AMC 10 · 2023 · #14

Grade 7 probability

probability-basicdivisibility-rulesdivisor-countprime-factorization identify-subproblemscaseworksystematic-enumeration ↑ Prerequisites: probability-basicdivisor-count

📏 Long solution 💡 3 insights

Problem

First pick n uniformly at random from {1, 2, …, 100}. Then pick a positive divisor d of n uniformly at random from n's divisors. Find the probability that the chosen divisor d is a multiple of 11.

Pick an answer.

(A)

$~\frac{4}{100}$

(B)

$~\frac{9}{200}$

(C)

$~\frac{1}{20}$

(D)

$~\frac{11}{200}$

(E)

$~\frac{3}{50}$

AMC 10 2023 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Identify Subproblems

Tool #7 (Identify Subproblems) splits the two-stage process into the two natural pieces: (A) probability that n is itself a multiple of 11 — otherwise the second stage can never succeed; and (B) given a specific multiple of 11, probability that a randomly chosen divisor is also a multiple of 11. Tool #2 (Systematic List) handles part (A) — count the multiples of 11 in [1, 100]. Tool #9 (Easier Related Problem) handles part (B): instead of analyzing divisor structures abstractly, test the easier case n = 11 (divisors 1, 11), see that exactly half of the divisors are multiples of 11, then check this is true for every multiple-of-11 in the range. Tool #5 (Look for a Pattern) confirms the $\frac{1}{2}$ pattern across all nine candidates.

1STEP 1

List the multiples of 11 up to 100: 11, 22, …, 99 — 9 numbers in all.

{11, 22, 33, 44, 55, 66, 77, 88, 99}, count = 9

2STEP 2

A divisor of a divisor is a divisor: if 11 ∣ d and d ∣ n then 11 ∣ n, so only a multiple-of-11 n can ever give an 11-multiple divisor.

11 ∣ d and d ∣ n → 11 ∣ n

3STEP 3

So only the 9 multiples of 11 matter; the chance the first pick n lands on one is $\frac{9}{100}$ .

P(n is a multiple of 11) =

\frac{9}{100}

4STEP 4

Test the easiest case n = 11: its divisors {1, 11} hold one multiple of 11, so the fraction that work is $\frac{1}{2}$ .

n = 11: divisors = {1, 11}, mult-of-11 = {11},

\frac{1}{2}

5STEP 5

Check n = 22 and n = 99: their divisors split $\frac{2}{4}$ and $\frac{3}{6}$ — every listed multiple of 11 gives fraction $\frac{1}{2}$ .

(#{divisors of n that are mult of 11})/(#{divisors of n}) =

\frac{1}{2}

6STEP 6

Why always $\frac{1}{2}$ ? Each n = 11k (k ≤ 9) has divisors that pair as d and 11d, splitting evenly between non-multiples and multiples of 11.

divisors(11k) = divisors(k) ∪ 11 · divisors(k); equal-size halves

7STEP 7

Law of total probability: only the 9 multiples of 11 contribute, each conditional $\frac{1}{2}$ weighted by $\frac{1}{100}$ , giving $\frac{9}{200}$ .

P = 9 ·

\frac{1}{2}

\frac{1}{100}

\frac{9}{200}

8STEP 8

The value $\frac{9}{200}$ is exactly choice (B); (A) $\frac{8}{200}$ and (D) $\frac{11}{200}$ come from miscounts, so rule them out.

P =

\frac{9}{200}

→ (B)

Answer

\frac{9}{200}

Magnitude check: only 9% of the 100 starting numbers can possibly yield a multiple-of-11 divisor, and among those the conditional success rate is exactly

\frac{1}{2}

, so the answer must be slightly under

\frac{1}{20}

\frac{10}{200}

. Indeed

\frac{9}{200}

= 0.045 ＜ 0.05. The pairing argument is iron-clad because 11² ＞ 100 guarantees each n has exactly one factor of 11. Direction check: the answer should be smaller than

\frac{9}{100}

(since not every first-stage success leads to a second-stage success), and

\frac{9}{200}

is exactly half — consistent.

💡Key takeaway

Only the nine multiples of 11 in 1-100 can possibly produce a divisor divisible by 11, and for each of those the divisors split exactly in half (those with a factor of 11 and those without). So the answer is $\frac{9}{100}$ · $\frac{1}{2}$ = $\frac{9}{200}$ .