problem 3 - Understanding the problem
A cricket hops between 4 leaves with equal probability. We need to find the probability that it returns to the starting leaf after 4 hops.

## Step 13: Problem 3 - Analyzing the hops
After the first hop, the cricket is on a different leaf. For it to return to the starting leaf after 4 hops, it must be on a different leaf after the third hop as well.

## Step 14: Problem 3 - Calculating the probability
Let's denote the probability of being on the starting leaf after n hops as $P_n$. $P_0 = 1$, $P_1 = 0$. For $n \geq 1$, $P_{n+1} = \frac{1}{3}(1 - P_n)$ because the probability of not being on the starting leaf after n hops is $1 - P_n$, and from any other leaf, there's a $\frac{1}{3}$ chance of hopping to the starting leaf.

## Step 15: Problem 3 - Computing $P_4$
$P_2 = \frac{1}{3}$, $P_3 = \frac{1}{3}(1 - \frac{1}{3}) = \frac{2}{9}$, $P_4 = \frac{1}{3}(1 - \frac{2}{9}) = \frac{1}{3} \times \frac{7}{9} = \frac{7}{27}$.

## Step 16: Problem 3 - Answer
ANSWER 3: E

## Step 17: Problem 4 - Understanding the problem
The letters and digits are cycled separately and combined. We need to find when "AJHSME 1989" appears again.

## Step 18: Problem 4 - Analyzing the cycles
The letters cycle every 6 steps, and the digits cycle every 4 steps (since there are 4 digits and they are distinct when considering their positions). The combined cycle will be the least common multiple (LCM) of 6 and 4.

## Step 19: Problem 4 - Calculating the LCM of 6 and 4
The LCM of 6 and 4 is 12.

## Step 20: Problem 4 - Answer
ANSWER 4: C

## Step 21: Problem 5 - Understanding the problem
A price is increased by a certain percent and then decreased by the same percent, resulting in 84% of the original price.

## Step 22: Problem 5 - Setting up the equation
Let $x$ be the original price and $p$ be the percentage change. The new price after increase is $x(1 + \frac{p}{100})$, and after decrease is $x(1 + \frac{p}{100})(1 - \frac{p}{100}) = 0.84x$.

## Step 23: Problem 5 - Solving for $p$
$x(1 - \frac{p^2}{10000}) = 0.84x$. Simplifying gives $1 - \frac{p^2}{10000} = 0.84$. Thus, $\frac{p^2}{10000} = 0.16$, so $p^2 = 1600$, and $p = 40$.

## Step 24: Problem 5 - Answer
ANSWER 5: E

## Step 25: Problem 6 - Understanding the problem
Harold tosses a nickel four times. We need to find the probability of getting at least as many heads as tails.

## Step 26: Problem 6 - Enumerating the outcomes
The possible outcomes where heads are at least as many as tails are: 4 heads, 3 heads and 1 tail, and 2 heads and 2 tails.

## Step 27: Problem 6 - Calculating the probabilities
The total number of outcomes is $2^4 = 16$. The number of ways to get 4 heads is 1, 3 heads and 1 tail is $\binom{4}{3} = 4$, and 2 heads and 2 tails is $\binom{4}{2} = 6$. So, the total favorable outcomes are $1 + 4 + 6 = 11$.

## Step 28: Problem 6 - Calculating the probability
The probability is $\frac{11}{16}$.

## Step 29: Problem 6 - Answer
ANSWER 6: E

## Step 30: Problem 7 - Understanding the problem
A machine applies a rule to a positive integer N: if N is even, output N/2; if N is odd, output 3N + 1. We need to find the sum of all N that result in 1 after 6 steps.

## Step 31: Problem 7 - Working backwards from 1
After 6 steps, the output is 1. So, before that, it was 2 (since 2 is even and 2/2 = 1). Before that, it could be 4. Continuing this process backwards: 4 -> 2 -> 1 (3 steps), 8 -> 4 -> 2 -> 1 (4 steps), 16 -> 8 -> 4 -> 2 -> 1 (5 steps), and so on, until we reach 6 steps.

## Step 32: Problem 7 - Enumerating the possible N
For 6 steps: $2^6 = 64$ is a possible N (since 64 -> 32 -> 16 -> 8 -> 4 -> 2 -> 1). Working bac

Problem 1:

We are asked what fraction of possible telephone numbers begin with 9 and end with 0.

A 7-digit telephone number cannot begin with 0 or 1.

Total possible telephone numbers:

- First digit: 2 through 9, so 8 choices.
- Remaining 6 digits: each has 10 choices.

So total numbers:

\[
8 \cdot 10^6
\]

Now count numbers that begin with 9 and end with 0:

- First digit fixed as 9: 1 choice.
- Last digit fixed as 0: 1 choice.
- Middle 5 digits: each has 10 choices.

So favorable numbers:

\[
10^5
\]

Thus the fraction is:

\[
\frac{10^5}{8\cdot 10^6}=\frac{1}{80}
\]

The tempting answer \(\frac{1}{100}\) would be correct only if all first digits were equally allowed, including 0 and 1, but they are not.

ANSWER 1: B

Problem 2:

We have the integers 1 through 25 separated into five groups of 5 numbers each. We find the median of each group, then take the median of those five medians. We want the least possible value of \(M\).

For a group of 5 numbers to have median at most \(m\), at least 3 numbers in that group must be at most \(m\).

Since \(M\) is the median of the five medians, at least 3 of the group medians must be at most \(M\).

Therefore, to make \(M\) as small as possible, we need at least:

\[
3 \times 3 = 9
\]

numbers that are at most \(M\).

The 9 smallest numbers are:

\[
1,2,3,4,5,6,7,8,9
\]

So \(M\) cannot be less than 9.

Now we show that \(M=9\) is possible. For example:

\[
\{1,2,3,22,23\} \quad \text{median } 3
\]

\[
\{4,5,6,24,25\} \quad \text{median } 6
\]

\[
\{7,8,9,20,21\} \quad \text{median } 9
\]

The remaining numbers can form two groups with medians larger than 9:

\[
\{10,11,12,13,14\} \quad \text{median } 12
\]

\[
\{15,16,17,18,19\} \quad \text{median } 17
\]

The five medians are:

\[
3,6,9,12,17
\]

Their median is 9.

So the least possible value is 9.

ANSWER 2: A

Problem 3:

The cricket starts on one leaf and hops to one of the other 3 leaves each time. We want the probability that after 4 hops it is back where it started.

Let \(p_n\) be the probability that the cricket is on its starting leaf after \(n\) hops.

Initially:

\[
p_0=1
\]

After one hop, it cannot be on the starting leaf:

\[
p_1=0
\]

If the cricket is not on the starting leaf, then on the next hop it has probability \(\frac13\) of hopping back to the starting leaf. Therefore:

\[
p_{n+1}=\frac{1-p_n}{3}
\]

Now compute:

\[
p_2=\frac{1-0}{3}=\frac13
\]

\[
p_3=\frac{1-\frac13}{3}=\frac{\frac23}{3}=\frac29
\]

\[
p_4=\frac{1-\frac29}{3}=\frac{\frac79}{3}=\frac7{27}
\]

So the probability is:

\[
\frac7{27}
\]

Choice A, \(\frac29\), is the probability after 3 hops, not 4 hops.

ANSWER 3: E

Problem 4:

The letters and digits are cycled separately.

The letters are:

\[
AJHSME
\]

There are 6 letters, so the letters return to their original order after 6 cycles.

The digits are:

\[
1989
\]

There are 4 digits, so the digits return to their original order after 4 cycles.

We need both the letters and the digits to return to their original arrangements at the same time.

So we need the least common multiple of 6 and 4:

\[
\operatorname{lcm}(6,4)=12
\]

Thus \(AJHSME\ 1989\) appears again for the first time on line 12.

Choice A, 6, would return only the letters, not necessarily the digits. Choice B, 10, returns neither. Choice E, 24, is a common multiple but not the first one.

ANSWER 4: C

Problem 5:

Let the original price be \(P\).

Suppose the price is increased by \(r\), where \(r\) is written as a decimal. Then the new price is:

\[
P(1+r)
\]

Then it is decreased by the same percent, so the final price is:

\[
P(1+r)(1-r)
\]

We are told this equals 84% of the original price:

\[
P(1+r)(1-r)=0.84P
\]

Cancel \(P\):

\[
(1+r)(1-r)=0.84
\]

\[
1-r^2=0.84
\]

\[
r^2=0.16
\]

\[
r=0.4
\]

So the percent is:

\[
40\%
\]

A common mistake is to think an increase and decrease by the same percent cancel each other, but they do not because the second percent is taken from a different price.

ANSWER 5: E

Problem 6:

Harold tosses a nickel 4 times. We want the probability that he gets at least as many heads as tails.

There are \(2^4=16\) equally likely outcomes.

“At least as many heads as tails” means:

\[
H \ge T
\]

Since there are 4 tosses total, this means Harold must get 2, 3, or 4 heads.

Count the outcomes:

\[
\binom42=6
\]

\[
\binom43=4
\]

\[
\binom44=1
\]

Total favorable outcomes:

\[
6+4+1=11
\]

So the probability is:

\[
\frac{11}{16}
\]

The tempting answer \(\frac12\) would count only strictly more heads than tails and would forget the tie case.

ANSWER 6: E

Problem 7:

The machine rule is:

- If \(N\) is even, output \(\frac N2\).
- If \(N\) is odd, output \(3N+1\).

We need all starting positive integers \(N\) that become 1 after exactly 6 steps.

It is easiest to work backward from 1.

If a number outputs \(y\), then possible previous numbers are:

1. \(2y\), because if the previous number was even, halving it gives \(y\).
2. \(\frac{y-1}{3}\), but only if this is a positive odd integer.

Now work backward 6 steps.

Step 0:

\[
\{1\}
\]

Step 1 predecessors of 1:

\[
\{2\}
\]

Step 2 predecessors of 2:

\[
\{4\}
\]

Step 3 predecessors of 4:

\[
2\cdot 4=8
\]

and

\[
\frac{4-1}{3}=1
\]

So:

\[
\{8,1\}
\]

Step 4:

Predecessors of 8: \(16\)

Predecessors of 1: \(2\)

So:

\[
\{16,2\}
\]

Step 5:

Predecessors of 16:

\[
32,\quad \frac{16-1}{3}=5
\]

Predecessors of 2:

\[
4
\]

So:

\[
\{32,5,4\}
\]

Step 6:

Predecessors of 32:

\[
64
\]

Predecessors of 5:

\[
10
\]

Predecessors of 4:

\[
8,\quad 1
\]

So the possible starting values are:

\[
64,10,8,1
\]

Their sum is:

\[
64+10+8+1=83
\]

ANSWER 7: E

Problem 8:

Let the list be:

\[
a_1,a_2,a_3,a_4,a_5,a_6,a_7,a_8
\]

Each new number is the product of the two previous numbers:

\[
a_3=a_1a_2
\]

\[
a_4=a_2a_3
\]

and so on.

We are told the last three numbers are:

\[
a_6=16,\quad a_7=64,\quad a_8=1024
\]

Check:

\[
16\cdot 64=1024
\]

Now work backward.

Since:

\[
a_7=a_5a_6
\]

we have:

\[
64=

Problem 1:
We are asked to find the fraction of possible 7-digit telephone numbers that begin with 9 and end with 0, given that a telephone number cannot begin with 0 or 1.

First, let's find the total number of possible telephone numbers:
- The first digit can be any number from 2 to 9, which gives 8 possibilities.
- The remaining 6 digits can be any number from 0 to 9, giving 10 possibilities each.
Total possible telephone numbers = 8 × 10 × 10 × 10 × 10 × 10 × 10 = 8,000,000.

Next, let's find the number of telephone numbers that begin with 9 and end with 0:
- The first digit must be 9 (1 possibility).
- The last digit must be 0 (1 possibility).
- The middle 5 digits can be any number from 0 to 9 (10 possibilities each).
Total such numbers = 1 × 10 × 10 × 10 × 10 × 10 × 1 = 100,000.

The fraction of telephone numbers that meet this condition is:
100,000 / 8,000,000 = 1 / 80.

ANSWER 1: B

Problem 2:
We are asked to find the least possible value of M, which is the median of the medians of five groups of 5 numbers each, chosen from the integers 1 through 25.

Let the 5 groups be G1, G2, G3, G4, and G5, and let their respective medians be m1, m2, m3, m4, and m5. 
Assume the medians are ordered such that m1 < m2 < m3 < m4 < m5. The median of these medians is M = m3.
To minimize m3, we must make the numbers in the groups as small as possible. 

For m3 to be a certain value, there must be at least 3 groups (G1, G2, G3) whose medians are less than or equal to m3. 
In each of these 3 groups, there are 3 numbers that are less than or equal to the group's median. 
Therefore, there are at least 3 groups × 3 numbers = 9 numbers that are less than or equal to m3.

Since the numbers are distinct integers from 1 to 25, the 9th smallest possible integer is 9. Thus, m3 must be at least 9. 
We can verify that 9 is possible with the following grouping:
G1: 1, 2, 7, 24, 25 (median 7)
G2: 3, 4, 8, 22, 23 (median 8)
G3: 5, 6, 9, 20, 21 (median 9)
G4: 10, 11, 12, 18, 19 (median 12)
G5: 13, 14, 15, 16, 17 (median 15)
The medians are 7, 8, 9, 12, and 15. Their median is 9.

ANSWER 2: A

Problem 3:
We want to find the probability that a cricket, which hops to one of 3 other leaves with equal probability, returns to its starting leaf after exactly 4 hops.

Let the leaves be A, B, C, and D, and assume the cricket starts at leaf A.
Let P(n) be the probability that the cricket is at leaf A after n hops.
- P(0) = 1 (it starts at A).
- For any hop n > 0, the cricket can only land on A if it was NOT on A in the previous hop. If it was on B, C, or D, it has a 1/3 chance of hopping to A.
Thus, the recursive formula is P(n) = (1 - P(n-1)) × (1/3).

Let's calculate step-by-step:
- P(1) = (1 - 1) × 1/3 = 0
- P(2) = (1 - 0) × 1/3 = 1/3
- P(3) = (1 - 1/3) × 1/3 = (2/3) × (1/3) = 2/9
- P(4) = (1 - 2/9) × 1/3 = (7/9) × (1/3) = 7/27

ANSWER 3: E

Problem 4:
We are given a sequence of letters "AJHSME" (6 characters) and a sequence of digits "1989" (4 characters). They are cycled separately line by line. We want to find the line number where the original combination "AJHSME 1989" appears again for the first time.

The letters cycle back to their original arrangement every 6 lines.
The digits cycle back to their original arrangement every 4 lines.
For both to return to their original arrangement simultaneously, the line number must be a common multiple of both 6 and 4. 

The least common multiple (LCM) of 6 and 4 is 12.
Therefore, on line 12, the letters will have completed exactly 2 full cycles and the digits will have completed exactly 3 full cycles, returning to "AJHSME 1989".

ANSWER 4: C

Problem 5:
Let the original price of the shirt be P, and let the percentage of increase and decrease be x.
When the price is increased by x percent, it becomes P(1 + x/100).
When this new price is decreased by x percent, it becomes P(1 + x/100)(1 - x/100).

We are given that the resulting price is 84% of the original price, so:
P(1 + x/100)(1 - x/100) = 0.84P

Divide both sides by P:
(1 + x/100)(1 - x/100) = 0.84
1 - (x/100)² = 0.84
(x/100)² = 1 - 0.84
(x/100)² = 0.16

Taking the square root of both sides:
x/100 = 0.4
x = 40

The price was increased and decreased by 40%.

ANSWER 5: E

Problem 6:
Harold tosses a nickel four times. We want to find the probability that he gets at least as many heads as tails.
"At least as many heads as tails" in 4 tosses means getting 2, 3, or 4 heads.

The total number of possible outcomes when tossing a coin 4 times is 2⁴ = 16.
Using combinations, we find the number of ways to get 2, 3, or 4 heads:
- 2 heads: 4C2 = 6 ways
- 3 heads: 4C3 = 4 ways
- 4 heads: 4C4 = 1 way

Total favorable outcomes = 6 + 4 + 1 = 11.
The probability is 11 / 16.

ANSWER 6: E

Problem 7:
The machine follows the Collatz-like rule:
- If N is even, output N/2
- If N is odd, output 3N + 1

We are told a 6-step process ends in 1. Let's work backward from Step 6 to find all possible starting positive integers N (Step 0).
- Step 6: 1
- Step 5: Must be 2 (since 3x + 1 = 1 gives x = 0, which is not a positive integer).
- Step 4: Must be 4 (since 3x + 1 = 2 gives x = 1/3, not an integer).
- Step 3: Can be 8 (since 8/2 = 4) or 1 (since 3(1) + 1 = 4).

Branch 1 (Step 3 is 8):
- Step 2: Must be 16 (since 3x + 1 = 8 gives x = 7/3).
- Step 1: Can be 32 (32/2 = 16) or 5 (3(5) + 1 = 16).
- Step 0 (from 32): Must be 64 (since 3x + 1 = 32 gives x = 31/3).
- Step 0 (from 5): Must be 10 (since 3x + 1 = 5 gives x = 4/3).

Branch 2 (Step 3 is 1):
- Step 2: Must be 2.
- Step 1: Must be 4.
- Step 0: Can be 8 (8/2 = 4) or 1 (3(1) + 1 = 4).

The possible starting integers N are 64, 10, 8, and 1.
The sum of all such integers is 64 + 10 + 8 + 1 = 83.

ANSWER 7: E

Problem 8:
We are given a sequence of 8 numbers where each number is the product of the two preceding numbers: a_n = a_{n-2} × a_{n-1}.
We know the last three numbers: a_6 = 16, a_7 = 64, and a_8 = 1024.
Let's work backward using the rule a_{n-2} = a_n / a_{n-1}:
- a_5 = a_7 / a_6 = 64 / 16 = 4
- a_4 

## Step 1: Problem 1 - Understanding the problem
We need to find the fraction of 7-digit telephone numbers that begin with 9 and end with 0, given that the numbers cannot start with 0 or 1.

## Step 2: Problem 1 - Calculating total possible telephone numbers
The first digit can be any number from 2 to 9, so there are 8 possibilities. The remaining 6 digits can be any number from 0 to 9, so there are 10 possibilities for each. Thus, the total number of telephone numbers is $8 \times 10^6$.

## Step 3: Problem 1 - Calculating the number of telephone numbers that begin with 9 and end with 0
For a number to begin with 9 and end with 0, the first digit must be 9 and the last digit must be 0. The first digit has 1 possibility (9), the last digit has 1 possibility (0), and the remaining 5 digits have 10 possibilities each. So, the total number of such telephone numbers is $1 \times 10^5 \times 1$.

## Step 4: Problem 1 - Calculating the fraction
The fraction of telephone numbers that begin with 9 and end with 0 is $\frac{1 \times 10^5 \times 1}{8 \times 10^6} = \frac{10^5}{8 \times 10^6} = \frac{1}{80}$.

## Step 5: Problem 1 - Answer
ANSWER 1: B

## Step 6: Problem 2 - Understanding the problem
We are given 25 integers divided into 5 groups of 5 numbers each. We need to find the least possible value of M, the median of the medians of these groups.

## Step 7: Problem 2 - Analyzing the medians
To minimize M, we should minimize the medians of the 5 groups. The integers are from 1 to 25. The median of each group will be the third number when the numbers are arranged in ascending order.

## Step 8: Problem 2 - Distributing numbers to minimize M
To minimize M, we can distribute the numbers such that the medians of the groups are as small as possible. One way to do this is by grouping the smallest numbers together and ensuring that the medians are as low as possible.

## Step 9: Problem 2 - Calculating the medians of the groups
Let's distribute the numbers: Group 1 = {1, 2, 3, 24, 25}, Group 2 = {4, 5, 6, 22, 23}, Group 3 = {7, 8, 9, 20, 21}, Group 4 = {10, 11, 12, 18, 19}, Group 5 = {13, 14, 15, 16, 17}. The medians are 3, 6, 9, 12, 15.

## Step 10: Problem 2 - Finding M
The medians of the groups are 3, 6, 9, 12, 15. The median of these medians (M) is 9.

## Step 11: Problem 2 - Answer
ANSWER 2: A

## Step 12: Problem 3 - Understanding the problem
A cricket hops between 4 leaves with equal probability. We need to find the probability that it returns to the starting leaf after 4 hops.

## Step 13: Problem 3 - Analyzing the hops
After the first hop, the cricket is on a different leaf. For it to return to the starting leaf after 4 hops, it must be on a different leaf after the third hop as well.

## Step 14: Problem 3 - Calculating the probability
Let's denote the probability of being on the starting leaf after n hops as $P_n$. $P_0 = 1$, $P_1 = 0$. For $n \geq 1$, $P_{n+1} = \frac{1}{3}(1 - P_n)$ because the probability of not being on the starting leaf after n hops is $1 - P_n$, and from any other leaf, there's a $\frac{1}{3}$ chance of hopping to the starting leaf.

## Step 15: Problem 3 - Computing $P_4$
$P_2 = \frac{1}{3}$, $P_3 = \frac{1}{3}(1 - \frac{1}{3}) = \frac{2}{9}$, $P_4 = \frac{1}{3}(1 - \frac{2}{9}) = \frac{1}{3} \times \frac{7}{9} = \frac{7}{27}$.

## Step 16: Problem 3 - Answer
ANSWER 3: E

## Step 17: Problem 4 - Understanding the problem
The letters and digits are cycled separately and combined. We need to find when "AJHSME 1989" appears again.

## Step 18: Problem 4 - Analyzing the cycles
The letters cycle every 6 steps, and the digits cycle every 4 steps (since there are 4 digits and they are distinct when considering their positions). The combined cycle will be the least common multiple (LCM) of 6 and 4.

## Step 19: Problem 4 - Calculating the LCM of 6 and 4
The LCM of 6 and 4 is 12.

## Step 20: Problem 4 - Answer
ANSWER 4: C

## Step 21: Problem 5 - Understanding the problem
A price is increased by a certain percent and then decreased by the same percent, resulting in 84% of the original price.

## Step 22: Problem 5 - Setting up the equation
Let $x$ be the original price and $p$ be the percentage change. The new price after increase is $x(1 + \frac{p}{100})$, and after decrease is $x(1 + \frac{p}{100})(1 - \frac{p}{100}) = 0.84x$.

## Step 23: Problem 5 - Solving for $p$
$x(1 - \frac{p^2}{10000}) = 0.84x$. Simplifying gives $1 - \frac{p^2}{10000} = 0.84$. Thus, $\frac{p^2}{10000} = 0.16$, so $p^2 = 1600$, and $p = 40$.

## Step 24: Problem 5 - Answer
ANSWER 5: E

## Step 25: Problem 6 - Understanding the problem
Harold tosses a nickel four times. We need to find the probability of getting at least as many heads as tails.

## Step 26: Problem 6 - Enumerating the outcomes
The possible outcomes where heads are at least as many as tails are: 4 heads, 3 heads and 1 tail, and 2 heads and 2 tails.

## Step 27: Problem 6 - Calculating the probabilities
The total number of outcomes is $2^4 = 16$. The number of ways to get 4 heads is 1, 3 heads and 1 tail is $\binom{4}{3} = 4$, and 2 heads and 2 tails is $\binom{4}{2} = 6$. So, the total favorable outcomes are $1 + 4 + 6 = 11$.

## Step 28: Problem 6 - Calculating the probability
The probability is $\frac{11}{16}$.

## Step 29: Problem 6 - Answer
ANSWER 6: E

## Step 30: Problem 7 - Understanding the problem
A machine applies a rule to a positive integer N: if N is even, output N/2; if N is odd, output 3N + 1. We need to find the sum of all N that result in 1 after 6 steps.

## Step 31: Problem 7 - Working backwards from 1
After 6 steps, the output is 1. So, before that, it was 2 (since 2 is even and 2/2 = 1). Before that, it could be 4. Continuing this process backwards: 4 -> 2 -> 1 (3 steps), 8 -> 4 -> 2 -> 1 (4 steps), 16 -> 8 -> 4 -> 2 -> 1 (5 steps), and so on, until we reach 6 steps.

## Step 32: Problem 7 - Enumerating the possible N
For 6 steps: $2^6 = 64$ is a possible N (since 64 -> 32 -> 16 -> 8 -> 4 -> 2 -> 1). Working bac

Problem 1:

We are asked what fraction of possible telephone numbers begin with 9 and end with 0.

A 7-digit telephone number cannot begin with 0 or 1.

Total possible telephone numbers:

- First digit: 2 through 9, so 8 choices.
- Remaining 6 digits: each has 10 choices.

So total numbers:

\[
8 \cdot 10^6
\]

Now count numbers that begin with 9 and end with 0:

- First digit fixed as 9: 1 choice.
- Last digit fixed as 0: 1 choice.
- Middle 5 digits: each has 10 choices.

So favorable numbers:

\[
10^5
\]

Thus the fraction is:

\[
\frac{10^5}{8\cdot 10^6}=\frac{1}{80}
\]

The tempting answer \(\frac{1}{100}\) would be correct only if all first digits were equally allowed, including 0 and 1, but they are not.

ANSWER 1: B

Problem 2:

We have the integers 1 through 25 separated into five groups of 5 numbers each. We find the median of each group, then take the median of those five medians. We want the least possible value of \(M\).

For a group of 5 numbers to have median at most \(m\), at least 3 numbers in that group must be at most \(m\).

Since \(M\) is the median of the five medians, at least 3 of the group medians must be at most \(M\).

Therefore, to make \(M\) as small as possible, we need at least:

\[
3 \times 3 = 9
\]

numbers that are at most \(M\).

The 9 smallest numbers are:

\[
1,2,3,4,5,6,7,8,9
\]

So \(M\) cannot be less than 9.

Now we show that \(M=9\) is possible. For example:

\[
\{1,2,3,22,23\} \quad \text{median } 3
\]

\[
\{4,5,6,24,25\} \quad \text{median } 6
\]

\[
\{7,8,9,20,21\} \quad \text{median } 9
\]

The remaining numbers can form two groups with medians larger than 9:

\[
\{10,11,12,13,14\} \quad \text{median } 12
\]

\[
\{15,16,17,18,19\} \quad \text{median } 17
\]

The five medians are:

\[
3,6,9,12,17
\]

Their median is 9.

So the least possible value is 9.

ANSWER 2: A

Problem 3:

The cricket starts on one leaf and hops to one of the other 3 leaves each time. We want the probability that after 4 hops it is back where it started.

Let \(p_n\) be the probability that the cricket is on its starting leaf after \(n\) hops.

Initially:

\[
p_0=1
\]

After one hop, it cannot be on the starting leaf:

\[
p_1=0
\]

If the cricket is not on the starting leaf, then on the next hop it has probability \(\frac13\) of hopping back to the starting leaf. Therefore:

\[
p_{n+1}=\frac{1-p_n}{3}
\]

Now compute:

\[
p_2=\frac{1-0}{3}=\frac13
\]

\[
p_3=\frac{1-\frac13}{3}=\frac{\frac23}{3}=\frac29
\]

\[
p_4=\frac{1-\frac29}{3}=\frac{\frac79}{3}=\frac7{27}
\]

So the probability is:

\[
\frac7{27}
\]

Choice A, \(\frac29\), is the probability after 3 hops, not 4 hops.

ANSWER 3: E

Problem 4:

The letters and digits are cycled separately.

The letters are:

\[
AJHSME
\]

There are 6 letters, so the letters return to their original order after 6 cycles.

The digits are:

\[
1989
\]

There are 4 digits, so the digits return to their original order after 4 cycles.

We need both the letters and the digits to return to their original arrangements at the same time.

So we need the least common multiple of 6 and 4:

\[
\operatorname{lcm}(6,4)=12
\]

Thus \(AJHSME\ 1989\) appears again for the first time on line 12.

Choice A, 6, would return only the letters, not necessarily the digits. Choice B, 10, returns neither. Choice E, 24, is a common multiple but not the first one.

ANSWER 4: C

Problem 5:

Let the original price be \(P\).

Suppose the price is increased by \(r\), where \(r\) is written as a decimal. Then the new price is:

\[
P(1+r)
\]

Then it is decreased by the same percent, so the final price is:

\[
P(1+r)(1-r)
\]

We are told this equals 84% of the original price:

\[
P(1+r)(1-r)=0.84P
\]

Cancel \(P\):

\[
(1+r)(1-r)=0.84
\]

\[
1-r^2=0.84
\]

\[
r^2=0.16
\]

\[
r=0.4
\]

So the percent is:

\[
40\%
\]

A common mistake is to think an increase and decrease by the same percent cancel each other, but they do not because the second percent is taken from a different price.

ANSWER 5: E

Problem 6:

Harold tosses a nickel 4 times. We want the probability that he gets at least as many heads as tails.

There are \(2^4=16\) equally likely outcomes.

“At least as many heads as tails” means:

\[
H \ge T
\]

Since there are 4 tosses total, this means Harold must get 2, 3, or 4 heads.

Count the outcomes:

\[
\binom42=6
\]

\[
\binom43=4
\]

\[
\binom44=1
\]

Total favorable outcomes:

\[
6+4+1=11
\]

So the probability is:

\[
\frac{11}{16}
\]

The tempting answer \(\frac12\) would count only strictly more heads than tails and would forget the tie case.

ANSWER 6: E

Problem 7:

The machine rule is:

- If \(N\) is even, output \(\frac N2\).
- If \(N\) is odd, output \(3N+1\).

We need all starting positive integers \(N\) that become 1 after exactly 6 steps.

It is easiest to work backward from 1.

If a number outputs \(y\), then possible previous numbers are:

1. \(2y\), because if the previous number was even, halving it gives \(y\).
2. \(\frac{y-1}{3}\), but only if this is a positive odd integer.

Now work backward 6 steps.

Step 0:

\[
\{1\}
\]

Step 1 predecessors of 1:

\[
\{2\}
\]

Step 2 predecessors of 2:

\[
\{4\}
\]

Step 3 predecessors of 4:

\[
2\cdot 4=8
\]

and

\[
\frac{4-1}{3}=1
\]

So:

\[
\{8,1\}
\]

Step 4:

Predecessors of 8: \(16\)

Predecessors of 1: \(2\)

So:

\[
\{16,2\}
\]

Step 5:

Predecessors of 16:

\[
32,\quad \frac{16-1}{3}=5
\]

Predecessors of 2:

\[
4
\]

So:

\[
\{32,5,4\}
\]

Step 6:

Predecessors of 32:

\[
64
\]

Predecessors of 5:

\[
10
\]

Predecessors of 4:

\[
8,\quad 1
\]

So the possible starting values are:

\[
64,10,8,1
\]

Their sum is:

\[
64+10+8+1=83
\]

ANSWER 7: E

Problem 8:

Let the list be:

\[
a_1,a_2,a_3,a_4,a_5,a_6,a_7,a_8
\]

Each new number is the product of the two previous numbers:

\[
a_3=a_1a_2
\]

\[
a_4=a_2a_3
\]

and so on.

We are told the last three numbers are:

\[
a_6=16,\quad a_7=64,\quad a_8=1024
\]

Check:

\[
16\cdot 64=1024
\]

Now work backward.

Since:

\[
a_7=a_5a_6
\]

we have:

\[
64=

Problem 1:
We are asked to find the fraction of possible 7-digit telephone numbers that begin with 9 and end with 0, given that a telephone number cannot begin with 0 or 1.

First, let's find the total number of possible telephone numbers:
- The first digit can be any number from 2 to 9, which gives 8 possibilities.
- The remaining 6 digits can be any number from 0 to 9, giving 10 possibilities each.
Total possible telephone numbers = 8 × 10 × 10 × 10 × 10 × 10 × 10 = 8,000,000.

Next, let's find the number of telephone numbers that begin with 9 and end with 0:
- The first digit must be 9 (1 possibility).
- The last digit must be 0 (1 possibility).
- The middle 5 digits can be any number from 0 to 9 (10 possibilities each).
Total such numbers = 1 × 10 × 10 × 10 × 10 × 10 × 1 = 100,000.

The fraction of telephone numbers that meet this condition is:
100,000 / 8,000,000 = 1 / 80.

ANSWER 1: B

Problem 2:
We are asked to find the least possible value of M, which is the median of the medians of five groups of 5 numbers each, chosen from the integers 1 through 25.

Let the 5 groups be G1, G2, G3, G4, and G5, and let their respective medians be m1, m2, m3, m4, and m5. 
Assume the medians are ordered such that m1 < m2 < m3 < m4 < m5. The median of these medians is M = m3.
To minimize m3, we must make the numbers in the groups as small as possible. 

For m3 to be a certain value, there must be at least 3 groups (G1, G2, G3) whose medians are less than or equal to m3. 
In each of these 3 groups, there are 3 numbers that are less than or equal to the group's median. 
Therefore, there are at least 3 groups × 3 numbers = 9 numbers that are less than or equal to m3.

Since the numbers are distinct integers from 1 to 25, the 9th smallest possible integer is 9. Thus, m3 must be at least 9. 
We can verify that 9 is possible with the following grouping:
G1: 1, 2, 7, 24, 25 (median 7)
G2: 3, 4, 8, 22, 23 (median 8)
G3: 5, 6, 9, 20, 21 (median 9)
G4: 10, 11, 12, 18, 19 (median 12)
G5: 13, 14, 15, 16, 17 (median 15)
The medians are 7, 8, 9, 12, and 15. Their median is 9.

ANSWER 2: A

Problem 3:
We want to find the probability that a cricket, which hops to one of 3 other leaves with equal probability, returns to its starting leaf after exactly 4 hops.

Let the leaves be A, B, C, and D, and assume the cricket starts at leaf A.
Let P(n) be the probability that the cricket is at leaf A after n hops.
- P(0) = 1 (it starts at A).
- For any hop n > 0, the cricket can only land on A if it was NOT on A in the previous hop. If it was on B, C, or D, it has a 1/3 chance of hopping to A.
Thus, the recursive formula is P(n) = (1 - P(n-1)) × (1/3).

Let's calculate step-by-step:
- P(1) = (1 - 1) × 1/3 = 0
- P(2) = (1 - 0) × 1/3 = 1/3
- P(3) = (1 - 1/3) × 1/3 = (2/3) × (1/3) = 2/9
- P(4) = (1 - 2/9) × 1/3 = (7/9) × (1/3) = 7/27

ANSWER 3: E

Problem 4:
We are given a sequence of letters "AJHSME" (6 characters) and a sequence of digits "1989" (4 characters). They are cycled separately line by line. We want to find the line number where the original combination "AJHSME 1989" appears again for the first time.

The letters cycle back to their original arrangement every 6 lines.
The digits cycle back to their original arrangement every 4 lines.
For both to return to their original arrangement simultaneously, the line number must be a common multiple of both 6 and 4. 

The least common multiple (LCM) of 6 and 4 is 12.
Therefore, on line 12, the letters will have completed exactly 2 full cycles and the digits will have completed exactly 3 full cycles, returning to "AJHSME 1989".

ANSWER 4: C

Problem 5:
Let the original price of the shirt be P, and let the percentage of increase and decrease be x.
When the price is increased by x percent, it becomes P(1 + x/100).
When this new price is decreased by x percent, it becomes P(1 + x/100)(1 - x/100).

We are given that the resulting price is 84% of the original price, so:
P(1 + x/100)(1 - x/100) = 0.84P

Divide both sides by P:
(1 + x/100)(1 - x/100) = 0.84
1 - (x/100)² = 0.84
(x/100)² = 1 - 0.84
(x/100)² = 0.16

Taking the square root of both sides:
x/100 = 0.4
x = 40

The price was increased and decreased by 40%.

ANSWER 5: E

Problem 6:
Harold tosses a nickel four times. We want to find the probability that he gets at least as many heads as tails.
"At least as many heads as tails" in 4 tosses means getting 2, 3, or 4 heads.

The total number of possible outcomes when tossing a coin 4 times is 2⁴ = 16.
Using combinations, we find the number of ways to get 2, 3, or 4 heads:
- 2 heads: 4C2 = 6 ways
- 3 heads: 4C3 = 4 ways
- 4 heads: 4C4 = 1 way

Total favorable outcomes = 6 + 4 + 1 = 11.
The probability is 11 / 16.

ANSWER 6: E

Problem 7:
The machine follows the Collatz-like rule:
- If N is even, output N/2
- If N is odd, output 3N + 1

We are told a 6-step process ends in 1. Let's work backward from Step 6 to find all possible starting positive integers N (Step 0).
- Step 6: 1
- Step 5: Must be 2 (since 3x + 1 = 1 gives x = 0, which is not a positive integer).
- Step 4: Must be 4 (since 3x + 1 = 2 gives x = 1/3, not an integer).
- Step 3: Can be 8 (since 8/2 = 4) or 1 (since 3(1) + 1 = 4).

Branch 1 (Step 3 is 8):
- Step 2: Must be 16 (since 3x + 1 = 8 gives x = 7/3).
- Step 1: Can be 32 (32/2 = 16) or 5 (3(5) + 1 = 16).
- Step 0 (from 32): Must be 64 (since 3x + 1 = 32 gives x = 31/3).
- Step 0 (from 5): Must be 10 (since 3x + 1 = 5 gives x = 4/3).

Branch 2 (Step 3 is 1):
- Step 2: Must be 2.
- Step 1: Must be 4.
- Step 0: Can be 8 (8/2 = 4) or 1 (3(1) + 1 = 4).

The possible starting integers N are 64, 10, 8, and 1.
The sum of all such integers is 64 + 10 + 8 + 1 = 83.

ANSWER 7: E

Problem 8:
We are given a sequence of 8 numbers where each number is the product of the two preceding numbers: a_n = a_{n-2} × a_{n-1}.
We know the last three numbers: a_6 = 16, a_7 = 64, and a_8 = 1024.
Let's work backward using the rule a_{n-2} = a_n / a_{n-1}:
- a_5 = a_7 / a_6 = 64 / 16 = 4
- a_4 

## Step 1: Problem 1 - Understanding the problem
We need to find the fraction of 7-digit telephone numbers that begin with 9 and end with 0, given that the numbers cannot start with 0 or 1.

## Step 2: Problem 1 - Calculating total possible telephone numbers
The first digit can be any number from 2 to 9, so there are 8 possibilities. The remaining 6 digits can be any number from 0 to 9, so there are 10 possibilities for each. Thus, the total number of telephone numbers is $8 \times 10^6$.

## Step 3: Problem 1 - Calculating the number of telephone numbers that begin with 9 and end with 0
For a number to begin with 9 and end with 0, the first digit must be 9 and the last digit must be 0. The first digit has 1 possibility (9), the last digit has 1 possibility (0), and the remaining 5 digits have 10 possibilities each. So, the total number of such telephone numbers is $1 \times 10^5 \times 1$.

## Step 4: Problem 1 - Calculating the fraction
The fraction of telephone numbers that begin with 9 and end with 0 is $\frac{1 \times 10^5 \times 1}{8 \times 10^6} = \frac{10^5}{8 \times 10^6} = \frac{1}{80}$.

## Step 5: Problem 1 - Answer
ANSWER 1: B

## Step 6: Problem 2 - Understanding the problem
We are given 25 integers divided into 5 groups of 5 numbers each. We need to find the least possible value of M, the median of the medians of these groups.

## Step 7: Problem 2 - Analyzing the medians
To minimize M, we should minimize the medians of the 5 groups. The integers are from 1 to 25. The median of each group will be the third number when the numbers are arranged in ascending order.

## Step 8: Problem 2 - Distributing numbers to minimize M
To minimize M, we can distribute the numbers such that the medians of the groups are as small as possible. One way to do this is by grouping the smallest numbers together and ensuring that the medians are as low as possible.

## Step 9: Problem 2 - Calculating the medians of the groups
Let's distribute the numbers: Group 1 = {1, 2, 3, 24, 25}, Group 2 = {4, 5, 6, 22, 23}, Group 3 = {7, 8, 9, 20, 21}, Group 4 = {10, 11, 12, 18, 19}, Group 5 = {13, 14, 15, 16, 17}. The medians are 3, 6, 9, 12, 15.

## Step 10: Problem 2 - Finding M
The medians of the groups are 3, 6, 9, 12, 15. The median of these medians (M) is 9.

## Step 11: Problem 2 - Answer
ANSWER 2: A

## Step 12: Problem 3 - Understanding the problem
A cricket hops between 4 leaves with equal probability. We need to find the probability that it returns to the starting leaf after 4 hops.

## Step 13: Problem 3 - Analyzing the hops
After the first hop, the cricket is on a different leaf. For it to return to the starting leaf after 4 hops, it must be on a different leaf after the third hop as well.

## Step 14: Problem 3 - Calculating the probability
Let's denote the probability of being on the starting leaf after n hops as $P_n$. $P_0 = 1$, $P_1 = 0$. For $n \geq 1$, $P_{n+1} = \frac{1}{3}(1 - P_n)$ because the probability of not being on the starting leaf after n hops is $1 - P_n$, and from any other leaf, there's a $\frac{1}{3}$ chance of hopping to the starting leaf.

## Step 15: Problem 3 - Computing $P_4$
$P_2 = \frac{1}{3}$, $P_3 = \frac{1}{3}(1 - \frac{1}{3}) = \frac{2}{9}$, $P_4 = \frac{1}{3}(1 - \frac{2}{9}) = \frac{1}{3} \times \frac{7}{9} = \frac{7}{27}$.

## Step 16: Problem 3 - Answer
ANSWER 3: E

## Step 17: Problem 4 - Understanding the problem
The letters and digits are cycled separately and combined. We need to find when "AJHSME 1989" appears again.

## Step 18: Problem 4 - Analyzing the cycles
The letters cycle every 6 steps, and the digits cycle every 4 steps (since there are 4 digits and they are distinct when considering their positions). The combined cycle will be the least common multiple (LCM) of 6 and 4.

## Step 19: Problem 4 - Calculating the LCM of 6 and 4
The LCM of 6 and 4 is 12.

## Step 20: Problem 4 - Answer
ANSWER 4: C

## Step 21: Problem 5 - Understanding the problem
A price is increased by a certain percent and then decreased by the same percent, resulting in 84% of the original price.

## Step 22: Problem 5 - Setting up the equation
Let $x$ be the original price and $p$ be the percentage change. The new price after increase is $x(1 + \frac{p}{100})$, and after decrease is $x(1 + \frac{p}{100})(1 - \frac{p}{100}) = 0.84x$.

## Step 23: Problem 5 - Solving for $p$
$x(1 - \frac{p^2}{10000}) = 0.84x$. Simplifying gives $1 - \frac{p^2}{10000} = 0.84$. Thus, $\frac{p^2}{10000} = 0.16$, so $p^2 = 1600$, and $p = 40$.

## Step 24: Problem 5 - Answer
ANSWER 5: E

## Step 25: Problem 6 - Understanding the problem
Harold tosses a nickel four times. We need to find the probability of getting at least as many heads as tails.

## Step 26: Problem 6 - Enumerating the outcomes
The possible outcomes where heads are at least as many as tails are: 4 heads, 3 heads and 1 tail, and 2 heads and 2 tails.

## Step 27: Problem 6 - Calculating the probabilities
The total number of outcomes is $2^4 = 16$. The number of ways to get 4 heads is 1, 3 heads and 1 tail is $\binom{4}{3} = 4$, and 2 heads and 2 tails is $\binom{4}{2} = 6$. So, the total favorable outcomes are $1 + 4 + 6 = 11$.

## Step 28: Problem 6 - Calculating the probability
The probability is $\frac{11}{16}$.

## Step 29: Problem 6 - Answer
ANSWER 6: E

## Step 30: Problem 7 - Understanding the problem
A machine applies a rule to a positive integer N: if N is even, output N/2; if N is odd, output 3N + 1. We need to find the sum of all N that result in 1 after 6 steps.

## Step 31: Problem 7 - Working backwards from 1
After 6 steps, the output is 1. So, before that, it was 2 (since 2 is even and 2/2 = 1). Before that, it could be 4. Continuing this process backwards: 4 -> 2 -> 1 (3 steps), 8 -> 4 -> 2 -> 1 (4 steps), 16 -> 8 -> 4 -> 2 -> 1 (5 steps), and so on, until we reach 6 steps.

## Step 32: Problem 7 - Enumerating the possible N
For 6 steps: $2^6 = 64$ is a possible N (since 64 -> 32 -> 16 -> 8 -> 4 -> 2 -> 1). Working bac

Problem 1:

We are asked what fraction of possible telephone numbers begin with 9 and end with 0.

A 7-digit telephone number cannot begin with 0 or 1.

Total possible telephone numbers:

- First digit: 2 through 9, so 8 choices.
- Remaining 6 digits: each has 10 choices.

So total numbers:

\[
8 \cdot 10^6
\]

Now count numbers that begin with 9 and end with 0:

- First digit fixed as 9: 1 choice.
- Last digit fixed as 0: 1 choice.
- Middle 5 digits: each has 10 choices.

So favorable numbers:

\[
10^5
\]

Thus the fraction is:

\[
\frac{10^5}{8\cdot 10^6}=\frac{1}{80}
\]

The tempting answer \(\frac{1}{100}\) would be correct only if all first digits were equally allowed, including 0 and 1, but they are not.

ANSWER 1: B

Problem 2:

We have the integers 1 through 25 separated into five groups of 5 numbers each. We find the median of each group, then take the median of those five medians. We want the least possible value of \(M\).

For a group of 5 numbers to have median at most \(m\), at least 3 numbers in that group must be at most \(m\).

Since \(M\) is the median of the five medians, at least 3 of the group medians must be at most \(M\).

Therefore, to make \(M\) as small as possible, we need at least:

\[
3 \times 3 = 9
\]

numbers that are at most \(M\).

The 9 smallest numbers are:

\[
1,2,3,4,5,6,7,8,9
\]

So \(M\) cannot be less than 9.

Now we show that \(M=9\) is possible. For example:

\[
\{1,2,3,22,23\} \quad \text{median } 3
\]

\[
\{4,5,6,24,25\} \quad \text{median } 6
\]

\[
\{7,8,9,20,21\} \quad \text{median } 9
\]

The remaining numbers can form two groups with medians larger than 9:

\[
\{10,11,12,13,14\} \quad \text{median } 12
\]

\[
\{15,16,17,18,19\} \quad \text{median } 17
\]

The five medians are:

\[
3,6,9,12,17
\]

Their median is 9.

So the least possible value is 9.

ANSWER 2: A

Problem 3:

The cricket starts on one leaf and hops to one of the other 3 leaves each time. We want the probability that after 4 hops it is back where it started.

Let \(p_n\) be the probability that the cricket is on its starting leaf after \(n\) hops.

Initially:

\[
p_0=1
\]

After one hop, it cannot be on the starting leaf:

\[
p_1=0
\]

If the cricket is not on the starting leaf, then on the next hop it has probability \(\frac13\) of hopping back to the starting leaf. Therefore:

\[
p_{n+1}=\frac{1-p_n}{3}
\]

Now compute:

\[
p_2=\frac{1-0}{3}=\frac13
\]

\[
p_3=\frac{1-\frac13}{3}=\frac{\frac23}{3}=\frac29
\]

\[
p_4=\frac{1-\frac29}{3}=\frac{\frac79}{3}=\frac7{27}
\]

So the probability is:

\[
\frac7{27}
\]

Choice A, \(\frac29\), is the probability after 3 hops, not 4 hops.

ANSWER 3: E

Problem 4:

The letters and digits are cycled separately.

The letters are:

\[
AJHSME
\]

There are 6 letters, so the letters return to their original order after 6 cycles.

The digits are:

\[
1989
\]

There are 4 digits, so the digits return to their original order after 4 cycles.

We need both the letters and the digits to return to their original arrangements at the same time.

So we need the least common multiple of 6 and 4:

\[
\operatorname{lcm}(6,4)=12
\]

Thus \(AJHSME\ 1989\) appears again for the first time on line 12.

Choice A, 6, would return only the letters, not necessarily the digits. Choice B, 10, returns neither. Choice E, 24, is a common multiple but not the first one.

ANSWER 4: C

Problem 5:

Let the original price be \(P\).

Suppose the price is increased by \(r\), where \(r\) is written as a decimal. Then the new price is:

\[
P(1+r)
\]

Then it is decreased by the same percent, so the final price is:

\[
P(1+r)(1-r)
\]

We are told this equals 84% of the original price:

\[
P(1+r)(1-r)=0.84P
\]

Cancel \(P\):

\[
(1+r)(1-r)=0.84
\]

\[
1-r^2=0.84
\]

\[
r^2=0.16
\]

\[
r=0.4
\]

So the percent is:

\[
40\%
\]

A common mistake is to think an increase and decrease by the same percent cancel each other, but they do not because the second percent is taken from a different price.

ANSWER 5: E

Problem 6:

Harold tosses a nickel 4 times. We want the probability that he gets at least as many heads as tails.

There are \(2^4=16\) equally likely outcomes.

“At least as many heads as tails” means:

\[
H \ge T
\]

Since there are 4 tosses total, this means Harold must get 2, 3, or 4 heads.

Count the outcomes:

\[
\binom42=6
\]

\[
\binom43=4
\]

\[
\binom44=1
\]

Total favorable outcomes:

\[
6+4+1=11
\]

So the probability is:

\[
\frac{11}{16}
\]

The tempting answer \(\frac12\) would count only strictly more heads than tails and would forget the tie case.

ANSWER 6: E

Problem 7:

The machine rule is:

- If \(N\) is even, output \(\frac N2\).
- If \(N\) is odd, output \(3N+1\).

We need all starting positive integers \(N\) that become 1 after exactly 6 steps.

It is easiest to work backward from 1.

If a number outputs \(y\), then possible previous numbers are:

1. \(2y\), because if the previous number was even, halving it gives \(y\).
2. \(\frac{y-1}{3}\), but only if this is a positive odd integer.

Now work backward 6 steps.

Step 0:

\[
\{1\}
\]

Step 1 predecessors of 1:

\[
\{2\}
\]

Step 2 predecessors of 2:

\[
\{4\}
\]

Step 3 predecessors of 4:

\[
2\cdot 4=8
\]

and

\[
\frac{4-1}{3}=1
\]

So:

\[
\{8,1\}
\]

Step 4:

Predecessors of 8: \(16\)

Predecessors of 1: \(2\)

So:

\[
\{16,2\}
\]

Step 5:

Predecessors of 16:

\[
32,\quad \frac{16-1}{3}=5
\]

Predecessors of 2:

\[
4
\]

So:

\[
\{32,5,4\}
\]

Step 6:

Predecessors of 32:

\[
64
\]

Predecessors of 5:

\[
10
\]

Predecessors of 4:

\[
8,\quad 1
\]

So the possible starting values are:

\[
64,10,8,1
\]

Their sum is:

\[
64+10+8+1=83
\]

ANSWER 7: E

Problem 8:

Let the list be:

\[
a_1,a_2,a_3,a_4,a_5,a_6,a_7,a_8
\]

Each new number is the product of the two previous numbers:

\[
a_3=a_1a_2
\]

\[
a_4=a_2a_3
\]

and so on.

We are told the last three numbers are:

\[
a_6=16,\quad a_7=64,\quad a_8=1024
\]

Check:

\[
16\cdot 64=1024
\]

Now work backward.

Since:

\[
a_7=a_5a_6
\]

we have:

\[
64=

Problem 1:
We are asked to find the fraction of possible 7-digit telephone numbers that begin with 9 and end with 0, given that a telephone number cannot begin with 0 or 1.

First, let's find the total number of possible telephone numbers:
- The first digit can be any number from 2 to 9, which gives 8 possibilities.
- The remaining 6 digits can be any number from 0 to 9, giving 10 possibilities each.
Total possible telephone numbers = 8 × 10 × 10 × 10 × 10 × 10 × 10 = 8,000,000.

Next, let's find the number of telephone numbers that begin with 9 and end with 0:
- The first digit must be 9 (1 possibility).
- The last digit must be 0 (1 possibility).
- The middle 5 digits can be any number from 0 to 9 (10 possibilities each).
Total such numbers = 1 × 10 × 10 × 10 × 10 × 10 × 1 = 100,000.

The fraction of telephone numbers that meet this condition is:
100,000 / 8,000,000 = 1 / 80.

ANSWER 1: B

Problem 2:
We are asked to find the least possible value of M, which is the median of the medians of five groups of 5 numbers each, chosen from the integers 1 through 25.

Let the 5 groups be G1, G2, G3, G4, and G5, and let their respective medians be m1, m2, m3, m4, and m5. 
Assume the medians are ordered such that m1 < m2 < m3 < m4 < m5. The median of these medians is M = m3.
To minimize m3, we must make the numbers in the groups as small as possible. 

For m3 to be a certain value, there must be at least 3 groups (G1, G2, G3) whose medians are less than or equal to m3. 
In each of these 3 groups, there are 3 numbers that are less than or equal to the group's median. 
Therefore, there are at least 3 groups × 3 numbers = 9 numbers that are less than or equal to m3.

Since the numbers are distinct integers from 1 to 25, the 9th smallest possible integer is 9. Thus, m3 must be at least 9. 
We can verify that 9 is possible with the following grouping:
G1: 1, 2, 7, 24, 25 (median 7)
G2: 3, 4, 8, 22, 23 (median 8)
G3: 5, 6, 9, 20, 21 (median 9)
G4: 10, 11, 12, 18, 19 (median 12)
G5: 13, 14, 15, 16, 17 (median 15)
The medians are 7, 8, 9, 12, and 15. Their median is 9.

ANSWER 2: A

Problem 3:
We want to find the probability that a cricket, which hops to one of 3 other leaves with equal probability, returns to its starting leaf after exactly 4 hops.

Let the leaves be A, B, C, and D, and assume the cricket starts at leaf A.
Let P(n) be the probability that the cricket is at leaf A after n hops.
- P(0) = 1 (it starts at A).
- For any hop n > 0, the cricket can only land on A if it was NOT on A in the previous hop. If it was on B, C, or D, it has a 1/3 chance of hopping to A.
Thus, the recursive formula is P(n) = (1 - P(n-1)) × (1/3).

Let's calculate step-by-step:
- P(1) = (1 - 1) × 1/3 = 0
- P(2) = (1 - 0) × 1/3 = 1/3
- P(3) = (1 - 1/3) × 1/3 = (2/3) × (1/3) = 2/9
- P(4) = (1 - 2/9) × 1/3 = (7/9) × (1/3) = 7/27

ANSWER 3: E

Problem 4:
We are given a sequence of letters "AJHSME" (6 characters) and a sequence of digits "1989" (4 characters). They are cycled separately line by line. We want to find the line number where the original combination "AJHSME 1989" appears again for the first time.

The letters cycle back to their original arrangement every 6 lines.
The digits cycle back to their original arrangement every 4 lines.
For both to return to their original arrangement simultaneously, the line number must be a common multiple of both 6 and 4. 

The least common multiple (LCM) of 6 and 4 is 12.
Therefore, on line 12, the letters will have completed exactly 2 full cycles and the digits will have completed exactly 3 full cycles, returning to "AJHSME 1989".

ANSWER 4: C

Problem 5:
Let the original price of the shirt be P, and let the percentage of increase and decrease be x.
When the price is increased by x percent, it becomes P(1 + x/100).
When this new price is decreased by x percent, it becomes P(1 + x/100)(1 - x/100).

We are given that the resulting price is 84% of the original price, so:
P(1 + x/100)(1 - x/100) = 0.84P

Divide both sides by P:
(1 + x/100)(1 - x/100) = 0.84
1 - (x/100)² = 0.84
(x/100)² = 1 - 0.84
(x/100)² = 0.16

Taking the square root of both sides:
x/100 = 0.4
x = 40

The price was increased and decreased by 40%.

ANSWER 5: E

Problem 6:
Harold tosses a nickel four times. We want to find the probability that he gets at least as many heads as tails.
"At least as many heads as tails" in 4 tosses means getting 2, 3, or 4 heads.

The total number of possible outcomes when tossing a coin 4 times is 2⁴ = 16.
Using combinations, we find the number of ways to get 2, 3, or 4 heads:
- 2 heads: 4C2 = 6 ways
- 3 heads: 4C3 = 4 ways
- 4 heads: 4C4 = 1 way

Total favorable outcomes = 6 + 4 + 1 = 11.
The probability is 11 / 16.

ANSWER 6: E

Problem 7:
The machine follows the Collatz-like rule:
- If N is even, output N/2
- If N is odd, output 3N + 1

We are told a 6-step process ends in 1. Let's work backward from Step 6 to find all possible starting positive integers N (Step 0).
- Step 6: 1
- Step 5: Must be 2 (since 3x + 1 = 1 gives x = 0, which is not a positive integer).
- Step 4: Must be 4 (since 3x + 1 = 2 gives x = 1/3, not an integer).
- Step 3: Can be 8 (since 8/2 = 4) or 1 (since 3(1) + 1 = 4).

Branch 1 (Step 3 is 8):
- Step 2: Must be 16 (since 3x + 1 = 8 gives x = 7/3).
- Step 1: Can be 32 (32/2 = 16) or 5 (3(5) + 1 = 16).
- Step 0 (from 32): Must be 64 (since 3x + 1 = 32 gives x = 31/3).
- Step 0 (from 5): Must be 10 (since 3x + 1 = 5 gives x = 4/3).

Branch 2 (Step 3 is 1):
- Step 2: Must be 2.
- Step 1: Must be 4.
- Step 0: Can be 8 (8/2 = 4) or 1 (3(1) + 1 = 4).

The possible starting integers N are 64, 10, 8, and 1.
The sum of all such integers is 64 + 10 + 8 + 1 = 83.

ANSWER 7: E

Problem 8:
We are given a sequence of 8 numbers where each number is the product of the two preceding numbers: a_n = a_{n-2} × a_{n-1}.
We know the last three numbers: a_6 = 16, a_7 = 64, and a_8 = 1024.
Let's work backward using the rule a_{n-2} = a_n / a_{n-1}:
- a_5 = a_7 / a_6 = 64 / 16 = 4
- a_4 

## Step 1: Problem 1 - Understanding the problem
We need to find the fraction of 7-digit telephone numbers that begin with 9 and end with 0, given that the numbers cannot start with 0 or 1.

## Step 2: Problem 1 - Calculating total possible telephone numbers
The first digit can be any number from 2 to 9, so there are 8 possibilities. The remaining 6 digits can be any number from 0 to 9, so there are 10 possibilities for each. Thus, the total number of telephone numbers is $8 \times 10^6$.

## Step 3: Problem 1 - Calculating the number of telephone numbers that begin with 9 and end with 0
For a number to begin with 9 and end with 0, the first digit must be 9 and the last digit must be 0. The first digit has 1 possibility (9), the last digit has 1 possibility (0), and the remaining 5 digits have 10 possibilities each. So, the total number of such telephone numbers is $1 \times 10^5 \times 1$.

## Step 4: Problem 1 - Calculating the fraction
The fraction of telephone numbers that begin with 9 and end with 0 is $\frac{1 \times 10^5 \times 1}{8 \times 10^6} = \frac{10^5}{8 \times 10^6} = \frac{1}{80}$.

## Step 5: Problem 1 - Answer
ANSWER 1: B

## Step 6: Problem 2 - Understanding the problem
We are given 25 integers divided into 5 groups of 5 numbers each. We need to find the least possible value of M, the median of the medians of these groups.

## Step 7: Problem 2 - Analyzing the medians
To minimize M, we should minimize the medians of the 5 groups. The integers are from 1 to 25. The median of each group will be the third number when the numbers are arranged in ascending order.

## Step 8: Problem 2 - Distributing numbers to minimize M
To minimize M, we can distribute the numbers such that the medians of the groups are as small as possible. One way to do this is by grouping the smallest numbers together and ensuring that the medians are as low as possible.

## Step 9: Problem 2 - Calculating the medians of the groups
Let's distribute the numbers: Group 1 = {1, 2, 3, 24, 25}, Group 2 = {4, 5, 6, 22, 23}, Group 3 = {7, 8, 9, 20, 21}, Group 4 = {10, 11, 12, 18, 19}, Group 5 = {13, 14, 15, 16, 17}. The medians are 3, 6, 9, 12, 15.

## Step 10: Problem 2 - Finding M
The medians of the groups are 3, 6, 9, 12, 15. The median of these medians (M) is 9.

## Step 11: Problem 2 - Answer
ANSWER 2: A

## Step 12: Problem 3 - Understanding the problem
A cricket hops between 4 leaves with equal probability. We need to find the probability that it returns to the starting leaf after 4 hops.

## Step 13: Problem 3 - Analyzing the hops
After the first hop, the cricket is on a different leaf. For it to return to the starting leaf after 4 hops, it must be on a different leaf after the third hop as well.

## Step 14: Problem 3 - Calculating the probability
Let's denote the probability of being on the starting leaf after n hops as $P_n$. $P_0 = 1$, $P_1 = 0$. For $n \geq 1$, $P_{n+1} = \frac{1}{3}(1 - P_n)$ because the probability of not being on the starting leaf after n hops is $1 - P_n$, and from any other leaf, there's a $\frac{1}{3}$ chance of hopping to the starting leaf.

## Step 15: Problem 3 - Computing $P_4$
$P_2 = \frac{1}{3}$, $P_3 = \frac{1}{3}(1 - \frac{1}{3}) = \frac{2}{9}$, $P_4 = \frac{1}{3}(1 - \frac{2}{9}) = \frac{1}{3} \times \frac{7}{9} = \frac{7}{27}$.

## Step 16: Problem 3 - Answer
ANSWER 3: E

## Step 17: Problem 4 - Understanding the problem
The letters and digits are cycled separately and combined. We need to find when "AJHSME 1989" appears again.

## Step 18: Problem 4 - Analyzing the cycles
The letters cycle every 6 steps, and the digits cycle every 4 steps (since there are 4 digits and they are distinct when considering their positions). The combined cycle will be the least common multiple (LCM) of 6 and 4.

## Step 19: Problem 4 - Calculating the LCM of 6 and 4
The LCM of 6 and 4 is 12.

## Step 20: Problem 4 - Answer
ANSWER 4: C

## Step 21: Problem 5 - Understanding the problem
A price is increased by a certain percent and then decreased by the same percent, resulting in 84% of the original price.

## Step 22: Problem 5 - Setting up the equation
Let $x$ be the original price and $p$ be the percentage change. The new price after increase is $x(1 + \frac{p}{100})$, and after decrease is $x(1 + \frac{p}{100})(1 - \frac{p}{100}) = 0.84x$.

## Step 23: Problem 5 - Solving for $p$
$x(1 - \frac{p^2}{10000}) = 0.84x$. Simplifying gives $1 - \frac{p^2}{10000} = 0.84$. Thus, $\frac{p^2}{10000} = 0.16$, so $p^2 = 1600$, and $p = 40$.

## Step 24: Problem 5 - Answer
ANSWER 5: E

## Step 25: Problem 6 - Understanding the problem
Harold tosses a nickel four times. We need to find the probability of getting at least as many heads as tails.

## Step 26: Problem 6 - Enumerating the outcomes
The possible outcomes where heads are at least as many as tails are: 4 heads, 3 heads and 1 tail, and 2 heads and 2 tails.

## Step 27: Problem 6 - Calculating the probabilities
The total number of outcomes is $2^4 = 16$. The number of ways to get 4 heads is 1, 3 heads and 1 tail is $\binom{4}{3} = 4$, and 2 heads and 2 tails is $\binom{4}{2} = 6$. So, the total favorable outcomes are $1 + 4 + 6 = 11$.

## Step 28: Problem 6 - Calculating the probability
The probability is $\frac{11}{16}$.

## Step 29: Problem 6 - Answer
ANSWER 6: E

## Step 30: Problem 7 - Understanding the problem
A machine applies a rule to a positive integer N: if N is even, output N/2; if N is odd, output 3N + 1. We need to find the sum of all N that result in 1 after 6 steps.

## Step 31: Problem 7 - Working backwards from 1
After 6 steps, the output is 1. So, before that, it was 2 (since 2 is even and 2/2 = 1). Before that, it could be 4. Continuing this process backwards: 4 -> 2 -> 1 (3 steps), 8 -> 4 -> 2 -> 1 (4 steps), 16 -> 8 -> 4 -> 2 -> 1 (5 steps), and so on, until we reach 6 steps.

## Step 32: Problem 7 - Enumerating the possible N
For 6 steps: $2^6 = 64$ is a possible N (since 64 -> 32 -> 16 -> 8 -> 4 -> 2 -> 1). Working bac

Problem 1:

We are asked what fraction of possible telephone numbers begin with 9 and end with 0.

A 7-digit telephone number cannot begin with 0 or 1.

Total possible telephone numbers:

- First digit: 2 through 9, so 8 choices.
- Remaining 6 digits: each has 10 choices.

So total numbers:

\[
8 \cdot 10^6
\]

Now count numbers that begin with 9 and end with 0:

- First digit fixed as 9: 1 choice.
- Last digit fixed as 0: 1 choice.
- Middle 5 digits: each has 10 choices.

So favorable numbers:

\[
10^5
\]

Thus the fraction is:

\[
\frac{10^5}{8\cdot 10^6}=\frac{1}{80}
\]

The tempting answer \(\frac{1}{100}\) would be correct only if all first digits were equally allowed, including 0 and 1, but they are not.

ANSWER 1: B

Problem 2:

We have the integers 1 through 25 separated into five groups of 5 numbers each. We find the median of each group, then take the median of those five medians. We want the least possible value of \(M\).

For a group of 5 numbers to have median at most \(m\), at least 3 numbers in that group must be at most \(m\).

Since \(M\) is the median of the five medians, at least 3 of the group medians must be at most \(M\).

Therefore, to make \(M\) as small as possible, we need at least:

\[
3 \times 3 = 9
\]

numbers that are at most \(M\).

The 9 smallest numbers are:

\[
1,2,3,4,5,6,7,8,9
\]

So \(M\) cannot be less than 9.

Now we show that \(M=9\) is possible. For example:

\[
\{1,2,3,22,23\} \quad \text{median } 3
\]

\[
\{4,5,6,24,25\} \quad \text{median } 6
\]

\[
\{7,8,9,20,21\} \quad \text{median } 9
\]

The remaining numbers can form two groups with medians larger than 9:

\[
\{10,11,12,13,14\} \quad \text{median } 12
\]

\[
\{15,16,17,18,19\} \quad \text{median } 17
\]

The five medians are:

\[
3,6,9,12,17
\]

Their median is 9.

So the least possible value is 9.

ANSWER 2: A

Problem 3:

The cricket starts on one leaf and hops to one of the other 3 leaves each time. We want the probability that after 4 hops it is back where it started.

Let \(p_n\) be the probability that the cricket is on its starting leaf after \(n\) hops.

Initially:

\[
p_0=1
\]

After one hop, it cannot be on the starting leaf:

\[
p_1=0
\]

If the cricket is not on the starting leaf, then on the next hop it has probability \(\frac13\) of hopping back to the starting leaf. Therefore:

\[
p_{n+1}=\frac{1-p_n}{3}
\]

Now compute:

\[
p_2=\frac{1-0}{3}=\frac13
\]

\[
p_3=\frac{1-\frac13}{3}=\frac{\frac23}{3}=\frac29
\]

\[
p_4=\frac{1-\frac29}{3}=\frac{\frac79}{3}=\frac7{27}
\]

So the probability is:

\[
\frac7{27}
\]

Choice A, \(\frac29\), is the probability after 3 hops, not 4 hops.

ANSWER 3: E

Problem 4:

The letters and digits are cycled separately.

The letters are:

\[
AJHSME
\]

There are 6 letters, so the letters return to their original order after 6 cycles.

The digits are:

\[
1989
\]

There are 4 digits, so the digits return to their original order after 4 cycles.

We need both the letters and the digits to return to their original arrangements at the same time.

So we need the least common multiple of 6 and 4:

\[
\operatorname{lcm}(6,4)=12
\]

Thus \(AJHSME\ 1989\) appears again for the first time on line 12.

Choice A, 6, would return only the letters, not necessarily the digits. Choice B, 10, returns neither. Choice E, 24, is a common multiple but not the first one.

ANSWER 4: C

Problem 5:

Let the original price be \(P\).

Suppose the price is increased by \(r\), where \(r\) is written as a decimal. Then the new price is:

\[
P(1+r)
\]

Then it is decreased by the same percent, so the final price is:

\[
P(1+r)(1-r)
\]

We are told this equals 84% of the original price:

\[
P(1+r)(1-r)=0.84P
\]

Cancel \(P\):

\[
(1+r)(1-r)=0.84
\]

\[
1-r^2=0.84
\]

\[
r^2=0.16
\]

\[
r=0.4
\]

So the percent is:

\[
40\%
\]

A common mistake is to think an increase and decrease by the same percent cancel each other, but they do not because the second percent is taken from a different price.

ANSWER 5: E

Problem 6:

Harold tosses a nickel 4 times. We want the probability that he gets at least as many heads as tails.

There are \(2^4=16\) equally likely outcomes.

“At least as many heads as tails” means:

\[
H \ge T
\]

Since there are 4 tosses total, this means Harold must get 2, 3, or 4 heads.

Count the outcomes:

\[
\binom42=6
\]

\[
\binom43=4
\]

\[
\binom44=1
\]

Total favorable outcomes:

\[
6+4+1=11
\]

So the probability is:

\[
\frac{11}{16}
\]

The tempting answer \(\frac12\) would count only strictly more heads than tails and would forget the tie case.

ANSWER 6: E

Problem 7:

The machine rule is:

- If \(N\) is even, output \(\frac N2\).
- If \(N\) is odd, output \(3N+1\).

We need all starting positive integers \(N\) that become 1 after exactly 6 steps.

It is easiest to work backward from 1.

If a number outputs \(y\), then possible previous numbers are:

1. \(2y\), because if the previous number was even, halving it gives \(y\).
2. \(\frac{y-1}{3}\), but only if this is a positive odd integer.

Now work backward 6 steps.

Step 0:

\[
\{1\}
\]

Step 1 predecessors of 1:

\[
\{2\}
\]

Step 2 predecessors of 2:

\[
\{4\}
\]

Step 3 predecessors of 4:

\[
2\cdot 4=8
\]

and

\[
\frac{4-1}{3}=1
\]

So:

\[
\{8,1\}
\]

Step 4:

Predecessors of 8: \(16\)

Predecessors of 1: \(2\)

So:

\[
\{16,2\}
\]

Step 5:

Predecessors of 16:

\[
32,\quad \frac{16-1}{3}=5
\]

Predecessors of 2:

\[
4
\]

So:

\[
\{32,5,4\}
\]

Step 6:

Predecessors of 32:

\[
64
\]

Predecessors of 5:

\[
10
\]

Predecessors of 4:

\[
8,\quad 1
\]

So the possible starting values are:

\[
64,10,8,1
\]

Their sum is:

\[
64+10+8+1=83
\]

ANSWER 7: E

Problem 8:

Let the list be:

\[
a_1,a_2,a_3,a_4,a_5,a_6,a_7,a_8
\]

Each new number is the product of the two previous numbers:

\[
a_3=a_1a_2
\]

\[
a_4=a_2a_3
\]

and so on.

We are told the last three numbers are:

\[
a_6=16,\quad a_7=64,\quad a_8=1024
\]

Check:

\[
16\cdot 64=1024
\]

Now work backward.

Since:

\[
a_7=a_5a_6
\]

we have:

\[
64=

Problem 1:
We are asked to find the fraction of possible 7-digit telephone numbers that begin with 9 and end with 0, given that a telephone number cannot begin with 0 or 1.

First, let's find the total number of possible telephone numbers:
- The first digit can be any number from 2 to 9, which gives 8 possibilities.
- The remaining 6 digits can be any number from 0 to 9, giving 10 possibilities each.
Total possible telephone numbers = 8 × 10 × 10 × 10 × 10 × 10 × 10 = 8,000,000.

Next, let's find the number of telephone numbers that begin with 9 and end with 0:
- The first digit must be 9 (1 possibility).
- The last digit must be 0 (1 possibility).
- The middle 5 digits can be any number from 0 to 9 (10 possibilities each).
Total such numbers = 1 × 10 × 10 × 10 × 10 × 10 × 1 = 100,000.

The fraction of telephone numbers that meet this condition is:
100,000 / 8,000,000 = 1 / 80.

ANSWER 1: B

Problem 2:
We are asked to find the least possible value of M, which is the median of the medians of five groups of 5 numbers each, chosen from the integers 1 through 25.

Let the 5 groups be G1, G2, G3, G4, and G5, and let their respective medians be m1, m2, m3, m4, and m5. 
Assume the medians are ordered such that m1 < m2 < m3 < m4 < m5. The median of these medians is M = m3.
To minimize m3, we must make the numbers in the groups as small as possible. 

For m3 to be a certain value, there must be at least 3 groups (G1, G2, G3) whose medians are less than or equal to m3. 
In each of these 3 groups, there are 3 numbers that are less than or equal to the group's median. 
Therefore, there are at least 3 groups × 3 numbers = 9 numbers that are less than or equal to m3.

Since the numbers are distinct integers from 1 to 25, the 9th smallest possible integer is 9. Thus, m3 must be at least 9. 
We can verify that 9 is possible with the following grouping:
G1: 1, 2, 7, 24, 25 (median 7)
G2: 3, 4, 8, 22, 23 (median 8)
G3: 5, 6, 9, 20, 21 (median 9)
G4: 10, 11, 12, 18, 19 (median 12)
G5: 13, 14, 15, 16, 17 (median 15)
The medians are 7, 8, 9, 12, and 15. Their median is 9.

ANSWER 2: A

Problem 3:
We want to find the probability that a cricket, which hops to one of 3 other leaves with equal probability, returns to its starting leaf after exactly 4 hops.

Let the leaves be A, B, C, and D, and assume the cricket starts at leaf A.
Let P(n) be the probability that the cricket is at leaf A after n hops.
- P(0) = 1 (it starts at A).
- For any hop n > 0, the cricket can only land on A if it was NOT on A in the previous hop. If it was on B, C, or D, it has a 1/3 chance of hopping to A.
Thus, the recursive formula is P(n) = (1 - P(n-1)) × (1/3).

Let's calculate step-by-step:
- P(1) = (1 - 1) × 1/3 = 0
- P(2) = (1 - 0) × 1/3 = 1/3
- P(3) = (1 - 1/3) × 1/3 = (2/3) × (1/3) = 2/9
- P(4) = (1 - 2/9) × 1/3 = (7/9) × (1/3) = 7/27

ANSWER 3: E

Problem 4:
We are given a sequence of letters "AJHSME" (6 characters) and a sequence of digits "1989" (4 characters). They are cycled separately line by line. We want to find the line number where the original combination "AJHSME 1989" appears again for the first time.

The letters cycle back to their original arrangement every 6 lines.
The digits cycle back to their original arrangement every 4 lines.
For both to return to their original arrangement simultaneously, the line number must be a common multiple of both 6 and 4. 

The least common multiple (LCM) of 6 and 4 is 12.
Therefore, on line 12, the letters will have completed exactly 2 full cycles and the digits will have completed exactly 3 full cycles, returning to "AJHSME 1989".

ANSWER 4: C

Problem 5:
Let the original price of the shirt be P, and let the percentage of increase and decrease be x.
When the price is increased by x percent, it becomes P(1 + x/100).
When this new price is decreased by x percent, it becomes P(1 + x/100)(1 - x/100).

We are given that the resulting price is 84% of the original price, so:
P(1 + x/100)(1 - x/100) = 0.84P

Divide both sides by P:
(1 + x/100)(1 - x/100) = 0.84
1 - (x/100)² = 0.84
(x/100)² = 1 - 0.84
(x/100)² = 0.16

Taking the square root of both sides:
x/100 = 0.4
x = 40

The price was increased and decreased by 40%.

ANSWER 5: E

Problem 6:
Harold tosses a nickel four times. We want to find the probability that he gets at least as many heads as tails.
"At least as many heads as tails" in 4 tosses means getting 2, 3, or 4 heads.

The total number of possible outcomes when tossing a coin 4 times is 2⁴ = 16.
Using combinations, we find the number of ways to get 2, 3, or 4 heads:
- 2 heads: 4C2 = 6 ways
- 3 heads: 4C3 = 4 ways
- 4 heads: 4C4 = 1 way

Total favorable outcomes = 6 + 4 + 1 = 11.
The probability is 11 / 16.

ANSWER 6: E

Problem 7:
The machine follows the Collatz-like rule:
- If N is even, output N/2
- If N is odd, output 3N + 1

We are told a 6-step process ends in 1. Let's work backward from Step 6 to find all possible starting positive integers N (Step 0).
- Step 6: 1
- Step 5: Must be 2 (since 3x + 1 = 1 gives x = 0, which is not a positive integer).
- Step 4: Must be 4 (since 3x + 1 = 2 gives x = 1/3, not an integer).
- Step 3: Can be 8 (since 8/2 = 4) or 1 (since 3(1) + 1 = 4).

Branch 1 (Step 3 is 8):
- Step 2: Must be 16 (since 3x + 1 = 8 gives x = 7/3).
- Step 1: Can be 32 (32/2 = 16) or 5 (3(5) + 1 = 16).
- Step 0 (from 32): Must be 64 (since 3x + 1 = 32 gives x = 31/3).
- Step 0 (from 5): Must be 10 (since 3x + 1 = 5 gives x = 4/3).

Branch 2 (Step 3 is 1):
- Step 2: Must be 2.
- Step 1: Must be 4.
- Step 0: Can be 8 (8/2 = 4) or 1 (3(1) + 1 = 4).

The possible starting integers N are 64, 10, 8, and 1.
The sum of all such integers is 64 + 10 + 8 + 1 = 83.

ANSWER 7: E

Problem 8:
We are given a sequence of 8 numbers where each number is the product of the two preceding numbers: a_n = a_{n-2} × a_{n-1}.
We know the last three numbers: a_6 = 16, a_7 = 64, and a_8 = 1024.
Let's work backward using the rule a_{n-2} = a_n / a_{n-1}:
- a_5 = a_7 / a_6 = 64 / 16 = 4
- a_4 

## Step 1: Problem 1 - Understanding the problem
We need to find the fraction of 7-digit telephone numbers that begin with 9 and end with 0, given that the numbers cannot start with 0 or 1.

## Step 2: Problem 1 - Calculating total possible telephone numbers
The first digit can be any number from 2 to 9, so there are 8 possibilities. The remaining 6 digits can be any number from 0 to 9, so there are 10 possibilities for each. Thus, the total number of telephone numbers is $8 \times 10^6$.

## Step 3: Problem 1 - Calculating the number of telephone numbers that begin with 9 and end with 0
For a number to begin with 9 and end with 0, the first digit must be 9 and the last digit must be 0. The first digit has 1 possibility (9), the last digit has 1 possibility (0), and the remaining 5 digits have 10 possibilities each. So, the total number of such telephone numbers is $1 \times 10^5 \times 1$.

## Step 4: Problem 1 - Calculating the fraction
The fraction of telephone numbers that begin with 9 and end with 0 is $\frac{1 \times 10^5 \times 1}{8 \times 10^6} = \frac{10^5}{8 \times 10^6} = \frac{1}{80}$.

## Step 5: Problem 1 - Answer
ANSWER 1: B

## Step 6: Problem 2 - Understanding the problem
We are given 25 integers divided into 5 groups of 5 numbers each. We need to find the least possible value of M, the median of the medians of these groups.

## Step 7: Problem 2 - Analyzing the medians
To minimize M, we should minimize the medians of the 5 groups. The integers are from 1 to 25. The median of each group will be the third number when the numbers are arranged in ascending order.

## Step 8: Problem 2 - Distributing numbers to minimize M
To minimize M, we can distribute the numbers such that the medians of the groups are as small as possible. One way to do this is by grouping the smallest numbers together and ensuring that the medians are as low as possible.

## Step 9: Problem 2 - Calculating the medians of the groups
Let's distribute the numbers: Group 1 = {1, 2, 3, 24, 25}, Group 2 = {4, 5, 6, 22, 23}, Group 3 = {7, 8, 9, 20, 21}, Group 4 = {10, 11, 12, 18, 19}, Group 5 = {13, 14, 15, 16, 17}. The medians are 3, 6, 9, 12, 15.

## Step 10: Problem 2 - Finding M
The medians of the groups are 3, 6, 9, 12, 15. The median of these medians (M) is 9.

## Step 11: Problem 2 - Answer
ANSWER 2: A

## Step 12: Problem 3 - Understanding the problem
A cricket hops between 4 leaves with equal probability. We need to find the probability that it returns to the starting leaf after 4 hops.

## Step 13: Problem 3 - Analyzing the hops
After the first hop, the cricket is on a different leaf. For it to return to the starting leaf after 4 hops, it must be on a different leaf after the third hop as well.

## Step 14: Problem 3 - Calculating the probability
Let's denote the probability of being on the starting leaf after n hops as $P_n$. $P_0 = 1$, $P_1 = 0$. For $n \geq 1$, $P_{n+1} = \frac{1}{3}(1 - P_n)$ because the probability of not being on the starting leaf after n hops is $1 - P_n$, and from any other leaf, there's a $\frac{1}{3}$ chance of hopping to the starting leaf.

## Step 15: Problem 3 - Computing $P_4$
$P_2 = \frac{1}{3}$, $P_3 = \frac{1}{3}(1 - \frac{1}{3}) = \frac{2}{9}$, $P_4 = \frac{1}{3}(1 - \frac{2}{9}) = \frac{1}{3} \times \frac{7}{9} = \frac{7}{27}$.

## Step 16: Problem 3 - Answer
ANSWER 3: E

## Step 17: Problem 4 - Understanding the problem
The letters and digits are cycled separately and combined. We need to find when "AJHSME 1989" appears again.

## Step 18: Problem 4 - Analyzing the cycles
The letters cycle every 6 steps, and the digits cycle every 4 steps (since there are 4 digits and they are distinct when considering their positions). The combined cycle will be the least common multiple (LCM) of 6 and 4.

## Step 19: Problem 4 - Calculating the LCM of 6 and 4
The LCM of 6 and 4 is 12.

## Step 20: Problem 4 - Answer
ANSWER 4: C

## Step 21: Problem 5 - Understanding the problem
A price is increased by a certain percent and then decreased by the same percent, resulting in 84% of the original price.

## Step 22: Problem 5 - Setting up the equation
Let $x$ be the original price and $p$ be the percentage change. The new price after increase is $x(1 + \frac{p}{100})$, and after decrease is $x(1 + \frac{p}{100})(1 - \frac{p}{100}) = 0.84x$.

## Step 23: Problem 5 - Solving for $p$
$x(1 - \frac{p^2}{10000}) = 0.84x$. Simplifying gives $1 - \frac{p^2}{10000} = 0.84$. Thus, $\frac{p^2}{10000} = 0.16$, so $p^2 = 1600$, and $p = 40$.

## Step 24: Problem 5 - Answer
ANSWER 5: E

## Step 25: Problem 6 - Understanding the problem
Harold tosses a nickel four times. We need to find the probability of getting at least as many heads as tails.

## Step 26: Problem 6 - Enumerating the outcomes
The possible outcomes where heads are at least as many as tails are: 4 heads, 3 heads and 1 tail, and 2 heads and 2 tails.

## Step 27: Problem 6 - Calculating the probabilities
The total number of outcomes is $2^4 = 16$. The number of ways to get 4 heads is 1, 3 heads and 1 tail is $\binom{4}{3} = 4$, and 2 heads and 2 tails is $\binom{4}{2} = 6$. So, the total favorable outcomes are $1 + 4 + 6 = 11$.

## Step 28: Problem 6 - Calculating the probability
The probability is $\frac{11}{16}$.

## Step 29: Problem 6 - Answer
ANSWER 6: E

## Step 30: Problem 7 - Understanding the problem
A machine applies a rule to a positive integer N: if N is even, output N/2; if N is odd, output 3N + 1. We need to find the sum of all N that result in 1 after 6 steps.

## Step 31: Problem 7 - Working backwards from 1
After 6 steps, the output is 1. So, before that, it was 2 (since 2 is even and 2/2 = 1). Before that, it could be 4. Continuing this process backwards: 4 -> 2 -> 1 (3 steps), 8 -> 4 -> 2 -> 1 (4 steps), 16 -> 8 -> 4 -> 2 -> 1 (5 steps), and so on, until we reach 6 steps.

## Step 32: Problem 7 - Enumerating the possible N
For 6 steps: $2^6 = 64$ is a possible N (since 64 -> 32 -> 16 -> 8 -> 4 -> 2 -> 1). Working bac

Problem 1:

We are asked what fraction of possible telephone numbers begin with 9 and end with 0.

A 7-digit telephone number cannot begin with 0 or 1.

Total possible telephone numbers:

- First digit: 2 through 9, so 8 choices.
- Remaining 6 digits: each has 10 choices.

So total numbers:

\[
8 \cdot 10^6
\]

Now count numbers that begin with 9 and end with 0:

- First digit fixed as 9: 1 choice.
- Last digit fixed as 0: 1 choice.
- Middle 5 digits: each has 10 choices.

So favorable numbers:

\[
10^5
\]

Thus the fraction is:

\[
\frac{10^5}{8\cdot 10^6}=\frac{1}{80}
\]

The tempting answer \(\frac{1}{100}\) would be correct only if all first digits were equally allowed, including 0 and 1, but they are not.

ANSWER 1: B

Problem 2:

We have the integers 1 through 25 separated into five groups of 5 numbers each. We find the median of each group, then take the median of those five medians. We want the least possible value of \(M\).

For a group of 5 numbers to have median at most \(m\), at least 3 numbers in that group must be at most \(m\).

Since \(M\) is the median of the five medians, at least 3 of the group medians must be at most \(M\).

Therefore, to make \(M\) as small as possible, we need at least:

\[
3 \times 3 = 9
\]

numbers that are at most \(M\).

The 9 smallest numbers are:

\[
1,2,3,4,5,6,7,8,9
\]

So \(M\) cannot be less than 9.

Now we show that \(M=9\) is possible. For example:

\[
\{1,2,3,22,23\} \quad \text{median } 3
\]

\[
\{4,5,6,24,25\} \quad \text{median } 6
\]

\[
\{7,8,9,20,21\} \quad \text{median } 9
\]

The remaining numbers can form two groups with medians larger than 9:

\[
\{10,11,12,13,14\} \quad \text{median } 12
\]

\[
\{15,16,17,18,19\} \quad \text{median } 17
\]

The five medians are:

\[
3,6,9,12,17
\]

Their median is 9.

So the least possible value is 9.

ANSWER 2: A

Problem 3:

The cricket starts on one leaf and hops to one of the other 3 leaves each time. We want the probability that after 4 hops it is back where it started.

Let \(p_n\) be the probability that the cricket is on its starting leaf after \(n\) hops.

Initially:

\[
p_0=1
\]

After one hop, it cannot be on the starting leaf:

\[
p_1=0
\]

If the cricket is not on the starting leaf, then on the next hop it has probability \(\frac13\) of hopping back to the starting leaf. Therefore:

\[
p_{n+1}=\frac{1-p_n}{3}
\]

Now compute:

\[
p_2=\frac{1-0}{3}=\frac13
\]

\[
p_3=\frac{1-\frac13}{3}=\frac{\frac23}{3}=\frac29
\]

\[
p_4=\frac{1-\frac29}{3}=\frac{\frac79}{3}=\frac7{27}
\]

So the probability is:

\[
\frac7{27}
\]

Choice A, \(\frac29\), is the probability after 3 hops, not 4 hops.

ANSWER 3: E

Problem 4:

The letters and digits are cycled separately.

The letters are:

\[
AJHSME
\]

There are 6 letters, so the letters return to their original order after 6 cycles.

The digits are:

\[
1989
\]

There are 4 digits, so the digits return to their original order after 4 cycles.

We need both the letters and the digits to return to their original arrangements at the same time.

So we need the least common multiple of 6 and 4:

\[
\operatorname{lcm}(6,4)=12
\]

Thus \(AJHSME\ 1989\) appears again for the first time on line 12.

Choice A, 6, would return only the letters, not necessarily the digits. Choice B, 10, returns neither. Choice E, 24, is a common multiple but not the first one.

ANSWER 4: C

Problem 5:

Let the original price be \(P\).

Suppose the price is increased by \(r\), where \(r\) is written as a decimal. Then the new price is:

\[
P(1+r)
\]

Then it is decreased by the same percent, so the final price is:

\[
P(1+r)(1-r)
\]

We are told this equals 84% of the original price:

\[
P(1+r)(1-r)=0.84P
\]

Cancel \(P\):

\[
(1+r)(1-r)=0.84
\]

\[
1-r^2=0.84
\]

\[
r^2=0.16
\]

\[
r=0.4
\]

So the percent is:

\[
40\%
\]

A common mistake is to think an increase and decrease by the same percent cancel each other, but they do not because the second percent is taken from a different price.

ANSWER 5: E

Problem 6:

Harold tosses a nickel 4 times. We want the probability that he gets at least as many heads as tails.

There are \(2^4=16\) equally likely outcomes.

“At least as many heads as tails” means:

\[
H \ge T
\]

Since there are 4 tosses total, this means Harold must get 2, 3, or 4 heads.

Count the outcomes:

\[
\binom42=6
\]

\[
\binom43=4
\]

\[
\binom44=1
\]

Total favorable outcomes:

\[
6+4+1=11
\]

So the probability is:

\[
\frac{11}{16}
\]

The tempting answer \(\frac12\) would count only strictly more heads than tails and would forget the tie case.

ANSWER 6: E

Problem 7:

The machine rule is:

- If \(N\) is even, output \(\frac N2\).
- If \(N\) is odd, output \(3N+1\).

We need all starting positive integers \(N\) that become 1 after exactly 6 steps.

It is easiest to work backward from 1.

If a number outputs \(y\), then possible previous numbers are:

1. \(2y\), because if the previous number was even, halving it gives \(y\).
2. \(\frac{y-1}{3}\), but only if this is a positive odd integer.

Now work backward 6 steps.

Step 0:

\[
\{1\}
\]

Step 1 predecessors of 1:

\[
\{2\}
\]

Step 2 predecessors of 2:

\[
\{4\}
\]

Step 3 predecessors of 4:

\[
2\cdot 4=8
\]

and

\[
\frac{4-1}{3}=1
\]

So:

\[
\{8,1\}
\]

Step 4:

Predecessors of 8: \(16\)

Predecessors of 1: \(2\)

So:

\[
\{16,2\}
\]

Step 5:

Predecessors of 16:

\[
32,\quad \frac{16-1}{3}=5
\]

Predecessors of 2:

\[
4
\]

So:

\[
\{32,5,4\}
\]

Step 6:

Predecessors of 32:

\[
64
\]

Predecessors of 5:

\[
10
\]

Predecessors of 4:

\[
8,\quad 1
\]

So the possible starting values are:

\[
64,10,8,1
\]

Their sum is:

\[
64+10+8+1=83
\]

ANSWER 7: E

Problem 8:

Let the list be:

\[
a_1,a_2,a_3,a_4,a_5,a_6,a_7,a_8
\]

Each new number is the product of the two previous numbers:

\[
a_3=a_1a_2
\]

\[
a_4=a_2a_3
\]

and so on.

We are told the last three numbers are:

\[
a_6=16,\quad a_7=64,\quad a_8=1024
\]

Check:

\[
16\cdot 64=1024
\]

Now work backward.

Since:

\[
a_7=a_5a_6
\]

we have:

\[
64=

Problem 1:
We are asked to find the fraction of possible 7-digit telephone numbers that begin with 9 and end with 0, given that a telephone number cannot begin with 0 or 1.

First, let's find the total number of possible telephone numbers:
- The first digit can be any number from 2 to 9, which gives 8 possibilities.
- The remaining 6 digits can be any number from 0 to 9, giving 10 possibilities each.
Total possible telephone numbers = 8 × 10 × 10 × 10 × 10 × 10 × 10 = 8,000,000.

Next, let's find the number of telephone numbers that begin with 9 and end with 0:
- The first digit must be 9 (1 possibility).
- The last digit must be 0 (1 possibility).
- The middle 5 digits can be any number from 0 to 9 (10 possibilities each).
Total such numbers = 1 × 10 × 10 × 10 × 10 × 10 × 1 = 100,000.

The fraction of telephone numbers that meet this condition is:
100,000 / 8,000,000 = 1 / 80.

ANSWER 1: B

Problem 2:
We are asked to find the least possible value of M, which is the median of the medians of five groups of 5 numbers each, chosen from the integers 1 through 25.

Let the 5 groups be G1, G2, G3, G4, and G5, and let their respective medians be m1, m2, m3, m4, and m5. 
Assume the medians are ordered such that m1 < m2 < m3 < m4 < m5. The median of these medians is M = m3.
To minimize m3, we must make the numbers in the groups as small as possible. 

For m3 to be a certain value, there must be at least 3 groups (G1, G2, G3) whose medians are less than or equal to m3. 
In each of these 3 groups, there are 3 numbers that are less than or equal to the group's median. 
Therefore, there are at least 3 groups × 3 numbers = 9 numbers that are less than or equal to m3.

Since the numbers are distinct integers from 1 to 25, the 9th smallest possible integer is 9. Thus, m3 must be at least 9. 
We can verify that 9 is possible with the following grouping:
G1: 1, 2, 7, 24, 25 (median 7)
G2: 3, 4, 8, 22, 23 (median 8)
G3: 5, 6, 9, 20, 21 (median 9)
G4: 10, 11, 12, 18, 19 (median 12)
G5: 13, 14, 15, 16, 17 (median 15)
The medians are 7, 8, 9, 12, and 15. Their median is 9.

ANSWER 2: A

Problem 3:
We want to find the probability that a cricket, which hops to one of 3 other leaves with equal probability, returns to its starting leaf after exactly 4 hops.

Let the leaves be A, B, C, and D, and assume the cricket starts at leaf A.
Let P(n) be the probability that the cricket is at leaf A after n hops.
- P(0) = 1 (it starts at A).
- For any hop n > 0, the cricket can only land on A if it was NOT on A in the previous hop. If it was on B, C, or D, it has a 1/3 chance of hopping to A.
Thus, the recursive formula is P(n) = (1 - P(n-1)) × (1/3).

Let's calculate step-by-step:
- P(1) = (1 - 1) × 1/3 = 0
- P(2) = (1 - 0) × 1/3 = 1/3
- P(3) = (1 - 1/3) × 1/3 = (2/3) × (1/3) = 2/9
- P(4) = (1 - 2/9) × 1/3 = (7/9) × (1/3) = 7/27

ANSWER 3: E

Problem 4:
We are given a sequence of letters "AJHSME" (6 characters) and a sequence of digits "1989" (4 characters). They are cycled separately line by line. We want to find the line number where the original combination "AJHSME 1989" appears again for the first time.

The letters cycle back to their original arrangement every 6 lines.
The digits cycle back to their original arrangement every 4 lines.
For both to return to their original arrangement simultaneously, the line number must be a common multiple of both 6 and 4. 

The least common multiple (LCM) of 6 and 4 is 12.
Therefore, on line 12, the letters will have completed exactly 2 full cycles and the digits will have completed exactly 3 full cycles, returning to "AJHSME 1989".

ANSWER 4: C

Problem 5:
Let the original price of the shirt be P, and let the percentage of increase and decrease be x.
When the price is increased by x percent, it becomes P(1 + x/100).
When this new price is decreased by x percent, it becomes P(1 + x/100)(1 - x/100).

We are given that the resulting price is 84% of the original price, so:
P(1 + x/100)(1 - x/100) = 0.84P

Divide both sides by P:
(1 + x/100)(1 - x/100) = 0.84
1 - (x/100)² = 0.84
(x/100)² = 1 - 0.84
(x/100)² = 0.16

Taking the square root of both sides:
x/100 = 0.4
x = 40

The price was increased and decreased by 40%.

ANSWER 5: E

Problem 6:
Harold tosses a nickel four times. We want to find the probability that he gets at least as many heads as tails.
"At least as many heads as tails" in 4 tosses means getting 2, 3, or 4 heads.

The total number of possible outcomes when tossing a coin 4 times is 2⁴ = 16.
Using combinations, we find the number of ways to get 2, 3, or 4 heads:
- 2 heads: 4C2 = 6 ways
- 3 heads: 4C3 = 4 ways
- 4 heads: 4C4 = 1 way

Total favorable outcomes = 6 + 4 + 1 = 11.
The probability is 11 / 16.

ANSWER 6: E

Problem 7:
The machine follows the Collatz-like rule:
- If N is even, output N/2
- If N is odd, output 3N + 1

We are told a 6-step process ends in 1. Let's work backward from Step 6 to find all possible starting positive integers N (Step 0).
- Step 6: 1
- Step 5: Must be 2 (since 3x + 1 = 1 gives x = 0, which is not a positive integer).
- Step 4: Must be 4 (since 3x + 1 = 2 gives x = 1/3, not an integer).
- Step 3: Can be 8 (since 8/2 = 4) or 1 (since 3(1) + 1 = 4).

Branch 1 (Step 3 is 8):
- Step 2: Must be 16 (since 3x + 1 = 8 gives x = 7/3).
- Step 1: Can be 32 (32/2 = 16) or 5 (3(5) + 1 = 16).
- Step 0 (from 32): Must be 64 (since 3x + 1 = 32 gives x = 31/3).
- Step 0 (from 5): Must be 10 (since 3x + 1 = 5 gives x = 4/3).

Branch 2 (Step 3 is 1):
- Step 2: Must be 2.
- Step 1: Must be 4.
- Step 0: Can be 8 (8/2 = 4) or 1 (3(1) + 1 = 4).

The possible starting integers N are 64, 10, 8, and 1.
The sum of all such integers is 64 + 10 + 8 + 1 = 83.

ANSWER 7: E

Problem 8:
We are given a sequence of 8 numbers where each number is the product of the two preceding numbers: a_n = a_{n-2} × a_{n-1}.
We know the last three numbers: a_6 = 16, a_7 = 64, and a_8 = 1024.
Let's work backward using the rule a_{n-2} = a_n / a_{n-1}:
- a_5 = a_7 / a_6 = 64 / 16 = 4
- a_4 

## Step 1: Problem 1 - Understanding the problem
We need to find the fraction of 7-digit telephone numbers that begin with 9 and end with 0, given that the numbers cannot start with 0 or 1.

## Step 2: Problem 1 - Calculating total possible telephone numbers
The first digit can be any number from 2 to 9, so there are 8 possibilities. The remaining 6 digits can be any number from 0 to 9, so there are 10 possibilities for each. Thus, the total number of telephone numbers is $8 \times 10^6$.

## Step 3: Problem 1 - Calculating the number of telephone numbers that begin with 9 and end with 0
For a number to begin with 9 and end with 0, the first digit must be 9 and the last digit must be 0. The first digit has 1 possibility (9), the last digit has 1 possibility (0), and the remaining 5 digits have 10 possibilities each. So, the total number of such telephone numbers is $1 \times 10^5 \times 1$.

## Step 4: Problem 1 - Calculating the fraction
The fraction of telephone numbers that begin with 9 and end with 0 is $\frac{1 \times 10^5 \times 1}{8 \times 10^6} = \frac{10^5}{8 \times 10^6} = \frac{1}{80}$.

## Step 5: Problem 1 - Answer
ANSWER 1: B

## Step 6: Problem 2 - Understanding the problem
We are given 25 integers divided into 5 groups of 5 numbers each. We need to find the least possible value of M, the median of the medians of these groups.

## Step 7: Problem 2 - Analyzing the medians
To minimize M, we should minimize the medians of the 5 groups. The integers are from 1 to 25. The median of each group will be the third number when the numbers are arranged in ascending order.

## Step 8: Problem 2 - Distributing numbers to minimize M
To minimize M, we can distribute the numbers such that the medians of the groups are as small as possible. One way to do this is by grouping the smallest numbers together and ensuring that the medians are as low as possible.

## Step 9: Problem 2 - Calculating the medians of the groups
Let's distribute the numbers: Group 1 = {1, 2, 3, 24, 25}, Group 2 = {4, 5, 6, 22, 23}, Group 3 = {7, 8, 9, 20, 21}, Group 4 = {10, 11, 12, 18, 19}, Group 5 = {13, 14, 15, 16, 17}. The medians are 3, 6, 9, 12, 15.

## Step 10: Problem 2 - Finding M
The medians of the groups are 3, 6, 9, 12, 15. The median of these medians (M) is 9.

## Step 11: Problem 2 - Answer
ANSWER 2: A

## Step 12: Problem 3 - Understanding the problem
A cricket hops between 4 leaves with equal probability. We need to find the probability that it returns to the starting leaf after 4 hops.

## Step 13: Problem 3 - Analyzing the hops
After the first hop, the cricket is on a different leaf. For it to return to the starting leaf after 4 hops, it must be on a different leaf after the third hop as well.

## Step 14: Problem 3 - Calculating the probability
Let's denote the probability of being on the starting leaf after n hops as $P_n$. $P_0 = 1$, $P_1 = 0$. For $n \geq 1$, $P_{n+1} = \frac{1}{3}(1 - P_n)$ because the probability of not being on the starting leaf after n hops is $1 - P_n$, and from any other leaf, there's a $\frac{1}{3}$ chance of hopping to the starting leaf.

## Step 15: Problem 3 - Computing $P_4$
$P_2 = \frac{1}{3}$, $P_3 = \frac{1}{3}(1 - \frac{1}{3}) = \frac{2}{9}$, $P_4 = \frac{1}{3}(1 - \frac{2}{9}) = \frac{1}{3} \times \frac{7}{9} = \frac{7}{27}$.

## Step 16: Problem 3 - Answer
ANSWER 3: E

## Step 17: Problem 4 - Understanding the problem
The letters and digits are cycled separately and combined. We need to find when "AJHSME 1989" appears again.

## Step 18: Problem 4 - Analyzing the cycles
The letters cycle every 6 steps, and the digits cycle every 4 steps (since there are 4 digits and they are distinct when considering their positions). The combined cycle will be the least common multiple (LCM) of 6 and 4.

## Step 19: Problem 4 - Calculating the LCM of 6 and 4
The LCM of 6 and 4 is 12.

## Step 20: Problem 4 - Answer
ANSWER 4: C

## Step 21: Problem 5 - Understanding the problem
A price is increased by a certain percent and then decreased by the same percent, resulting in 84% of the original price.

## Step 22: Problem 5 - Setting up the equation
Let $x$ be the original price and $p$ be the percentage change. The new price after increase is $x(1 + \frac{p}{100})$, and after decrease is $x(1 + \frac{p}{100})(1 - \frac{p}{100}) = 0.84x$.

## Step 23: Problem 5 - Solving for $p$
$x(1 - \frac{p^2}{10000}) = 0.84x$. Simplifying gives $1 - \frac{p^2}{10000} = 0.84$. Thus, $\frac{p^2}{10000} = 0.16$, so $p^2 = 1600$, and $p = 40$.

## Step 24: Problem 5 - Answer
ANSWER 5: E

## Step 25: Problem 6 - Understanding the problem
Harold tosses a nickel four times. We need to find the probability of getting at least as many heads as tails.

## Step 26: Problem 6 - Enumerating the outcomes
The possible outcomes where heads are at least as many as tails are: 4 heads, 3 heads and 1 tail, and 2 heads and 2 tails.

## Step 27: Problem 6 - Calculating the probabilities
The total number of outcomes is $2^4 = 16$. The number of ways to get 4 heads is 1, 3 heads and 1 tail is $\binom{4}{3} = 4$, and 2 heads and 2 tails is $\binom{4}{2} = 6$. So, the total favorable outcomes are $1 + 4 + 6 = 11$.

## Step 28: Problem 6 - Calculating the probability
The probability is $\frac{11}{16}$.

## Step 29: Problem 6 - Answer
ANSWER 6: E

## Step 30: Problem 7 - Understanding the problem
A machine applies a rule to a positive integer N: if N is even, output N/2; if N is odd, output 3N + 1. We need to find the sum of all N that result in 1 after 6 steps.

## Step 31: Problem 7 - Working backwards from 1
After 6 steps, the output is 1. So, before that, it was 2 (since 2 is even and 2/2 = 1). Before that, it could be 4. Continuing this process backwards: 4 -> 2 -> 1 (3 steps), 8 -> 4 -> 2 -> 1 (4 steps), 16 -> 8 -> 4 -> 2 -> 1 (5 steps), and so on, until we reach 6 steps.

## Step 32: Problem 7 - Enumerating the possible N
For 6 steps: $2^6 = 64$ is a possible N (since 64 -> 32 -> 16 -> 8 -> 4 -> 2 -> 1). Working bac

Problem 1:

We are asked what fraction of possible telephone numbers begin with 9 and end with 0.

A 7-digit telephone number cannot begin with 0 or 1.

Total possible telephone numbers:

- First digit: 2 through 9, so 8 choices.
- Remaining 6 digits: each has 10 choices.

So total numbers:

\[
8 \cdot 10^6
\]

Now count numbers that begin with 9 and end with 0:

- First digit fixed as 9: 1 choice.
- Last digit fixed as 0: 1 choice.
- Middle 5 digits: each has 10 choices.

So favorable numbers:

\[
10^5
\]

Thus the fraction is:

\[
\frac{10^5}{8\cdot 10^6}=\frac{1}{80}
\]

The tempting answer \(\frac{1}{100}\) would be correct only if all first digits were equally allowed, including 0 and 1, but they are not.

ANSWER 1: B

Problem 2:

We have the integers 1 through 25 separated into five groups of 5 numbers each. We find the median of each group, then take the median of those five medians. We want the least possible value of \(M\).

For a group of 5 numbers to have median at most \(m\), at least 3 numbers in that group must be at most \(m\).

Since \(M\) is the median of the five medians, at least 3 of the group medians must be at most \(M\).

Therefore, to make \(M\) as small as possible, we need at least:

\[
3 \times 3 = 9
\]

numbers that are at most \(M\).

The 9 smallest numbers are:

\[
1,2,3,4,5,6,7,8,9
\]

So \(M\) cannot be less than 9.

Now we show that \(M=9\) is possible. For example:

\[
\{1,2,3,22,23\} \quad \text{median } 3
\]

\[
\{4,5,6,24,25\} \quad \text{median } 6
\]

\[
\{7,8,9,20,21\} \quad \text{median } 9
\]

The remaining numbers can form two groups with medians larger than 9:

\[
\{10,11,12,13,14\} \quad \text{median } 12
\]

\[
\{15,16,17,18,19\} \quad \text{median } 17
\]

The five medians are:

\[
3,6,9,12,17
\]

Their median is 9.

So the least possible value is 9.

ANSWER 2: A

Problem 3:

The cricket starts on one leaf and hops to one of the other 3 leaves each time. We want the probability that after 4 hops it is back where it started.

Let \(p_n\) be the probability that the cricket is on its starting leaf after \(n\) hops.

Initially:

\[
p_0=1
\]

After one hop, it cannot be on the starting leaf:

\[
p_1=0
\]

If the cricket is not on the starting leaf, then on the next hop it has probability \(\frac13\) of hopping back to the starting leaf. Therefore:

\[
p_{n+1}=\frac{1-p_n}{3}
\]

Now compute:

\[
p_2=\frac{1-0}{3}=\frac13
\]

\[
p_3=\frac{1-\frac13}{3}=\frac{\frac23}{3}=\frac29
\]

\[
p_4=\frac{1-\frac29}{3}=\frac{\frac79}{3}=\frac7{27}
\]

So the probability is:

\[
\frac7{27}
\]

Choice A, \(\frac29\), is the probability after 3 hops, not 4 hops.

ANSWER 3: E

Problem 4:

The letters and digits are cycled separately.

The letters are:

\[
AJHSME
\]

There are 6 letters, so the letters return to their original order after 6 cycles.

The digits are:

\[
1989
\]

There are 4 digits, so the digits return to their original order after 4 cycles.

We need both the letters and the digits to return to their original arrangements at the same time.

So we need the least common multiple of 6 and 4:

\[
\operatorname{lcm}(6,4)=12
\]

Thus \(AJHSME\ 1989\) appears again for the first time on line 12.

Choice A, 6, would return only the letters, not necessarily the digits. Choice B, 10, returns neither. Choice E, 24, is a common multiple but not the first one.

ANSWER 4: C

Problem 5:

Let the original price be \(P\).

Suppose the price is increased by \(r\), where \(r\) is written as a decimal. Then the new price is:

\[
P(1+r)
\]

Then it is decreased by the same percent, so the final price is:

\[
P(1+r)(1-r)
\]

We are told this equals 84% of the original price:

\[
P(1+r)(1-r)=0.84P
\]

Cancel \(P\):

\[
(1+r)(1-r)=0.84
\]

\[
1-r^2=0.84
\]

\[
r^2=0.16
\]

\[
r=0.4
\]

So the percent is:

\[
40\%
\]

A common mistake is to think an increase and decrease by the same percent cancel each other, but they do not because the second percent is taken from a different price.

ANSWER 5: E

Problem 6:

Harold tosses a nickel 4 times. We want the probability that he gets at least as many heads as tails.

There are \(2^4=16\) equally likely outcomes.

“At least as many heads as tails” means:

\[
H \ge T
\]

Since there are 4 tosses total, this means Harold must get 2, 3, or 4 heads.

Count the outcomes:

\[
\binom42=6
\]

\[
\binom43=4
\]

\[
\binom44=1
\]

Total favorable outcomes:

\[
6+4+1=11
\]

So the probability is:

\[
\frac{11}{16}
\]

The tempting answer \(\frac12\) would count only strictly more heads than tails and would forget the tie case.

ANSWER 6: E

Problem 7:

The machine rule is:

- If \(N\) is even, output \(\frac N2\).
- If \(N\) is odd, output \(3N+1\).

We need all starting positive integers \(N\) that become 1 after exactly 6 steps.

It is easiest to work backward from 1.

If a number outputs \(y\), then possible previous numbers are:

1. \(2y\), because if the previous number was even, halving it gives \(y\).
2. \(\frac{y-1}{3}\), but only if this is a positive odd integer.

Now work backward 6 steps.

Step 0:

\[
\{1\}
\]

Step 1 predecessors of 1:

\[
\{2\}
\]

Step 2 predecessors of 2:

\[
\{4\}
\]

Step 3 predecessors of 4:

\[
2\cdot 4=8
\]

and

\[
\frac{4-1}{3}=1
\]

So:

\[
\{8,1\}
\]

Step 4:

Predecessors of 8: \(16\)

Predecessors of 1: \(2\)

So:

\[
\{16,2\}
\]

Step 5:

Predecessors of 16:

\[
32,\quad \frac{16-1}{3}=5
\]

Predecessors of 2:

\[
4
\]

So:

\[
\{32,5,4\}
\]

Step 6:

Predecessors of 32:

\[
64
\]

Predecessors of 5:

\[
10
\]

Predecessors of 4:

\[
8,\quad 1
\]

So the possible starting values are:

\[
64,10,8,1
\]

Their sum is:

\[
64+10+8+1=83
\]

ANSWER 7: E

Problem 8:

Let the list be:

\[
a_1,a_2,a_3,a_4,a_5,a_6,a_7,a_8
\]

Each new number is the product of the two previous numbers:

\[
a_3=a_1a_2
\]

\[
a_4=a_2a_3
\]

and so on.

We are told the last three numbers are:

\[
a_6=16,\quad a_7=64,\quad a_8=1024
\]

Check:

\[
16\cdot 64=1024
\]

Now work backward.

Since:

\[
a_7=a_5a_6
\]

we have:

\[
64=

Problem 1:
We are asked to find the fraction of possible 7-digit telephone numbers that begin with 9 and end with 0, given that a telephone number cannot begin with 0 or 1.

First, let's find the total number of possible telephone numbers:
- The first digit can be any number from 2 to 9, which gives 8 possibilities.
- The remaining 6 digits can be any number from 0 to 9, giving 10 possibilities each.
Total possible telephone numbers = 8 × 10 × 10 × 10 × 10 × 10 × 10 = 8,000,000.

Next, let's find the number of telephone numbers that begin with 9 and end with 0:
- The first digit must be 9 (1 possibility).
- The last digit must be 0 (1 possibility).
- The middle 5 digits can be any number from 0 to 9 (10 possibilities each).
Total such numbers = 1 × 10 × 10 × 10 × 10 × 10 × 1 = 100,000.

The fraction of telephone numbers that meet this condition is:
100,000 / 8,000,000 = 1 / 80.

ANSWER 1: B

Problem 2:
We are asked to find the least possible value of M, which is the median of the medians of five groups of 5 numbers each, chosen from the integers 1 through 25.

Let the 5 groups be G1, G2, G3, G4, and G5, and let their respective medians be m1, m2, m3, m4, and m5. 
Assume the medians are ordered such that m1 < m2 < m3 < m4 < m5. The median of these medians is M = m3.
To minimize m3, we must make the numbers in the groups as small as possible. 

For m3 to be a certain value, there must be at least 3 groups (G1, G2, G3) whose medians are less than or equal to m3. 
In each of these 3 groups, there are 3 numbers that are less than or equal to the group's median. 
Therefore, there are at least 3 groups × 3 numbers = 9 numbers that are less than or equal to m3.

Since the numbers are distinct integers from 1 to 25, the 9th smallest possible integer is 9. Thus, m3 must be at least 9. 
We can verify that 9 is possible with the following grouping:
G1: 1, 2, 7, 24, 25 (median 7)
G2: 3, 4, 8, 22, 23 (median 8)
G3: 5, 6, 9, 20, 21 (median 9)
G4: 10, 11, 12, 18, 19 (median 12)
G5: 13, 14, 15, 16, 17 (median 15)
The medians are 7, 8, 9, 12, and 15. Their median is 9.

ANSWER 2: A

Problem 3:
We want to find the probability that a cricket, which hops to one of 3 other leaves with equal probability, returns to its starting leaf after exactly 4 hops.

Let the leaves be A, B, C, and D, and assume the cricket starts at leaf A.
Let P(n) be the probability that the cricket is at leaf A after n hops.
- P(0) = 1 (it starts at A).
- For any hop n > 0, the cricket can only land on A if it was NOT on A in the previous hop. If it was on B, C, or D, it has a 1/3 chance of hopping to A.
Thus, the recursive formula is P(n) = (1 - P(n-1)) × (1/3).

Let's calculate step-by-step:
- P(1) = (1 - 1) × 1/3 = 0
- P(2) = (1 - 0) × 1/3 = 1/3
- P(3) = (1 - 1/3) × 1/3 = (2/3) × (1/3) = 2/9
- P(4) = (1 - 2/9) × 1/3 = (7/9) × (1/3) = 7/27

ANSWER 3: E

Problem 4:
We are given a sequence of letters "AJHSME" (6 characters) and a sequence of digits "1989" (4 characters). They are cycled separately line by line. We want to find the line number where the original combination "AJHSME 1989" appears again for the first time.

The letters cycle back to their original arrangement every 6 lines.
The digits cycle back to their original arrangement every 4 lines.
For both to return to their original arrangement simultaneously, the line number must be a common multiple of both 6 and 4. 

The least common multiple (LCM) of 6 and 4 is 12.
Therefore, on line 12, the letters will have completed exactly 2 full cycles and the digits will have completed exactly 3 full cycles, returning to "AJHSME 1989".

ANSWER 4: C

Problem 5:
Let the original price of the shirt be P, and let the percentage of increase and decrease be x.
When the price is increased by x percent, it becomes P(1 + x/100).
When this new price is decreased by x percent, it becomes P(1 + x/100)(1 - x/100).

We are given that the resulting price is 84% of the original price, so:
P(1 + x/100)(1 - x/100) = 0.84P

Divide both sides by P:
(1 + x/100)(1 - x/100) = 0.84
1 - (x/100)² = 0.84
(x/100)² = 1 - 0.84
(x/100)² = 0.16

Taking the square root of both sides:
x/100 = 0.4
x = 40

The price was increased and decreased by 40%.

ANSWER 5: E

Problem 6:
Harold tosses a nickel four times. We want to find the probability that he gets at least as many heads as tails.
"At least as many heads as tails" in 4 tosses means getting 2, 3, or 4 heads.

The total number of possible outcomes when tossing a coin 4 times is 2⁴ = 16.
Using combinations, we find the number of ways to get 2, 3, or 4 heads:
- 2 heads: 4C2 = 6 ways
- 3 heads: 4C3 = 4 ways
- 4 heads: 4C4 = 1 way

Total favorable outcomes = 6 + 4 + 1 = 11.
The probability is 11 / 16.

ANSWER 6: E

Problem 7:
The machine follows the Collatz-like rule:
- If N is even, output N/2
- If N is odd, output 3N + 1

We are told a 6-step process ends in 1. Let's work backward from Step 6 to find all possible starting positive integers N (Step 0).
- Step 6: 1
- Step 5: Must be 2 (since 3x + 1 = 1 gives x = 0, which is not a positive integer).
- Step 4: Must be 4 (since 3x + 1 = 2 gives x = 1/3, not an integer).
- Step 3: Can be 8 (since 8/2 = 4) or 1 (since 3(1) + 1 = 4).

Branch 1 (Step 3 is 8):
- Step 2: Must be 16 (since 3x + 1 = 8 gives x = 7/3).
- Step 1: Can be 32 (32/2 = 16) or 5 (3(5) + 1 = 16).
- Step 0 (from 32): Must be 64 (since 3x + 1 = 32 gives x = 31/3).
- Step 0 (from 5): Must be 10 (since 3x + 1 = 5 gives x = 4/3).

Branch 2 (Step 3 is 1):
- Step 2: Must be 2.
- Step 1: Must be 4.
- Step 0: Can be 8 (8/2 = 4) or 1 (3(1) + 1 = 4).

The possible starting integers N are 64, 10, 8, and 1.
The sum of all such integers is 64 + 10 + 8 + 1 = 83.

ANSWER 7: E

Problem 8:
We are given a sequence of 8 numbers where each number is the product of the two preceding numbers: a_n = a_{n-2} × a_{n-1}.
We know the last three numbers: a_6 = 16, a_7 = 64, and a_8 = 1024.
Let's work backward using the rule a_{n-2} = a_n / a_{n-1}:
- a_5 = a_7 / a_6 = 64 / 16 = 4
- a_4 

## Step 1: Problem 1 - Understanding the problem
We need to find the fraction of 7-digit telephone numbers that begin with 9 and end with 0, given that the numbers cannot start with 0 or 1.

## Step 2: Problem 1 - Calculating total possible telephone numbers
The first digit can be any number from 2 to 9, so there are 8 possibilities. The remaining 6 digits can be any number from 0 to 9, so there are 10 possibilities for each. Thus, the total number of telephone numbers is $8 \times 10^6$.

## Step 3: Problem 1 - Calculating the number of telephone numbers that begin with 9 and end with 0
For a number to begin with 9 and end with 0, the first digit must be 9 and the last digit must be 0. The first digit has 1 possibility (9), the last digit has 1 possibility (0), and the remaining 5 digits have 10 possibilities each. So, the total number of such telephone numbers is $1 \times 10^5 \times 1$.

## Step 4: Problem 1 - Calculating the fraction
The fraction of telephone numbers that begin with 9 and end with 0 is $\frac{1 \times 10^5 \times 1}{8 \times 10^6} = \frac{10^5}{8 \times 10^6} = \frac{1}{80}$.

## Step 5: Problem 1 - Answer
ANSWER 1: B

## Step 6: Problem 2 - Understanding the problem
We are given 25 integers divided into 5 groups of 5 numbers each. We need to find the least possible value of M, the median of the medians of these groups.

## Step 7: Problem 2 - Analyzing the medians
To minimize M, we should minimize the medians of the 5 groups. The integers are from 1 to 25. The median of each group will be the third number when the numbers are arranged in ascending order.

## Step 8: Problem 2 - Distributing numbers to minimize M
To minimize M, we can distribute the numbers such that the medians of the groups are as small as possible. One way to do this is by grouping the smallest numbers together and ensuring that the medians are as low as possible.

## Step 9: Problem 2 - Calculating the medians of the groups
Let's distribute the numbers: Group 1 = {1, 2, 3, 24, 25}, Group 2 = {4, 5, 6, 22, 23}, Group 3 = {7, 8, 9, 20, 21}, Group 4 = {10, 11, 12, 18, 19}, Group 5 = {13, 14, 15, 16, 17}. The medians are 3, 6, 9, 12, 15.

## Step 10: Problem 2 - Finding M
The medians of the groups are 3, 6, 9, 12, 15. The median of these medians (M) is 9.

## Step 11: Problem 2 - Answer
ANSWER 2: A

## Step 12: Problem 3 - Understanding the problem
A cricket hops between 4 leaves with equal probability. We need to find the probability that it returns to the starting leaf after 4 hops.

## Step 13: Problem 3 - Analyzing the hops
After the first hop, the cricket is on a different leaf. For it to return to the starting leaf after 4 hops, it must be on a different leaf after the third hop as well.

## Step 14: Problem 3 - Calculating the probability
Let's denote the probability of being on the starting leaf after n hops as $P_n$. $P_0 = 1$, $P_1 = 0$. For $n \geq 1$, $P_{n+1} = \frac{1}{3}(1 - P_n)$ because the probability of not being on the starting leaf after n hops is $1 - P_n$, and from any other leaf, there's a $\frac{1}{3}$ chance of hopping to the starting leaf.

## Step 15: Problem 3 - Computing $P_4$
$P_2 = \frac{1}{3}$, $P_3 = \frac{1}{3}(1 - \frac{1}{3}) = \frac{2}{9}$, $P_4 = \frac{1}{3}(1 - \frac{2}{9}) = \frac{1}{3} \times \frac{7}{9} = \frac{7}{27}$.

## Step 16: Problem 3 - Answer
ANSWER 3: E

## Step 17: Problem 4 - Understanding the problem
The letters and digits are cycled separately and combined. We need to find when "AJHSME 1989" appears again.

## Step 18: Problem 4 - Analyzing the cycles
The letters cycle every 6 steps, and the digits cycle every 4 steps (since there are 4 digits and they are distinct when considering their positions). The combined cycle will be the least common multiple (LCM) of 6 and 4.

## Step 19: Problem 4 - Calculating the LCM of 6 and 4
The LCM of 6 and 4 is 12.

## Step 20: Problem 4 - Answer
ANSWER 4: C

## Step 21: Problem 5 - Understanding the problem
A price is increased by a certain percent and then decreased by the same percent, resulting in 84% of the original price.

## Step 22: Problem 5 - Setting up the equation
Let $x$ be the original price and $p$ be the percentage change. The new price after increase is $x(1 + \frac{p}{100})$, and after decrease is $x(1 + \frac{p}{100})(1 - \frac{p}{100}) = 0.84x$.

## Step 23: Problem 5 - Solving for $p$
$x(1 - \frac{p^2}{10000}) = 0.84x$. Simplifying gives $1 - \frac{p^2}{10000} = 0.84$. Thus, $\frac{p^2}{10000} = 0.16$, so $p^2 = 1600$, and $p = 40$.

## Step 24: Problem 5 - Answer
ANSWER 5: E

## Step 25: Problem 6 - Understanding the problem
Harold tosses a nickel four times. We need to find the probability of getting at least as many heads as tails.

## Step 26: Problem 6 - Enumerating the outcomes
The possible outcomes where heads are at least as many as tails are: 4 heads, 3 heads and 1 tail, and 2 heads and 2 tails.

## Step 27: Problem 6 - Calculating the probabilities
The total number of outcomes is $2^4 = 16$. The number of ways to get 4 heads is 1, 3 heads and 1 tail is $\binom{4}{3} = 4$, and 2 heads and 2 tails is $\binom{4}{2} = 6$. So, the total favorable outcomes are $1 + 4 + 6 = 11$.

## Step 28: Problem 6 - Calculating the probability
The probability is $\frac{11}{16}$.

## Step 29: Problem 6 - Answer
ANSWER 6: E

## Step 30: Problem 7 - Understanding the problem
A machine applies a rule to a positive integer N: if N is even, output N/2; if N is odd, output 3N + 1. We need to find the sum of all N that result in 1 after 6 steps.

## Step 31: Problem 7 - Working backwards from 1
After 6 steps, the output is 1. So, before that, it was 2 (since 2 is even and 2/2 = 1). Before that, it could be 4. Continuing this process backwards: 4 -> 2 -> 1 (3 steps), 8 -> 4 -> 2 -> 1 (4 steps), 16 -> 8 -> 4 -> 2 -> 1 (5 steps), and so on, until we reach 6 steps.

## Step 32: Problem 7 - Enumerating the possible N
For 6 steps: $2^6 = 64$ is a possible N (since 64 -> 32 -> 16 -> 8 -> 4 -> 2 -> 1). Working bac

Problem 1:

We are asked what fraction of possible telephone numbers begin with 9 and end with 0.

A 7-digit telephone number cannot begin with 0 or 1.

Total possible telephone numbers:

- First digit: 2 through 9, so 8 choices.
- Remaining 6 digits: each has 10 choices.

So total numbers:

\[
8 \cdot 10^6
\]

Now count numbers that begin with 9 and end with 0:

- First digit fixed as 9: 1 choice.
- Last digit fixed as 0: 1 choice.
- Middle 5 digits: each has 10 choices.

So favorable numbers:

\[
10^5
\]

Thus the fraction is:

\[
\frac{10^5}{8\cdot 10^6}=\frac{1}{80}
\]

The tempting answer \(\frac{1}{100}\) would be correct only if all first digits were equally allowed, including 0 and 1, but they are not.

ANSWER 1: B

Problem 2:

We have the integers 1 through 25 separated into five groups of 5 numbers each. We find the median of each group, then take the median of those five medians. We want the least possible value of \(M\).

For a group of 5 numbers to have median at most \(m\), at least 3 numbers in that group must be at most \(m\).

Since \(M\) is the median of the five medians, at least 3 of the group medians must be at most \(M\).

Therefore, to make \(M\) as small as possible, we need at least:

\[
3 \times 3 = 9
\]

numbers that are at most \(M\).

The 9 smallest numbers are:

\[
1,2,3,4,5,6,7,8,9
\]

So \(M\) cannot be less than 9.

Now we show that \(M=9\) is possible. For example:

\[
\{1,2,3,22,23\} \quad \text{median } 3
\]

\[
\{4,5,6,24,25\} \quad \text{median } 6
\]

\[
\{7,8,9,20,21\} \quad \text{median } 9
\]

The remaining numbers can form two groups with medians larger than 9:

\[
\{10,11,12,13,14\} \quad \text{median } 12
\]

\[
\{15,16,17,18,19\} \quad \text{median } 17
\]

The five medians are:

\[
3,6,9,12,17
\]

Their median is 9.

So the least possible value is 9.

ANSWER 2: A

Problem 3:

The cricket starts on one leaf and hops to one of the other 3 leaves each time. We want the probability that after 4 hops it is back where it started.

Let \(p_n\) be the probability that the cricket is on its starting leaf after \(n\) hops.

Initially:

\[
p_0=1
\]

After one hop, it cannot be on the starting leaf:

\[
p_1=0
\]

If the cricket is not on the starting leaf, then on the next hop it has probability \(\frac13\) of hopping back to the starting leaf. Therefore:

\[
p_{n+1}=\frac{1-p_n}{3}
\]

Now compute:

\[
p_2=\frac{1-0}{3}=\frac13
\]

\[
p_3=\frac{1-\frac13}{3}=\frac{\frac23}{3}=\frac29
\]

\[
p_4=\frac{1-\frac29}{3}=\frac{\frac79}{3}=\frac7{27}
\]

So the probability is:

\[
\frac7{27}
\]

Choice A, \(\frac29\), is the probability after 3 hops, not 4 hops.

ANSWER 3: E

Problem 4:

The letters and digits are cycled separately.

The letters are:

\[
AJHSME
\]

There are 6 letters, so the letters return to their original order after 6 cycles.

The digits are:

\[
1989
\]

There are 4 digits, so the digits return to their original order after 4 cycles.

We need both the letters and the digits to return to their original arrangements at the same time.

So we need the least common multiple of 6 and 4:

\[
\operatorname{lcm}(6,4)=12
\]

Thus \(AJHSME\ 1989\) appears again for the first time on line 12.

Choice A, 6, would return only the letters, not necessarily the digits. Choice B, 10, returns neither. Choice E, 24, is a common multiple but not the first one.

ANSWER 4: C

Problem 5:

Let the original price be \(P\).

Suppose the price is increased by \(r\), where \(r\) is written as a decimal. Then the new price is:

\[
P(1+r)
\]

Then it is decreased by the same percent, so the final price is:

\[
P(1+r)(1-r)
\]

We are told this equals 84% of the original price:

\[
P(1+r)(1-r)=0.84P
\]

Cancel \(P\):

\[
(1+r)(1-r)=0.84
\]

\[
1-r^2=0.84
\]

\[
r^2=0.16
\]

\[
r=0.4
\]

So the percent is:

\[
40\%
\]

A common mistake is to think an increase and decrease by the same percent cancel each other, but they do not because the second percent is taken from a different price.

ANSWER 5: E

Problem 6:

Harold tosses a nickel 4 times. We want the probability that he gets at least as many heads as tails.

There are \(2^4=16\) equally likely outcomes.

“At least as many heads as tails” means:

\[
H \ge T
\]

Since there are 4 tosses total, this means Harold must get 2, 3, or 4 heads.

Count the outcomes:

\[
\binom42=6
\]

\[
\binom43=4
\]

\[
\binom44=1
\]

Total favorable outcomes:

\[
6+4+1=11
\]

So the probability is:

\[
\frac{11}{16}
\]

The tempting answer \(\frac12\) would count only strictly more heads than tails and would forget the tie case.

ANSWER 6: E

Problem 7:

The machine rule is:

- If \(N\) is even, output \(\frac N2\).
- If \(N\) is odd, output \(3N+1\).

We need all starting positive integers \(N\) that become 1 after exactly 6 steps.

It is easiest to work backward from 1.

If a number outputs \(y\), then possible previous numbers are:

1. \(2y\), because if the previous number was even, halving it gives \(y\).
2. \(\frac{y-1}{3}\), but only if this is a positive odd integer.

Now work backward 6 steps.

Step 0:

\[
\{1\}
\]

Step 1 predecessors of 1:

\[
\{2\}
\]

Step 2 predecessors of 2:

\[
\{4\}
\]

Step 3 predecessors of 4:

\[
2\cdot 4=8
\]

and

\[
\frac{4-1}{3}=1
\]

So:

\[
\{8,1\}
\]

Step 4:

Predecessors of 8: \(16\)

Predecessors of 1: \(2\)

So:

\[
\{16,2\}
\]

Step 5:

Predecessors of 16:

\[
32,\quad \frac{16-1}{3}=5
\]

Predecessors of 2:

\[
4
\]

So:

\[
\{32,5,4\}
\]

Step 6:

Predecessors of 32:

\[
64
\]

Predecessors of 5:

\[
10
\]

Predecessors of 4:

\[
8,\quad 1
\]

So the possible starting values are:

\[
64,10,8,1
\]

Their sum is:

\[
64+10+8+1=83
\]

ANSWER 7: E

Problem 8:

Let the list be:

\[
a_1,a_2,a_3,a_4,a_5,a_6,a_7,a_8
\]

Each new number is the product of the two previous numbers:

\[
a_3=a_1a_2
\]

\[
a_4=a_2a_3
\]

and so on.

We are told the last three numbers are:

\[
a_6=16,\quad a_7=64,\quad a_8=1024
\]

Check:

\[
16\cdot 64=1024
\]

Now work backward.

Since:

\[
a_7=a_5a_6
\]

we have:

\[
64=

Problem 1:
We are asked to find the fraction of possible 7-digit telephone numbers that begin with 9 and end with 0, given that a telephone number cannot begin with 0 or 1.

First, let's find the total number of possible telephone numbers:
- The first digit can be any number from 2 to 9, which gives 8 possibilities.
- The remaining 6 digits can be any number from 0 to 9, giving 10 possibilities each.
Total possible telephone numbers = 8 × 10 × 10 × 10 × 10 × 10 × 10 = 8,000,000.

Next, let's find the number of telephone numbers that begin with 9 and end with 0:
- The first digit must be 9 (1 possibility).
- The last digit must be 0 (1 possibility).
- The middle 5 digits can be any number from 0 to 9 (10 possibilities each).
Total such numbers = 1 × 10 × 10 × 10 × 10 × 10 × 1 = 100,000.

The fraction of telephone numbers that meet this condition is:
100,000 / 8,000,000 = 1 / 80.

ANSWER 1: B

Problem 2:
We are asked to find the least possible value of M, which is the median of the medians of five groups of 5 numbers each, chosen from the integers 1 through 25.

Let the 5 groups be G1, G2, G3, G4, and G5, and let their respective medians be m1, m2, m3, m4, and m5. 
Assume the medians are ordered such that m1 < m2 < m3 < m4 < m5. The median of these medians is M = m3.
To minimize m3, we must make the numbers in the groups as small as possible. 

For m3 to be a certain value, there must be at least 3 groups (G1, G2, G3) whose medians are less than or equal to m3. 
In each of these 3 groups, there are 3 numbers that are less than or equal to the group's median. 
Therefore, there are at least 3 groups × 3 numbers = 9 numbers that are less than or equal to m3.

Since the numbers are distinct integers from 1 to 25, the 9th smallest possible integer is 9. Thus, m3 must be at least 9. 
We can verify that 9 is possible with the following grouping:
G1: 1, 2, 7, 24, 25 (median 7)
G2: 3, 4, 8, 22, 23 (median 8)
G3: 5, 6, 9, 20, 21 (median 9)
G4: 10, 11, 12, 18, 19 (median 12)
G5: 13, 14, 15, 16, 17 (median 15)
The medians are 7, 8, 9, 12, and 15. Their median is 9.

ANSWER 2: A

Problem 3:
We want to find the probability that a cricket, which hops to one of 3 other leaves with equal probability, returns to its starting leaf after exactly 4 hops.

Let the leaves be A, B, C, and D, and assume the cricket starts at leaf A.
Let P(n) be the probability that the cricket is at leaf A after n hops.
- P(0) = 1 (it starts at A).
- For any hop n > 0, the cricket can only land on A if it was NOT on A in the previous hop. If it was on B, C, or D, it has a 1/3 chance of hopping to A.
Thus, the recursive formula is P(n) = (1 - P(n-1)) × (1/3).

Let's calculate step-by-step:
- P(1) = (1 - 1) × 1/3 = 0
- P(2) = (1 - 0) × 1/3 = 1/3
- P(3) = (1 - 1/3) × 1/3 = (2/3) × (1/3) = 2/9
- P(4) = (1 - 2/9) × 1/3 = (7/9) × (1/3) = 7/27

ANSWER 3: E

Problem 4:
We are given a sequence of letters "AJHSME" (6 characters) and a sequence of digits "1989" (4 characters). They are cycled separately line by line. We want to find the line number where the original combination "AJHSME 1989" appears again for the first time.

The letters cycle back to their original arrangement every 6 lines.
The digits cycle back to their original arrangement every 4 lines.
For both to return to their original arrangement simultaneously, the line number must be a common multiple of both 6 and 4. 

The least common multiple (LCM) of 6 and 4 is 12.
Therefore, on line 12, the letters will have completed exactly 2 full cycles and the digits will have completed exactly 3 full cycles, returning to "AJHSME 1989".

ANSWER 4: C

Problem 5:
Let the original price of the shirt be P, and let the percentage of increase and decrease be x.
When the price is increased by x percent, it becomes P(1 + x/100).
When this new price is decreased by x percent, it becomes P(1 + x/100)(1 - x/100).

We are given that the resulting price is 84% of the original price, so:
P(1 + x/100)(1 - x/100) = 0.84P

Divide both sides by P:
(1 + x/100)(1 - x/100) = 0.84
1 - (x/100)² = 0.84
(x/100)² = 1 - 0.84
(x/100)² = 0.16

Taking the square root of both sides:
x/100 = 0.4
x = 40

The price was increased and decreased by 40%.

ANSWER 5: E

Problem 6:
Harold tosses a nickel four times. We want to find the probability that he gets at least as many heads as tails.
"At least as many heads as tails" in 4 tosses means getting 2, 3, or 4 heads.

The total number of possible outcomes when tossing a coin 4 times is 2⁴ = 16.
Using combinations, we find the number of ways to get 2, 3, or 4 heads:
- 2 heads: 4C2 = 6 ways
- 3 heads: 4C3 = 4 ways
- 4 heads: 4C4 = 1 way

Total favorable outcomes = 6 + 4 + 1 = 11.
The probability is 11 / 16.

ANSWER 6: E

Problem 7:
The machine follows the Collatz-like rule:
- If N is even, output N/2
- If N is odd, output 3N + 1

We are told a 6-step process ends in 1. Let's work backward from Step 6 to find all possible starting positive integers N (Step 0).
- Step 6: 1
- Step 5: Must be 2 (since 3x + 1 = 1 gives x = 0, which is not a positive integer).
- Step 4: Must be 4 (since 3x + 1 = 2 gives x = 1/3, not an integer).
- Step 3: Can be 8 (since 8/2 = 4) or 1 (since 3(1) + 1 = 4).

Branch 1 (Step 3 is 8):
- Step 2: Must be 16 (since 3x + 1 = 8 gives x = 7/3).
- Step 1: Can be 32 (32/2 = 16) or 5 (3(5) + 1 = 16).
- Step 0 (from 32): Must be 64 (since 3x + 1 = 32 gives x = 31/3).
- Step 0 (from 5): Must be 10 (since 3x + 1 = 5 gives x = 4/3).

Branch 2 (Step 3 is 1):
- Step 2: Must be 2.
- Step 1: Must be 4.
- Step 0: Can be 8 (8/2 = 4) or 1 (3(1) + 1 = 4).

The possible starting integers N are 64, 10, 8, and 1.
The sum of all such integers is 64 + 10 + 8 + 1 = 83.

ANSWER 7: E

Problem 8:
We are given a sequence of 8 numbers where each number is the product of the two preceding numbers: a_n = a_{n-2} × a_{n-1}.
We know the last three numbers: a_6 = 16, a_7 = 64, and a_8 = 1024.
Let's work backward using the rule a_{n-2} = a_n / a_{n-1}:
- a_5 = a_7 / a_6 = 64 / 16 = 4
- a_4 

## Step 1: Problem 1 - Understanding the problem
We need to find the fraction of 7-digit telephone numbers that begin with 9 and end with 0, given that the numbers cannot start with 0 or 1.

## Step 2: Problem 1 - Calculating total possible telephone numbers
The first digit can be any number from 2 to 9, so there are 8 possibilities. The remaining 6 digits can be any number from 0 to 9, so there are 10 possibilities for each. Thus, the total number of telephone numbers is $8 \times 10^6$.

## Step 3: Problem 1 - Calculating the number of telephone numbers that begin with 9 and end with 0
For a number to begin with 9 and end with 0, the first digit must be 9 and the last digit must be 0. The first digit has 1 possibility (9), the last digit has 1 possibility (0), and the remaining 5 digits have 10 possibilities each. So, the total number of such telephone numbers is $1 \times 10^5 \times 1$.

## Step 4: Problem 1 - Calculating the fraction
The fraction of telephone numbers that begin with 9 and end with 0 is $\frac{1 \times 10^5 \times 1}{8 \times 10^6} = \frac{10^5}{8 \times 10^6} = \frac{1}{80}$.

## Step 5: Problem 1 - Answer
ANSWER 1: B

## Step 6: Problem 2 - Understanding the problem
We are given 25 integers divided into 5 groups of 5 numbers each. We need to find the least possible value of M, the median of the medians of these groups.

## Step 7: Problem 2 - Analyzing the medians
To minimize M, we should minimize the medians of the 5 groups. The integers are from 1 to 25. The median of each group will be the third number when the numbers are arranged in ascending order.

## Step 8: Problem 2 - Distributing numbers to minimize M
To minimize M, we can distribute the numbers such that the medians of the groups are as small as possible. One way to do this is by grouping the smallest numbers together and ensuring that the medians are as low as possible.

## Step 9: Problem 2 - Calculating the medians of the groups
Let's distribute the numbers: Group 1 = {1, 2, 3, 24, 25}, Group 2 = {4, 5, 6, 22, 23}, Group 3 = {7, 8, 9, 20, 21}, Group 4 = {10, 11, 12, 18, 19}, Group 5 = {13, 14, 15, 16, 17}. The medians are 3, 6, 9, 12, 15.

## Step 10: Problem 2 - Finding M
The medians of the groups are 3, 6, 9, 12, 15. The median of these medians (M) is 9.

## Step 11: Problem 2 - Answer
ANSWER 2: A

## Step 12: Problem 3 - Understanding the problem
A cricket hops between 4 leaves with equal probability. We need to find the probability that it returns to the starting leaf after 4 hops.

## Step 13: Problem 3 - Analyzing the hops
After the first hop, the cricket is on a different leaf. For it to return to the starting leaf after 4 hops, it must be on a different leaf after the third hop as well.

## Step 14: Problem 3 - Calculating the probability
Let's denote the probability of being on the starting leaf after n hops as $P_n$. $P_0 = 1$, $P_1 = 0$. For $n \geq 1$, $P_{n+1} = \frac{1}{3}(1 - P_n)$ because the probability of not being on the starting leaf after n hops is $1 - P_n$, and from any other leaf, there's a $\frac{1}{3}$ chance of hopping to the starting leaf.

## Step 15: Problem 3 - Computing $P_4$
$P_2 = \frac{1}{3}$, $P_3 = \frac{1}{3}(1 - \frac{1}{3}) = \frac{2}{9}$, $P_4 = \frac{1}{3}(1 - \frac{2}{9}) = \frac{1}{3} \times \frac{7}{9} = \frac{7}{27}$.

## Step 16: Problem 3 - Answer
ANSWER 3: E

## Step 17: Problem 4 - Understanding the problem
The letters and digits are cycled separately and combined. We need to find when "AJHSME 1989" appears again.

## Step 18: Problem 4 - Analyzing the cycles
The letters cycle every 6 steps, and the digits cycle every 4 steps (since there are 4 digits and they are distinct when considering their positions). The combined cycle will be the least common multiple (LCM) of 6 and 4.

## Step 19: Problem 4 - Calculating the LCM of 6 and 4
The LCM of 6 and 4 is 12.

## Step 20: Problem 4 - Answer
ANSWER 4: C

## Step 21: Problem 5 - Understanding the problem
A price is increased by a certain percent and then decreased by the same percent, resulting in 84% of the original price.

## Step 22: Problem 5 - Setting up the equation
Let $x$ be the original price and $p$ be the percentage change. The new price after increase is $x(1 + \frac{p}{100})$, and after decrease is $x(1 + \frac{p}{100})(1 - \frac{p}{100}) = 0.84x$.

## Step 23: Problem 5 - Solving for $p$
$x(1 - \frac{p^2}{10000}) = 0.84x$. Simplifying gives $1 - \frac{p^2}{10000} = 0.84$. Thus, $\frac{p^2}{10000} = 0.16$, so $p^2 = 1600$, and $p = 40$.

## Step 24: Problem 5 - Answer
ANSWER 5: E

## Step 25: Problem 6 - Understanding the problem
Harold tosses a nickel four times. We need to find the probability of getting at least as many heads as tails.

## Step 26: Problem 6 - Enumerating the outcomes
The possible outcomes where heads are at least as many as tails are: 4 heads, 3 heads and 1 tail, and 2 heads and 2 tails.

## Step 27: Problem 6 - Calculating the probabilities
The total number of outcomes is $2^4 = 16$. The number of ways to get 4 heads is 1, 3 heads and 1 tail is $\binom{4}{3} = 4$, and 2 heads and 2 tails is $\binom{4}{2} = 6$. So, the total favorable outcomes are $1 + 4 + 6 = 11$.

## Step 28: Problem 6 - Calculating the probability
The probability is $\frac{11}{16}$.

## Step 29: Problem 6 - Answer
ANSWER 6: E

## Step 30: Problem 7 - Understanding the problem
A machine applies a rule to a positive integer N: if N is even, output N/2; if N is odd, output 3N + 1. We need to find the sum of all N that result in 1 after 6 steps.

## Step 31: Problem 7 - Working backwards from 1
After 6 steps, the output is 1. So, before that, it was 2 (since 2 is even and 2/2 = 1). Before that, it could be 4. Continuing this process backwards: 4 -> 2 -> 1 (3 steps), 8 -> 4 -> 2 -> 1 (4 steps), 16 -> 8 -> 4 -> 2 -> 1 (5 steps), and so on, until we reach 6 steps.

## Step 32: Problem 7 - Enumerating the possible N
For 6 steps: $2^6 = 64$ is a possible N (since 64 -> 32 -> 16 -> 8 -> 4 -> 2 -> 1). Working bac

Problem 1:

We are asked what fraction of possible telephone numbers begin with 9 and end with 0.

A 7-digit telephone number cannot begin with 0 or 1.

Total possible telephone numbers:

- First digit: 2 through 9, so 8 choices.
- Remaining 6 digits: each has 10 choices.

So total numbers:

\[
8 \cdot 10^6
\]

Now count numbers that begin with 9 and end with 0:

- First digit fixed as 9: 1 choice.
- Last digit fixed as 0: 1 choice.
- Middle 5 digits: each has 10 choices.

So favorable numbers:

\[
10^5
\]

Thus the fraction is:

\[
\frac{10^5}{8\cdot 10^6}=\frac{1}{80}
\]

The tempting answer \(\frac{1}{100}\) would be correct only if all first digits were equally allowed, including 0 and 1, but they are not.

ANSWER 1: B

Problem 2:

We have the integers 1 through 25 separated into five groups of 5 numbers each. We find the median of each group, then take the median of those five medians. We want the least possible value of \(M\).

For a group of 5 numbers to have median at most \(m\), at least 3 numbers in that group must be at most \(m\).

Since \(M\) is the median of the five medians, at least 3 of the group medians must be at most \(M\).

Therefore, to make \(M\) as small as possible, we need at least:

\[
3 \times 3 = 9
\]

numbers that are at most \(M\).

The 9 smallest numbers are:

\[
1,2,3,4,5,6,7,8,9
\]

So \(M\) cannot be less than 9.

Now we show that \(M=9\) is possible. For example:

\[
\{1,2,3,22,23\} \quad \text{median } 3
\]

\[
\{4,5,6,24,25\} \quad \text{median } 6
\]

\[
\{7,8,9,20,21\} \quad \text{median } 9
\]

The remaining numbers can form two groups with medians larger than 9:

\[
\{10,11,12,13,14\} \quad \text{median } 12
\]

\[
\{15,16,17,18,19\} \quad \text{median } 17
\]

The five medians are:

\[
3,6,9,12,17
\]

Their median is 9.

So the least possible value is 9.

ANSWER 2: A

Problem 3:

The cricket starts on one leaf and hops to one of the other 3 leaves each time. We want the probability that after 4 hops it is back where it started.

Let \(p_n\) be the probability that the cricket is on its starting leaf after \(n\) hops.

Initially:

\[
p_0=1
\]

After one hop, it cannot be on the starting leaf:

\[
p_1=0
\]

If the cricket is not on the starting leaf, then on the next hop it has probability \(\frac13\) of hopping back to the starting leaf. Therefore:

\[
p_{n+1}=\frac{1-p_n}{3}
\]

Now compute:

\[
p_2=\frac{1-0}{3}=\frac13
\]

\[
p_3=\frac{1-\frac13}{3}=\frac{\frac23}{3}=\frac29
\]

\[
p_4=\frac{1-\frac29}{3}=\frac{\frac79}{3}=\frac7{27}
\]

So the probability is:

\[
\frac7{27}
\]

Choice A, \(\frac29\), is the probability after 3 hops, not 4 hops.

ANSWER 3: E

Problem 4:

The letters and digits are cycled separately.

The letters are:

\[
AJHSME
\]

There are 6 letters, so the letters return to their original order after 6 cycles.

The digits are:

\[
1989
\]

There are 4 digits, so the digits return to their original order after 4 cycles.

We need both the letters and the digits to return to their original arrangements at the same time.

So we need the least common multiple of 6 and 4:

\[
\operatorname{lcm}(6,4)=12
\]

Thus \(AJHSME\ 1989\) appears again for the first time on line 12.

Choice A, 6, would return only the letters, not necessarily the digits. Choice B, 10, returns neither. Choice E, 24, is a common multiple but not the first one.

ANSWER 4: C

Problem 5:

Let the original price be \(P\).

Suppose the price is increased by \(r\), where \(r\) is written as a decimal. Then the new price is:

\[
P(1+r)
\]

Then it is decreased by the same percent, so the final price is:

\[
P(1+r)(1-r)
\]

We are told this equals 84% of the original price:

\[
P(1+r)(1-r)=0.84P
\]

Cancel \(P\):

\[
(1+r)(1-r)=0.84
\]

\[
1-r^2=0.84
\]

\[
r^2=0.16
\]

\[
r=0.4
\]

So the percent is:

\[
40\%
\]

A common mistake is to think an increase and decrease by the same percent cancel each other, but they do not because the second percent is taken from a different price.

ANSWER 5: E

Problem 6:

Harold tosses a nickel 4 times. We want the probability that he gets at least as many heads as tails.

There are \(2^4=16\) equally likely outcomes.

“At least as many heads as tails” means:

\[
H \ge T
\]

Since there are 4 tosses total, this means Harold must get 2, 3, or 4 heads.

Count the outcomes:

\[
\binom42=6
\]

\[
\binom43=4
\]

\[
\binom44=1
\]

Total favorable outcomes:

\[
6+4+1=11
\]

So the probability is:

\[
\frac{11}{16}
\]

The tempting answer \(\frac12\) would count only strictly more heads than tails and would forget the tie case.

ANSWER 6: E

Problem 7:

The machine rule is:

- If \(N\) is even, output \(\frac N2\).
- If \(N\) is odd, output \(3N+1\).

We need all starting positive integers \(N\) that become 1 after exactly 6 steps.

It is easiest to work backward from 1.

If a number outputs \(y\), then possible previous numbers are:

1. \(2y\), because if the previous number was even, halving it gives \(y\).
2. \(\frac{y-1}{3}\), but only if this is a positive odd integer.

Now work backward 6 steps.

Step 0:

\[
\{1\}
\]

Step 1 predecessors of 1:

\[
\{2\}
\]

Step 2 predecessors of 2:

\[
\{4\}
\]

Step 3 predecessors of 4:

\[
2\cdot 4=8
\]

and

\[
\frac{4-1}{3}=1
\]

So:

\[
\{8,1\}
\]

Step 4:

Predecessors of 8: \(16\)

Predecessors of 1: \(2\)

So:

\[
\{16,2\}
\]

Step 5:

Predecessors of 16:

\[
32,\quad \frac{16-1}{3}=5
\]

Predecessors of 2:

\[
4
\]

So:

\[
\{32,5,4\}
\]

Step 6:

Predecessors of 32:

\[
64
\]

Predecessors of 5:

\[
10
\]

Predecessors of 4:

\[
8,\quad 1
\]

So the possible starting values are:

\[
64,10,8,1
\]

Their sum is:

\[
64+10+8+1=83
\]

ANSWER 7: E

Problem 8:

Let the list be:

\[
a_1,a_2,a_3,a_4,a_5,a_6,a_7,a_8
\]

Each new number is the product of the two previous numbers:

\[
a_3=a_1a_2
\]

\[
a_4=a_2a_3
\]

and so on.

We are told the last three numbers are:

\[
a_6=16,\quad a_7=64,\quad a_8=1024
\]

Check:

\[
16\cdot 64=1024
\]

Now work backward.

Since:

\[
a_7=a_5a_6
\]

we have:

\[
64=

Problem 1:
We are asked to find the fraction of possible 7-digit telephone numbers that begin with 9 and end with 0, given that a telephone number cannot begin with 0 or 1.

First, let's find the total number of possible telephone numbers:
- The first digit can be any number from 2 to 9, which gives 8 possibilities.
- The remaining 6 digits can be any number from 0 to 9, giving 10 possibilities each.
Total possible telephone numbers = 8 × 10 × 10 × 10 × 10 × 10 × 10 = 8,000,000.

Next, let's find the number of telephone numbers that begin with 9 and end with 0:
- The first digit must be 9 (1 possibility).
- The last digit must be 0 (1 possibility).
- The middle 5 digits can be any number from 0 to 9 (10 possibilities each).
Total such numbers = 1 × 10 × 10 × 10 × 10 × 10 × 1 = 100,000.

The fraction of telephone numbers that meet this condition is:
100,000 / 8,000,000 = 1 / 80.

ANSWER 1: B

Problem 2:
We are asked to find the least possible value of M, which is the median of the medians of five groups of 5 numbers each, chosen from the integers 1 through 25.

Let the 5 groups be G1, G2, G3, G4, and G5, and let their respective medians be m1, m2, m3, m4, and m5. 
Assume the medians are ordered such that m1 < m2 < m3 < m4 < m5. The median of these medians is M = m3.
To minimize m3, we must make the numbers in the groups as small as possible. 

For m3 to be a certain value, there must be at least 3 groups (G1, G2, G3) whose medians are less than or equal to m3. 
In each of these 3 groups, there are 3 numbers that are less than or equal to the group's median. 
Therefore, there are at least 3 groups × 3 numbers = 9 numbers that are less than or equal to m3.

Since the numbers are distinct integers from 1 to 25, the 9th smallest possible integer is 9. Thus, m3 must be at least 9. 
We can verify that 9 is possible with the following grouping:
G1: 1, 2, 7, 24, 25 (median 7)
G2: 3, 4, 8, 22, 23 (median 8)
G3: 5, 6, 9, 20, 21 (median 9)
G4: 10, 11, 12, 18, 19 (median 12)
G5: 13, 14, 15, 16, 17 (median 15)
The medians are 7, 8, 9, 12, and 15. Their median is 9.

ANSWER 2: A

Problem 3:
We want to find the probability that a cricket, which hops to one of 3 other leaves with equal probability, returns to its starting leaf after exactly 4 hops.

Let the leaves be A, B, C, and D, and assume the cricket starts at leaf A.
Let P(n) be the probability that the cricket is at leaf A after n hops.
- P(0) = 1 (it starts at A).
- For any hop n > 0, the cricket can only land on A if it was NOT on A in the previous hop. If it was on B, C, or D, it has a 1/3 chance of hopping to A.
Thus, the recursive formula is P(n) = (1 - P(n-1)) × (1/3).

Let's calculate step-by-step:
- P(1) = (1 - 1) × 1/3 = 0
- P(2) = (1 - 0) × 1/3 = 1/3
- P(3) = (1 - 1/3) × 1/3 = (2/3) × (1/3) = 2/9
- P(4) = (1 - 2/9) × 1/3 = (7/9) × (1/3) = 7/27

ANSWER 3: E

Problem 4:
We are given a sequence of letters "AJHSME" (6 characters) and a sequence of digits "1989" (4 characters). They are cycled separately line by line. We want to find the line number where the original combination "AJHSME 1989" appears again for the first time.

The letters cycle back to their original arrangement every 6 lines.
The digits cycle back to their original arrangement every 4 lines.
For both to return to their original arrangement simultaneously, the line number must be a common multiple of both 6 and 4. 

The least common multiple (LCM) of 6 and 4 is 12.
Therefore, on line 12, the letters will have completed exactly 2 full cycles and the digits will have completed exactly 3 full cycles, returning to "AJHSME 1989".

ANSWER 4: C

Problem 5:
Let the original price of the shirt be P, and let the percentage of increase and decrease be x.
When the price is increased by x percent, it becomes P(1 + x/100).
When this new price is decreased by x percent, it becomes P(1 + x/100)(1 - x/100).

We are given that the resulting price is 84% of the original price, so:
P(1 + x/100)(1 - x/100) = 0.84P

Divide both sides by P:
(1 + x/100)(1 - x/100) = 0.84
1 - (x/100)² = 0.84
(x/100)² = 1 - 0.84
(x/100)² = 0.16

Taking the square root of both sides:
x/100 = 0.4
x = 40

The price was increased and decreased by 40%.

ANSWER 5: E

Problem 6:
Harold tosses a nickel four times. We want to find the probability that he gets at least as many heads as tails.
"At least as many heads as tails" in 4 tosses means getting 2, 3, or 4 heads.

The total number of possible outcomes when tossing a coin 4 times is 2⁴ = 16.
Using combinations, we find the number of ways to get 2, 3, or 4 heads:
- 2 heads: 4C2 = 6 ways
- 3 heads: 4C3 = 4 ways
- 4 heads: 4C4 = 1 way

Total favorable outcomes = 6 + 4 + 1 = 11.
The probability is 11 / 16.

ANSWER 6: E

Problem 7:
The machine follows the Collatz-like rule:
- If N is even, output N/2
- If N is odd, output 3N + 1

We are told a 6-step process ends in 1. Let's work backward from Step 6 to find all possible starting positive integers N (Step 0).
- Step 6: 1
- Step 5: Must be 2 (since 3x + 1 = 1 gives x = 0, which is not a positive integer).
- Step 4: Must be 4 (since 3x + 1 = 2 gives x = 1/3, not an integer).
- Step 3: Can be 8 (since 8/2 = 4) or 1 (since 3(1) + 1 = 4).

Branch 1 (Step 3 is 8):
- Step 2: Must be 16 (since 3x + 1 = 8 gives x = 7/3).
- Step 1: Can be 32 (32/2 = 16) or 5 (3(5) + 1 = 16).
- Step 0 (from 32): Must be 64 (since 3x + 1 = 32 gives x = 31/3).
- Step 0 (from 5): Must be 10 (since 3x + 1 = 5 gives x = 4/3).

Branch 2 (Step 3 is 1):
- Step 2: Must be 2.
- Step 1: Must be 4.
- Step 0: Can be 8 (8/2 = 4) or 1 (3(1) + 1 = 4).

The possible starting integers N are 64, 10, 8, and 1.
The sum of all such integers is 64 + 10 + 8 + 1 = 83.

ANSWER 7: E

Problem 8:
We are given a sequence of 8 numbers where each number is the product of the two preceding numbers: a_n = a_{n-2} × a_{n-1}.
We know the last three numbers: a_6 = 16, a_7 = 64, and a_8 = 1024.
Let's work backward using the rule a_{n-2} = a_n / a_{n-1}:
- a_5 = a_7 / a_6 = 64 / 16 = 4
- a_4 

## Step 1: Problem 1 - Understanding the problem
We need to find the fraction of 7-digit telephone numbers that begin with 9 and end with 0, given that the numbers cannot start with 0 or 1.

## Step 2: Problem 1 - Calculating total possible telephone numbers
The first digit can be any number from 2 to 9, so there are 8 possibilities. The remaining 6 digits can be any number from 0 to 9, so there are 10 possibilities for each. Thus, the total number of telephone numbers is $8 \times 10^6$.

## Step 3: Problem 1 - Calculating the number of telephone numbers that begin with 9 and end with 0
For a number to begin with 9 and end with 0, the first digit must be 9 and the last digit must be 0. The first digit has 1 possibility (9), the last digit has 1 possibility (0), and the remaining 5 digits have 10 possibilities each. So, the total number of such telephone numbers is $1 \times 10^5 \times 1$.

## Step 4: Problem 1 - Calculating the fraction
The fraction of telephone numbers that begin with 9 and end with 0 is $\frac{1 \times 10^5 \times 1}{8 \times 10^6} = \frac{10^5}{8 \times 10^6} = \frac{1}{80}$.

## Step 5: Problem 1 - Answer
ANSWER 1: B

## Step 6: Problem 2 - Understanding the problem
We are given 25 integers divided into 5 groups of 5 numbers each. We need to find the least possible value of M, the median of the medians of these groups.

## Step 7: Problem 2 - Analyzing the medians
To minimize M, we should minimize the medians of the 5 groups. The integers are from 1 to 25. The median of each group will be the third number when the numbers are arranged in ascending order.

## Step 8: Problem 2 - Distributing numbers to minimize M
To minimize M, we can distribute the numbers such that the medians of the groups are as small as possible. One way to do this is by grouping the smallest numbers together and ensuring that the medians are as low as possible.

## Step 9: Problem 2 - Calculating the medians of the groups
Let's distribute the numbers: Group 1 = {1, 2, 3, 24, 25}, Group 2 = {4, 5, 6, 22, 23}, Group 3 = {7, 8, 9, 20, 21}, Group 4 = {10, 11, 12, 18, 19}, Group 5 = {13, 14, 15, 16, 17}. The medians are 3, 6, 9, 12, 15.

## Step 10: Problem 2 - Finding M
The medians of the groups are 3, 6, 9, 12, 15. The median of these medians (M) is 9.

## Step 11: Problem 2 - Answer
ANSWER 2: A

## Step 12: Problem 3 - Understanding the problem
A cricket hops between 4 leaves with equal probability. We need to find the probability that it returns to the starting leaf after 4 hops.

## Step 13: Problem 3 - Analyzing the hops
After the first hop, the cricket is on a different leaf. For it to return to the starting leaf after 4 hops, it must be on a different leaf after the third hop as well.

## Step 14: Problem 3 - Calculating the probability
Let's denote the probability of being on the starting leaf after n hops as $P_n$. $P_0 = 1$, $P_1 = 0$. For $n \geq 1$, $P_{n+1} = \frac{1}{3}(1 - P_n)$ because the probability of not being on the starting leaf after n hops is $1 - P_n$, and from any other leaf, there's a $\frac{1}{3}$ chance of hopping to the starting leaf.

## Step 15: Problem 3 - Computing $P_4$
$P_2 = \frac{1}{3}$, $P_3 = \frac{1}{3}(1 - \frac{1}{3}) = \frac{2}{9}$, $P_4 = \frac{1}{3}(1 - \frac{2}{9}) = \frac{1}{3} \times \frac{7}{9} = \frac{7}{27}$.

## Step 16: Problem 3 - Answer
ANSWER 3: E

## Step 17: Problem 4 - Understanding the problem
The letters and digits are cycled separately and combined. We need to find when "AJHSME 1989" appears again.

## Step 18: Problem 4 - Analyzing the cycles
The letters cycle every 6 steps, and the digits cycle every 4 steps (since there are 4 digits and they are distinct when considering their positions). The combined cycle will be the least common multiple (LCM) of 6 and 4.

## Step 19: Problem 4 - Calculating the LCM of 6 and 4
The LCM of 6 and 4 is 12.

## Step 20: Problem 4 - Answer
ANSWER 4: C

## Step 21: Problem 5 - Understanding the problem
A price is increased by a certain percent and then decreased by the same percent, resulting in 84% of the original price.

## Step 22: Problem 5 - Setting up the equation
Let $x$ be the original price and $p$ be the percentage change. The new price after increase is $x(1 + \frac{p}{100})$, and after decrease is $x(1 + \frac{p}{100})(1 - \frac{p}{100}) = 0.84x$.

## Step 23: Problem 5 - Solving for $p$
$x(1 - \frac{p^2}{10000}) = 0.84x$. Simplifying gives $1 - \frac{p^2}{10000} = 0.84$. Thus, $\frac{p^2}{10000} = 0.16$, so $p^2 = 1600$, and $p = 40$.

## Step 24: Problem 5 - Answer
ANSWER 5: E

## Step 25: Problem 6 - Understanding the problem
Harold tosses a nickel four times. We need to find the probability of getting at least as many heads as tails.

## Step 26: Problem 6 - Enumerating the outcomes
The possible outcomes where heads are at least as many as tails are: 4 heads, 3 heads and 1 tail, and 2 heads and 2 tails.

## Step 27: Problem 6 - Calculating the probabilities
The total number of outcomes is $2^4 = 16$. The number of ways to get 4 heads is 1, 3 heads and 1 tail is $\binom{4}{3} = 4$, and 2 heads and 2 tails is $\binom{4}{2} = 6$. So, the total favorable outcomes are $1 + 4 + 6 = 11$.

## Step 28: Problem 6 - Calculating the probability
The probability is $\frac{11}{16}$.

## Step 29: Problem 6 - Answer
ANSWER 6: E

## Step 30: Problem 7 - Understanding the problem
A machine applies a rule to a positive integer N: if N is even, output N/2; if N is odd, output 3N + 1. We need to find the sum of all N that result in 1 after 6 steps.

## Step 31: Problem 7 - Working backwards from 1
After 6 steps, the output is 1. So, before that, it was 2 (since 2 is even and 2/2 = 1). Before that, it could be 4. Continuing this process backwards: 4 -> 2 -> 1 (3 steps), 8 -> 4 -> 2 -> 1 (4 steps), 16 -> 8 -> 4 -> 2 -> 1 (5 steps), and so on, until we reach 6 steps.

## Step 32: Problem 7 - Enumerating the possible N
For 6 steps: $2^6 = 64$ is a possible N (since 64 -> 32 -> 16 -> 8 -> 4 -> 2 -> 1). Working bac

Problem 1:

We are asked what fraction of possible telephone numbers begin with 9 and end with 0.

A 7-digit telephone number cannot begin with 0 or 1.

Total possible telephone numbers:

- First digit: 2 through 9, so 8 choices.
- Remaining 6 digits: each has 10 choices.

So total numbers:

\[
8 \cdot 10^6
\]

Now count numbers that begin with 9 and end with 0:

- First digit fixed as 9: 1 choice.
- Last digit fixed as 0: 1 choice.
- Middle 5 digits: each has 10 choices.

So favorable numbers:

\[
10^5
\]

Thus the fraction is:

\[
\frac{10^5}{8\cdot 10^6}=\frac{1}{80}
\]

The tempting answer \(\frac{1}{100}\) would be correct only if all first digits were equally allowed, including 0 and 1, but they are not.

ANSWER 1: B

Problem 2:

We have the integers 1 through 25 separated into five groups of 5 numbers each. We find the median of each group, then take the median of those five medians. We want the least possible value of \(M\).

For a group of 5 numbers to have median at most \(m\), at least 3 numbers in that group must be at most \(m\).

Since \(M\) is the median of the five medians, at least 3 of the group medians must be at most \(M\).

Therefore, to make \(M\) as small as possible, we need at least:

\[
3 \times 3 = 9
\]

numbers that are at most \(M\).

The 9 smallest numbers are:

\[
1,2,3,4,5,6,7,8,9
\]

So \(M\) cannot be less than 9.

Now we show that \(M=9\) is possible. For example:

\[
\{1,2,3,22,23\} \quad \text{median } 3
\]

\[
\{4,5,6,24,25\} \quad \text{median } 6
\]

\[
\{7,8,9,20,21\} \quad \text{median } 9
\]

The remaining numbers can form two groups with medians larger than 9:

\[
\{10,11,12,13,14\} \quad \text{median } 12
\]

\[
\{15,16,17,18,19\} \quad \text{median } 17
\]

The five medians are:

\[
3,6,9,12,17
\]

Their median is 9.

So the least possible value is 9.

ANSWER 2: A

Problem 3:

The cricket starts on one leaf and hops to one of the other 3 leaves each time. We want the probability that after 4 hops it is back where it started.

Let \(p_n\) be the probability that the cricket is on its starting leaf after \(n\) hops.

Initially:

\[
p_0=1
\]

After one hop, it cannot be on the starting leaf:

\[
p_1=0
\]

If the cricket is not on the starting leaf, then on the next hop it has probability \(\frac13\) of hopping back to the starting leaf. Therefore:

\[
p_{n+1}=\frac{1-p_n}{3}
\]

Now compute:

\[
p_2=\frac{1-0}{3}=\frac13
\]

\[
p_3=\frac{1-\frac13}{3}=\frac{\frac23}{3}=\frac29
\]

\[
p_4=\frac{1-\frac29}{3}=\frac{\frac79}{3}=\frac7{27}
\]

So the probability is:

\[
\frac7{27}
\]

Choice A, \(\frac29\), is the probability after 3 hops, not 4 hops.

ANSWER 3: E

Problem 4:

The letters and digits are cycled separately.

The letters are:

\[
AJHSME
\]

There are 6 letters, so the letters return to their original order after 6 cycles.

The digits are:

\[
1989
\]

There are 4 digits, so the digits return to their original order after 4 cycles.

We need both the letters and the digits to return to their original arrangements at the same time.

So we need the least common multiple of 6 and 4:

\[
\operatorname{lcm}(6,4)=12
\]

Thus \(AJHSME\ 1989\) appears again for the first time on line 12.

Choice A, 6, would return only the letters, not necessarily the digits. Choice B, 10, returns neither. Choice E, 24, is a common multiple but not the first one.

ANSWER 4: C

Problem 5:

Let the original price be \(P\).

Suppose the price is increased by \(r\), where \(r\) is written as a decimal. Then the new price is:

\[
P(1+r)
\]

Then it is decreased by the same percent, so the final price is:

\[
P(1+r)(1-r)
\]

We are told this equals 84% of the original price:

\[
P(1+r)(1-r)=0.84P
\]

Cancel \(P\):

\[
(1+r)(1-r)=0.84
\]

\[
1-r^2=0.84
\]

\[
r^2=0.16
\]

\[
r=0.4
\]

So the percent is:

\[
40\%
\]

A common mistake is to think an increase and decrease by the same percent cancel each other, but they do not because the second percent is taken from a different price.

ANSWER 5: E

Problem 6:

Harold tosses a nickel 4 times. We want the probability that he gets at least as many heads as tails.

There are \(2^4=16\) equally likely outcomes.

“At least as many heads as tails” means:

\[
H \ge T
\]

Since there are 4 tosses total, this means Harold must get 2, 3, or 4 heads.

Count the outcomes:

\[
\binom42=6
\]

\[
\binom43=4
\]

\[
\binom44=1
\]

Total favorable outcomes:

\[
6+4+1=11
\]

So the probability is:

\[
\frac{11}{16}
\]

The tempting answer \(\frac12\) would count only strictly more heads than tails and would forget the tie case.

ANSWER 6: E

Problem 7:

The machine rule is:

- If \(N\) is even, output \(\frac N2\).
- If \(N\) is odd, output \(3N+1\).

We need all starting positive integers \(N\) that become 1 after exactly 6 steps.

It is easiest to work backward from 1.

If a number outputs \(y\), then possible previous numbers are:

1. \(2y\), because if the previous number was even, halving it gives \(y\).
2. \(\frac{y-1}{3}\), but only if this is a positive odd integer.

Now work backward 6 steps.

Step 0:

\[
\{1\}
\]

Step 1 predecessors of 1:

\[
\{2\}
\]

Step 2 predecessors of 2:

\[
\{4\}
\]

Step 3 predecessors of 4:

\[
2\cdot 4=8
\]

and

\[
\frac{4-1}{3}=1
\]

So:

\[
\{8,1\}
\]

Step 4:

Predecessors of 8: \(16\)

Predecessors of 1: \(2\)

So:

\[
\{16,2\}
\]

Step 5:

Predecessors of 16:

\[
32,\quad \frac{16-1}{3}=5
\]

Predecessors of 2:

\[
4
\]

So:

\[
\{32,5,4\}
\]

Step 6:

Predecessors of 32:

\[
64
\]

Predecessors of 5:

\[
10
\]

Predecessors of 4:

\[
8,\quad 1
\]

So the possible starting values are:

\[
64,10,8,1
\]

Their sum is:

\[
64+10+8+1=83
\]

ANSWER 7: E

Problem 8:

Let the list be:

\[
a_1,a_2,a_3,a_4,a_5,a_6,a_7,a_8
\]

Each new number is the product of the two previous numbers:

\[
a_3=a_1a_2
\]

\[
a_4=a_2a_3
\]

and so on.

We are told the last three numbers are:

\[
a_6=16,\quad a_7=64,\quad a_8=1024
\]

Check:

\[
16\cdot 64=1024
\]

Now work backward.

Since:

\[
a_7=a_5a_6
\]

we have:

\[
64=

Problem 1:
We are asked to find the fraction of possible 7-digit telephone numbers that begin with 9 and end with 0, given that a telephone number cannot begin with 0 or 1.

First, let's find the total number of possible telephone numbers:
- The first digit can be any number from 2 to 9, which gives 8 possibilities.
- The remaining 6 digits can be any number from 0 to 9, giving 10 possibilities each.
Total possible telephone numbers = 8 × 10 × 10 × 10 × 10 × 10 × 10 = 8,000,000.

Next, let's find the number of telephone numbers that begin with 9 and end with 0:
- The first digit must be 9 (1 possibility).
- The last digit must be 0 (1 possibility).
- The middle 5 digits can be any number from 0 to 9 (10 possibilities each).
Total such numbers = 1 × 10 × 10 × 10 × 10 × 10 × 1 = 100,000.

The fraction of telephone numbers that meet this condition is:
100,000 / 8,000,000 = 1 / 80.

ANSWER 1: B

Problem 2:
We are asked to find the least possible value of M, which is the median of the medians of five groups of 5 numbers each, chosen from the integers 1 through 25.

Let the 5 groups be G1, G2, G3, G4, and G5, and let their respective medians be m1, m2, m3, m4, and m5. 
Assume the medians are ordered such that m1 < m2 < m3 < m4 < m5. The median of these medians is M = m3.
To minimize m3, we must make the numbers in the groups as small as possible. 

For m3 to be a certain value, there must be at least 3 groups (G1, G2, G3) whose medians are less than or equal to m3. 
In each of these 3 groups, there are 3 numbers that are less than or equal to the group's median. 
Therefore, there are at least 3 groups × 3 numbers = 9 numbers that are less than or equal to m3.

Since the numbers are distinct integers from 1 to 25, the 9th smallest possible integer is 9. Thus, m3 must be at least 9. 
We can verify that 9 is possible with the following grouping:
G1: 1, 2, 7, 24, 25 (median 7)
G2: 3, 4, 8, 22, 23 (median 8)
G3: 5, 6, 9, 20, 21 (median 9)
G4: 10, 11, 12, 18, 19 (median 12)
G5: 13, 14, 15, 16, 17 (median 15)
The medians are 7, 8, 9, 12, and 15. Their median is 9.

ANSWER 2: A

Problem 3:
We want to find the probability that a cricket, which hops to one of 3 other leaves with equal probability, returns to its starting leaf after exactly 4 hops.

Let the leaves be A, B, C, and D, and assume the cricket starts at leaf A.
Let P(n) be the probability that the cricket is at leaf A after n hops.
- P(0) = 1 (it starts at A).
- For any hop n > 0, the cricket can only land on A if it was NOT on A in the previous hop. If it was on B, C, or D, it has a 1/3 chance of hopping to A.
Thus, the recursive formula is P(n) = (1 - P(n-1)) × (1/3).

Let's calculate step-by-step:
- P(1) = (1 - 1) × 1/3 = 0
- P(2) = (1 - 0) × 1/3 = 1/3
- P(3) = (1 - 1/3) × 1/3 = (2/3) × (1/3) = 2/9
- P(4) = (1 - 2/9) × 1/3 = (7/9) × (1/3) = 7/27

ANSWER 3: E

Problem 4:
We are given a sequence of letters "AJHSME" (6 characters) and a sequence of digits "1989" (4 characters). They are cycled separately line by line. We want to find the line number where the original combination "AJHSME 1989" appears again for the first time.

The letters cycle back to their original arrangement every 6 lines.
The digits cycle back to their original arrangement every 4 lines.
For both to return to their original arrangement simultaneously, the line number must be a common multiple of both 6 and 4. 

The least common multiple (LCM) of 6 and 4 is 12.
Therefore, on line 12, the letters will have completed exactly 2 full cycles and the digits will have completed exactly 3 full cycles, returning to "AJHSME 1989".

ANSWER 4: C

Problem 5:
Let the original price of the shirt be P, and let the percentage of increase and decrease be x.
When the price is increased by x percent, it becomes P(1 + x/100).
When this new price is decreased by x percent, it becomes P(1 + x/100)(1 - x/100).

We are given that the resulting price is 84% of the original price, so:
P(1 + x/100)(1 - x/100) = 0.84P

Divide both sides by P:
(1 + x/100)(1 - x/100) = 0.84
1 - (x/100)² = 0.84
(x/100)² = 1 - 0.84
(x/100)² = 0.16

Taking the square root of both sides:
x/100 = 0.4
x = 40

The price was increased and decreased by 40%.

ANSWER 5: E

Problem 6:
Harold tosses a nickel four times. We want to find the probability that he gets at least as many heads as tails.
"At least as many heads as tails" in 4 tosses means getting 2, 3, or 4 heads.

The total number of possible outcomes when tossing a coin 4 times is 2⁴ = 16.
Using combinations, we find the number of ways to get 2, 3, or 4 heads:
- 2 heads: 4C2 = 6 ways
- 3 heads: 4C3 = 4 ways
- 4 heads: 4C4 = 1 way

Total favorable outcomes = 6 + 4 + 1 = 11.
The probability is 11 / 16.

ANSWER 6: E

Problem 7:
The machine follows the Collatz-like rule:
- If N is even, output N/2
- If N is odd, output 3N + 1

We are told a 6-step process ends in 1. Let's work backward from Step 6 to find all possible starting positive integers N (Step 0).
- Step 6: 1
- Step 5: Must be 2 (since 3x + 1 = 1 gives x = 0, which is not a positive integer).
- Step 4: Must be 4 (since 3x + 1 = 2 gives x = 1/3, not an integer).
- Step 3: Can be 8 (since 8/2 = 4) or 1 (since 3(1) + 1 = 4).

Branch 1 (Step 3 is 8):
- Step 2: Must be 16 (since 3x + 1 = 8 gives x = 7/3).
- Step 1: Can be 32 (32/2 = 16) or 5 (3(5) + 1 = 16).
- Step 0 (from 32): Must be 64 (since 3x + 1 = 32 gives x = 31/3).
- Step 0 (from 5): Must be 10 (since 3x + 1 = 5 gives x = 4/3).

Branch 2 (Step 3 is 1):
- Step 2: Must be 2.
- Step 1: Must be 4.
- Step 0: Can be 8 (8/2 = 4) or 1 (3(1) + 1 = 4).

The possible starting integers N are 64, 10, 8, and 1.
The sum of all such integers is 64 + 10 + 8 + 1 = 83.

ANSWER 7: E

Problem 8:
We are given a sequence of 8 numbers where each number is the product of the two preceding numbers: a_n = a_{n-2} × a_{n-1}.
We know the last three numbers: a_6 = 16, a_7 = 64, and a_8 = 1024.
Let's work backward using the rule a_{n-2} = a_n / a_{n-1}:
- a_5 = a_7 / a_6 = 64 / 16 = 4
- a_4 

## Step 1: Problem 1 - Understanding the problem
We need to find the fraction of 7-digit telephone numbers that begin with 9 and end with 0, given that the numbers cannot start with 0 or 1.

## Step 2: Problem 1 - Calculating total possible telephone numbers
The first digit can be any number from 2 to 9, so there are 8 possibilities. The remaining 6 digits can be any number from 0 to 9, so there are 10 possibilities for each. Thus, the total number of telephone numbers is $8 \times 10^6$.

## Step 3: Problem 1 - Calculating the number of telephone numbers that begin with 9 and end with 0
For a number to begin with 9 and end with 0, the first digit must be 9 and the last digit must be 0. The first digit has 1 possibility (9), the last digit has 1 possibility (0), and the remaining 5 digits have 10 possibilities each. So, the total number of such telephone numbers is $1 \times 10^5 \times 1$.

## Step 4: Problem 1 - Calculating the fraction
The fraction of telephone numbers that begin with 9 and end with 0 is $\frac{1 \times 10^5 \times 1}{8 \times 10^6} = \frac{10^5}{8 \times 10^6} = \frac{1}{80}$.

## Step 5: Problem 1 - Answer
ANSWER 1: B

## Step 6: Problem 2 - Understanding the problem
We are given 25 integers divided into 5 groups of 5 numbers each. We need to find the least possible value of M, the median of the medians of these groups.

## Step 7: Problem 2 - Analyzing the medians
To minimize M, we should minimize the medians of the 5 groups. The integers are from 1 to 25. The median of each group will be the third number when the numbers are arranged in ascending order.

## Step 8: Problem 2 - Distributing numbers to minimize M
To minimize M, we can distribute the numbers such that the medians of the groups are as small as possible. One way to do this is by grouping the smallest numbers together and ensuring that the medians are as low as possible.

## Step 9: Problem 2 - Calculating the medians of the groups
Let's distribute the numbers: Group 1 = {1, 2, 3, 24, 25}, Group 2 = {4, 5, 6, 22, 23}, Group 3 = {7, 8, 9, 20, 21}, Group 4 = {10, 11, 12, 18, 19}, Group 5 = {13, 14, 15, 16, 17}. The medians are 3, 6, 9, 12, 15.

## Step 10: Problem 2 - Finding M
The medians of the groups are 3, 6, 9, 12, 15. The median of these medians (M) is 9.

## Step 11: Problem 2 - Answer
ANSWER 2: A

## Step 12: Problem 3 - Understanding the problem
A cricket hops between 4 leaves with equal probability. We need to find the probability that it returns to the starting leaf after 4 hops.

## Step 13: Problem 3 - Analyzing the hops
After the first hop, the cricket is on a different leaf. For it to return to the starting leaf after 4 hops, it must be on a different leaf after the third hop as well.

## Step 14: Problem 3 - Calculating the probability
Let's denote the probability of being on the starting leaf after n hops as $P_n$. $P_0 = 1$, $P_1 = 0$. For $n \geq 1$, $P_{n+1} = \frac{1}{3}(1 - P_n)$ because the probability of not being on the starting leaf after n hops is $1 - P_n$, and from any other leaf, there's a $\frac{1}{3}$ chance of hopping to the starting leaf.

## Step 15: Problem 3 - Computing $P_4$
$P_2 = \frac{1}{3}$, $P_3 = \frac{1}{3}(1 - \frac{1}{3}) = \frac{2}{9}$, $P_4 = \frac{1}{3}(1 - \frac{2}{9}) = \frac{1}{3} \times \frac{7}{9} = \frac{7}{27}$.

## Step 16: Problem 3 - Answer
ANSWER 3: E

## Step 17: Problem 4 - Understanding the problem
The letters and digits are cycled separately and combined. We need to find when "AJHSME 1989" appears again.

## Step 18: Problem 4 - Analyzing the cycles
The letters cycle every 6 steps, and the digits cycle every 4 steps (since there are 4 digits and they are distinct when considering their positions). The combined cycle will be the least common multiple (LCM) of 6 and 4.

## Step 19: Problem 4 - Calculating the LCM of 6 and 4
The LCM of 6 and 4 is 12.

## Step 20: Problem 4 - Answer
ANSWER 4: C

## Step 21: Problem 5 - Understanding the problem
A price is increased by a certain percent and then decreased by the same percent, resulting in 84% of the original price.

## Step 22: Problem 5 - Setting up the equation
Let $x$ be the original price and $p$ be the percentage change. The new price after increase is $x(1 + \frac{p}{100})$, and after decrease is $x(1 + \frac{p}{100})(1 - \frac{p}{100}) = 0.84x$.

## Step 23: Problem 5 - Solving for $p$
$x(1 - \frac{p^2}{10000}) = 0.84x$. Simplifying gives $1 - \frac{p^2}{10000} = 0.84$. Thus, $\frac{p^2}{10000} = 0.16$, so $p^2 = 1600$, and $p = 40$.

## Step 24: Problem 5 - Answer
ANSWER 5: E

## Step 25: Problem 6 - Understanding the problem
Harold tosses a nickel four times. We need to find the probability of getting at least as many heads as tails.

## Step 26: Problem 6 - Enumerating the outcomes
The possible outcomes where heads are at least as many as tails are: 4 heads, 3 heads and 1 tail, and 2 heads and 2 tails.

## Step 27: Problem 6 - Calculating the probabilities
The total number of outcomes is $2^4 = 16$. The number of ways to get 4 heads is 1, 3 heads and 1 tail is $\binom{4}{3} = 4$, and 2 heads and 2 tails is $\binom{4}{2} = 6$. So, the total favorable outcomes are $1 + 4 + 6 = 11$.

## Step 28: Problem 6 - Calculating the probability
The probability is $\frac{11}{16}$.

## Step 29: Problem 6 - Answer
ANSWER 6: E

## Step 30: Problem 7 - Understanding the problem
A machine applies a rule to a positive integer N: if N is even, output N/2; if N is odd, output 3N + 1. We need to find the sum of all N that result in 1 after 6 steps.

## Step 31: Problem 7 - Working backwards from 1
After 6 steps, the output is 1. So, before that, it was 2 (since 2 is even and 2/2 = 1). Before that, it could be 4. Continuing this process backwards: 4 -> 2 -> 1 (3 steps), 8 -> 4 -> 2 -> 1 (4 steps), 16 -> 8 -> 4 -> 2 -> 1 (5 steps), and so on, until we reach 6 steps.

## Step 32: Problem 7 - Enumerating the possible N
For 6 steps: $2^6 = 64$ is a possible N (since 64 -> 32 -> 16 -> 8 -> 4 -> 2 -> 1). Working bac

Problem 1:

We are asked what fraction of possible telephone numbers begin with 9 and end with 0.

A 7-digit telephone number cannot begin with 0 or 1.

Total possible telephone numbers:

- First digit: 2 through 9, so 8 choices.
- Remaining 6 digits: each has 10 choices.

So total numbers:

\[
8 \cdot 10^6
\]

Now count numbers that begin with 9 and end with 0:

- First digit fixed as 9: 1 choice.
- Last digit fixed as 0: 1 choice.
- Middle 5 digits: each has 10 choices.

So favorable numbers:

\[
10^5
\]

Thus the fraction is:

\[
\frac{10^5}{8\cdot 10^6}=\frac{1}{80}
\]

The tempting answer \(\frac{1}{100}\) would be correct only if all first digits were equally allowed, including 0 and 1, but they are not.

ANSWER 1: B

Problem 2:

We have the integers 1 through 25 separated into five groups of 5 numbers each. We find the median of each group, then take the median of those five medians. We want the least possible value of \(M\).

For a group of 5 numbers to have median at most \(m\), at least 3 numbers in that group must be at most \(m\).

Since \(M\) is the median of the five medians, at least 3 of the group medians must be at most \(M\).

Therefore, to make \(M\) as small as possible, we need at least:

\[
3 \times 3 = 9
\]

numbers that are at most \(M\).

The 9 smallest numbers are:

\[
1,2,3,4,5,6,7,8,9
\]

So \(M\) cannot be less than 9.

Now we show that \(M=9\) is possible. For example:

\[
\{1,2,3,22,23\} \quad \text{median } 3
\]

\[
\{4,5,6,24,25\} \quad \text{median } 6
\]

\[
\{7,8,9,20,21\} \quad \text{median } 9
\]

The remaining numbers can form two groups with medians larger than 9:

\[
\{10,11,12,13,14\} \quad \text{median } 12
\]

\[
\{15,16,17,18,19\} \quad \text{median } 17
\]

The five medians are:

\[
3,6,9,12,17
\]

Their median is 9.

So the least possible value is 9.

ANSWER 2: A

Problem 3:

The cricket starts on one leaf and hops to one of the other 3 leaves each time. We want the probability that after 4 hops it is back where it started.

Let \(p_n\) be the probability that the cricket is on its starting leaf after \(n\) hops.

Initially:

\[
p_0=1
\]

After one hop, it cannot be on the starting leaf:

\[
p_1=0
\]

If the cricket is not on the starting leaf, then on the next hop it has probability \(\frac13\) of hopping back to the starting leaf. Therefore:

\[
p_{n+1}=\frac{1-p_n}{3}
\]

Now compute:

\[
p_2=\frac{1-0}{3}=\frac13
\]

\[
p_3=\frac{1-\frac13}{3}=\frac{\frac23}{3}=\frac29
\]

\[
p_4=\frac{1-\frac29}{3}=\frac{\frac79}{3}=\frac7{27}
\]

So the probability is:

\[
\frac7{27}
\]

Choice A, \(\frac29\), is the probability after 3 hops, not 4 hops.

ANSWER 3: E

Problem 4:

The letters and digits are cycled separately.

The letters are:

\[
AJHSME
\]

There are 6 letters, so the letters return to their original order after 6 cycles.

The digits are:

\[
1989
\]

There are 4 digits, so the digits return to their original order after 4 cycles.

We need both the letters and the digits to return to their original arrangements at the same time.

So we need the least common multiple of 6 and 4:

\[
\operatorname{lcm}(6,4)=12
\]

Thus \(AJHSME\ 1989\) appears again for the first time on line 12.

Choice A, 6, would return only the letters, not necessarily the digits. Choice B, 10, returns neither. Choice E, 24, is a common multiple but not the first one.

ANSWER 4: C

Problem 5:

Let the original price be \(P\).

Suppose the price is increased by \(r\), where \(r\) is written as a decimal. Then the new price is:

\[
P(1+r)
\]

Then it is decreased by the same percent, so the final price is:

\[
P(1+r)(1-r)
\]

We are told this equals 84% of the original price:

\[
P(1+r)(1-r)=0.84P
\]

Cancel \(P\):

\[
(1+r)(1-r)=0.84
\]

\[
1-r^2=0.84
\]

\[
r^2=0.16
\]

\[
r=0.4
\]

So the percent is:

\[
40\%
\]

A common mistake is to think an increase and decrease by the same percent cancel each other, but they do not because the second percent is taken from a different price.

ANSWER 5: E

Problem 6:

Harold tosses a nickel 4 times. We want the probability that he gets at least as many heads as tails.

There are \(2^4=16\) equally likely outcomes.

“At least as many heads as tails” means:

\[
H \ge T
\]

Since there are 4 tosses total, this means Harold must get 2, 3, or 4 heads.

Count the outcomes:

\[
\binom42=6
\]

\[
\binom43=4
\]

\[
\binom44=1
\]

Total favorable outcomes:

\[
6+4+1=11
\]

So the probability is:

\[
\frac{11}{16}
\]

The tempting answer \(\frac12\) would count only strictly more heads than tails and would forget the tie case.

ANSWER 6: E

Problem 7:

The machine rule is:

- If \(N\) is even, output \(\frac N2\).
- If \(N\) is odd, output \(3N+1\).

We need all starting positive integers \(N\) that become 1 after exactly 6 steps.

It is easiest to work backward from 1.

If a number outputs \(y\), then possible previous numbers are:

1. \(2y\), because if the previous number was even, halving it gives \(y\).
2. \(\frac{y-1}{3}\), but only if this is a positive odd integer.

Now work backward 6 steps.

Step 0:

\[
\{1\}
\]

Step 1 predecessors of 1:

\[
\{2\}
\]

Step 2 predecessors of 2:

\[
\{4\}
\]

Step 3 predecessors of 4:

\[
2\cdot 4=8
\]

and

\[
\frac{4-1}{3}=1
\]

So:

\[
\{8,1\}
\]

Step 4:

Predecessors of 8: \(16\)

Predecessors of 1: \(2\)

So:

\[
\{16,2\}
\]

Step 5:

Predecessors of 16:

\[
32,\quad \frac{16-1}{3}=5
\]

Predecessors of 2:

\[
4
\]

So:

\[
\{32,5,4\}
\]

Step 6:

Predecessors of 32:

\[
64
\]

Predecessors of 5:

\[
10
\]

Predecessors of 4:

\[
8,\quad 1
\]

So the possible starting values are:

\[
64,10,8,1
\]

Their sum is:

\[
64+10+8+1=83
\]

ANSWER 7: E

Problem 8:

Let the list be:

\[
a_1,a_2,a_3,a_4,a_5,a_6,a_7,a_8
\]

Each new number is the product of the two previous numbers:

\[
a_3=a_1a_2
\]

\[
a_4=a_2a_3
\]

and so on.

We are told the last three numbers are:

\[
a_6=16,\quad a_7=64,\quad a_8=1024
\]

Check:

\[
16\cdot 64=1024
\]

Now work backward.

Since:

\[
a_7=a_5a_6
\]

we have:

\[
64=

Problem 1:
We are asked to find the fraction of possible 7-digit telephone numbers that begin with 9 and end with 0, given that a telephone number cannot begin with 0 or 1.

First, let's find the total number of possible telephone numbers:
- The first digit can be any number from 2 to 9, which gives 8 possibilities.
- The remaining 6 digits can be any number from 0 to 9, giving 10 possibilities each.
Total possible telephone numbers = 8 × 10 × 10 × 10 × 10 × 10 × 10 = 8,000,000.

Next, let's find the number of telephone numbers that begin with 9 and end with 0:
- The first digit must be 9 (1 possibility).
- The last digit must be 0 (1 possibility).
- The middle 5 digits can be any number from 0 to 9 (10 possibilities each).
Total such numbers = 1 × 10 × 10 × 10 × 10 × 10 × 1 = 100,000.

The fraction of telephone numbers that meet this condition is:
100,000 / 8,000,000 = 1 / 80.

ANSWER 1: B

Problem 2:
We are asked to find the least possible value of M, which is the median of the medians of five groups of 5 numbers each, chosen from the integers 1 through 25.

Let the 5 groups be G1, G2, G3, G4, and G5, and let their respective medians be m1, m2, m3, m4, and m5. 
Assume the medians are ordered such that m1 < m2 < m3 < m4 < m5. The median of these medians is M = m3.
To minimize m3, we must make the numbers in the groups as small as possible. 

For m3 to be a certain value, there must be at least 3 groups (G1, G2, G3) whose medians are less than or equal to m3. 
In each of these 3 groups, there are 3 numbers that are less than or equal to the group's median. 
Therefore, there are at least 3 groups × 3 numbers = 9 numbers that are less than or equal to m3.

Since the numbers are distinct integers from 1 to 25, the 9th smallest possible integer is 9. Thus, m3 must be at least 9. 
We can verify that 9 is possible with the following grouping:
G1: 1, 2, 7, 24, 25 (median 7)
G2: 3, 4, 8, 22, 23 (median 8)
G3: 5, 6, 9, 20, 21 (median 9)
G4: 10, 11, 12, 18, 19 (median 12)
G5: 13, 14, 15, 16, 17 (median 15)
The medians are 7, 8, 9, 12, and 15. Their median is 9.

ANSWER 2: A

Problem 3:
We want to find the probability that a cricket, which hops to one of 3 other leaves with equal probability, returns to its starting leaf after exactly 4 hops.

Let the leaves be A, B, C, and D, and assume the cricket starts at leaf A.
Let P(n) be the probability that the cricket is at leaf A after n hops.
- P(0) = 1 (it starts at A).
- For any hop n > 0, the cricket can only land on A if it was NOT on A in the previous hop. If it was on B, C, or D, it has a 1/3 chance of hopping to A.
Thus, the recursive formula is P(n) = (1 - P(n-1)) × (1/3).

Let's calculate step-by-step:
- P(1) = (1 - 1) × 1/3 = 0
- P(2) = (1 - 0) × 1/3 = 1/3
- P(3) = (1 - 1/3) × 1/3 = (2/3) × (1/3) = 2/9
- P(4) = (1 - 2/9) × 1/3 = (7/9) × (1/3) = 7/27

ANSWER 3: E

Problem 4:
We are given a sequence of letters "AJHSME" (6 characters) and a sequence of digits "1989" (4 characters). They are cycled separately line by line. We want to find the line number where the original combination "AJHSME 1989" appears again for the first time.

The letters cycle back to their original arrangement every 6 lines.
The digits cycle back to their original arrangement every 4 lines.
For both to return to their original arrangement simultaneously, the line number must be a common multiple of both 6 and 4. 

The least common multiple (LCM) of 6 and 4 is 12.
Therefore, on line 12, the letters will have completed exactly 2 full cycles and the digits will have completed exactly 3 full cycles, returning to "AJHSME 1989".

ANSWER 4: C

Problem 5:
Let the original price of the shirt be P, and let the percentage of increase and decrease be x.
When the price is increased by x percent, it becomes P(1 + x/100).
When this new price is decreased by x percent, it becomes P(1 + x/100)(1 - x/100).

We are given that the resulting price is 84% of the original price, so:
P(1 + x/100)(1 - x/100) = 0.84P

Divide both sides by P:
(1 + x/100)(1 - x/100) = 0.84
1 - (x/100)² = 0.84
(x/100)² = 1 - 0.84
(x/100)² = 0.16

Taking the square root of both sides:
x/100 = 0.4
x = 40

The price was increased and decreased by 40%.

ANSWER 5: E

Problem 6:
Harold tosses a nickel four times. We want to find the probability that he gets at least as many heads as tails.
"At least as many heads as tails" in 4 tosses means getting 2, 3, or 4 heads.

The total number of possible outcomes when tossing a coin 4 times is 2⁴ = 16.
Using combinations, we find the number of ways to get 2, 3, or 4 heads:
- 2 heads: 4C2 = 6 ways
- 3 heads: 4C3 = 4 ways
- 4 heads: 4C4 = 1 way

Total favorable outcomes = 6 + 4 + 1 = 11.
The probability is 11 / 16.

ANSWER 6: E

Problem 7:
The machine follows the Collatz-like rule:
- If N is even, output N/2
- If N is odd, output 3N + 1

We are told a 6-step process ends in 1. Let's work backward from Step 6 to find all possible starting positive integers N (Step 0).
- Step 6: 1
- Step 5: Must be 2 (since 3x + 1 = 1 gives x = 0, which is not a positive integer).
- Step 4: Must be 4 (since 3x + 1 = 2 gives x = 1/3, not an integer).
- Step 3: Can be 8 (since 8/2 = 4) or 1 (since 3(1) + 1 = 4).

Branch 1 (Step 3 is 8):
- Step 2: Must be 16 (since 3x + 1 = 8 gives x = 7/3).
- Step 1: Can be 32 (32/2 = 16) or 5 (3(5) + 1 = 16).
- Step 0 (from 32): Must be 64 (since 3x + 1 = 32 gives x = 31/3).
- Step 0 (from 5): Must be 10 (since 3x + 1 = 5 gives x = 4/3).

Branch 2 (Step 3 is 1):
- Step 2: Must be 2.
- Step 1: Must be 4.
- Step 0: Can be 8 (8/2 = 4) or 1 (3(1) + 1 = 4).

The possible starting integers N are 64, 10, 8, and 1.
The sum of all such integers is 64 + 10 + 8 + 1 = 83.

ANSWER 7: E

Problem 8:
We are given a sequence of 8 numbers where each number is the product of the two preceding numbers: a_n = a_{n-2} × a_{n-1}.
We know the last three numbers: a_6 = 16, a_7 = 64, and a_8 = 1024.
Let's work backward using the rule a_{n-2} = a_n / a_{n-1}:
- a_5 = a_7 / a_6 = 64 / 16 = 4
- a_4