AMC 10 · 2023 · #14

Grade 7 probability

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

📏 Long solution 💡 3 insights

Problem

A number is chosen at random from among the first $100$ positive integers, and a positive integer divisor of that number is then chosen at random. What is the probability that the chosen divisor is divisible by $11$ ?

Pick an answer.

(A)

$~rac{4}{100}$

(B)

$~rac{9}{200}$

(C)

$~rac{1}{20}$

(D)

$~rac{11}{200}$

(E)

$~rac{3}{50}$

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Toolkit + CCSS Solution

Understand

Restated: First pick $n$ uniformly at random from $\{1, 2, \dots, 100\}$. 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 $11$.

Givens: Stage 1: $n$ chosen uniformly from $1, 2, \dots, 100$ (each has probability $\tfrac{1}{100}$); Stage 2: given $n$, a divisor $d$ chosen uniformly from the divisors of $n$; We want $P(d \text{ is a multiple of } 11)$; Answer choices: (A) $\tfrac{4}{100}$, (B) $\tfrac{9}{200}$, (C) $\tfrac{1}{20}$, (D) $\tfrac{11}{200}$, (E) $\tfrac{3}{50}$

Unknowns: The total probability that the final $d$ is divisible by $11$

Understand

Plan

Primary tool: #7 Identify Subproblems

Secondary: #2 Systematic List, #5 Look for a Pattern, #9 Easier Related Problem, #3 Eliminate Possibilities

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 $\tfrac12$ pattern across all nine candidates.

Execute — Answer: B

#2 Systematic List 4.OA.B.4 Step 1

Subproblem A: count the multiples of $11$ in $\{1, 2, \dots, 100\}$.
List them in order: $11, 22, 33, 44, 55, 66, 77, 88, 99$ — that is $9$ numbers.
(Next would be $110 > 100$.)

$$\{11, 22, 33, 44, 55, 66, 77, 88, 99\}, \;\text{count} = 9$$

💡 Listing multiples of $11$ in order is the Grade 4 "recognize multiples" move — fast and exhaustive.

#7 Identify Subproblems 4.OA.B.4 Step 2

Why does only $n$ being a multiple of $11$ matter?
A divisor of $n$ divides $n$.
If $11$ divides some divisor $d$, then since $d \mid n$, the chain gives $11 \mid n$.
So if $n$ is NOT a multiple of $11$, none of $n$'s divisors can be a multiple of $11$, and the second-stage probability is $0$.

$$11 \mid d \text{ and } d \mid n \;\Rightarrow\; 11 \mid n$$

💡 "Divisor of a divisor is a divisor" — the Grade 4 chain rule for divisibility decides which $n$'s can contribute at all.

#7 Identify Subproblems 7.SP.C.7 Step 3

So we only need to look at the $9$ multiples of $11$.
Probability that the first-stage $n$ is one of these: $\tfrac{9}{100}$.

$$P(n \text{ is a multiple of } 11) = \dfrac{9}{100}$$

💡 Grade 7 probability: count favorable outcomes ($9$) divided by total outcomes ($100$) in a uniform-random pick.

#9 Easier Related Problem 4.OA.B.4 Step 4

Subproblem B (Tool #9 Easier Related Problem): for each multiple of $11$ in the list, what fraction of its divisors are themselves multiples of $11$?
Try the smallest, $n = 11$: divisors are $\{1, 11\}$.
Multiples of $11$ among them: $\{11\}$.
Fraction $= \tfrac{1}{2}$.

$$n = 11: \;\text{divisors} = \{1, 11\}, \;\text{mult-of-}11 = \{11\}, \;\tfrac{1}{2}$$

💡 The simplest case shows the structure: pair each non-$11$-multiple divisor $k$ with the multiple-of-$11$ divisor $11k$ — exactly half are multiples of $11$.

#5 Look for a Pattern 4.OA.B.4 Step 5

Look for the pattern (Tool #5).
Try $n = 22 = 2 \cdot 11$: divisors $\{1, 2, 11, 22\}$ — multiples of $11$ are $\{11, 22\}$, fraction $\tfrac{2}{4} = \tfrac{1}{2}$.
Try $n = 99 = 3^2 \cdot 11$: divisors $\{1, 3, 9, 11, 33, 99\}$ — multiples of $11$ are $\{11, 33, 99\}$, fraction $\tfrac{3}{6} = \tfrac{1}{2}$.
Pattern: every multiple of $11$ on our list gives fraction $\tfrac{1}{2}$.

$$\dfrac{\#\{\text{divisors of } n \text{ that are mult of } 11\}}{\#\{\text{divisors of } n\}} = \dfrac{1}{2}$$

💡 Three test cases all give $\tfrac{1}{2}$ — confident the pattern holds for the remaining six.

#7 Identify Subproblems 6.NS.B.4 Step 6

Why is the fraction always $\tfrac{1}{2}$?
Each of our nine $n$'s is $11 \cdot k$ with $k \in \{1, 2, \dots, 9\}$ and $\gcd(k, 11) = 1$ (since $11^2 > 100$).
The divisors of $n$ pair up: every divisor of $k$, call it $d$, corresponds to two divisors of $n$, namely $d$ (not a multiple of $11$) and $11d$ (a multiple of $11$).
So divisors of $n$ split evenly between multiples and non-multiples of $11$.

$$\text{divisors}(11k) = \text{divisors}(k) \,\cup\, 11 \cdot \text{divisors}(k); \quad \text{equal-size halves}$$

💡 Pairing every divisor $d$ with $11d$ partitions the divisors of $n$ into two equal piles — the Grade 6 GCF-and-factor view of divisors.

#7 Identify Subproblems 7.SP.C.8 Step 7

Combine the two stages (law of total probability).
Probability the final $d$ is a multiple of $11$ is $\sum_n P(d \text{ mult of } 11 \mid n) \cdot P(n)$.
For the $91$ non-multiples of $11$, the conditional probability is $0$; for the $9$ multiples of $11$, it is $\tfrac{1}{2}$, each weighted by $\tfrac{1}{100}$.

$$P = 9 \cdot \dfrac{1}{2} \cdot \dfrac{1}{100} = \dfrac{9}{200}$$

💡 Two-stage uniform pick → multiply stage probabilities and add over outcomes — the Grade 7 compound-probability move.

#3 Eliminate Possibilities 7.SP.C.7 Step 8

Match against the choices: $\tfrac{9}{200}$ is exactly answer (B).
Sanity check (A) $\tfrac{4}{100} = \tfrac{8}{200}$: close but slightly off, would correspond to $8$ multiples instead of $9$.
(D) $\tfrac{11}{200}$: would correspond to a different conditional probability.

$$P = \dfrac{9}{200} \;\Rightarrow\; \textbf{(B)}$$

💡 Match the closed-form fraction to one of the five choices — standard multiple-choice closeout.

[1] #2 4.OA.B.4 Subproblem A: count the multiples of $11$ in $\{1, 2, \dots, 100\}$. List them i

[2] #7 4.OA.B.4 Why does only $n$ being a multiple of $11$ matter? A divisor of $n$ divides $n$.

[3] #7 7.SP.C.7 So we only need to look at the $9$ multiples of $11$. Probability that the first

[4] #9 4.OA.B.4 Subproblem B (Tool #9 Easier Related Problem): for each multiple of $11$ in the

[5] #5 4.OA.B.4 Look for the pattern (Tool #5). Try $n = 22 = 2 \cdot 11$: divisors ${1, 2, 11,

[6] #7 6.NS.B.4 Why is the fraction always $\tfrac{1}{2}$? Each of our nine $n$'s is $11 \cdot k

[7] #7 7.SP.C.8 Combine the two stages (law of total probability). Probability the final $d$ is

[8] #3 7.SP.C.7 Match against the choices: $\tfrac{9}{200}$ is exactly answer (B). Sanity check

Review

Reasonableness: 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 $\tfrac12$, so the answer must be slightly under $\tfrac{1}{20} = \tfrac{10}{200}$. Indeed $\tfrac{9}{200} = 0.045 < 0.05$. The pairing argument is iron-clad because $11^2 > 100$ guarantees each $n$ has exactly one factor of $11$. Direction check: the answer should be smaller than $\tfrac{9}{100}$ (since not every first-stage success leads to a second-stage success), and $\tfrac{9}{200}$ is exactly half — consistent.

Alternative: Tool #2 (Systematic List) brute force: count over all $100$ numbers, for each list every divisor, mark which are multiples of $11$, then form the unconditional fraction by averaging. Most numbers contribute $0$. Only the nine multiples of $11$ contribute — and each contributes $\tfrac12$. Averaging: $\tfrac{1}{100} \cdot 9 \cdot \tfrac12 = \tfrac{9}{200}$. Identical result, more grunt work, no probability law needed.

CCSS standards used (min grade 7)

4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite (Listing the nine multiples of $11$ in $[1, 100]$, listing divisors of sample $n$ values ($11, 22, 99$), and applying the divisor-chain fact $11 \mid d \mid n \Rightarrow 11 \mid n$.)
6.NS.B.4 Find greatest common factor and least common multiple of two numbers (Pairing every divisor $d$ of $11k$ with $11d$ to show the divisors split into two equal halves (multiples and non-multiples of $11$), once we know $\gcd(k, 11) = 1$.)
7.SP.C.7 Develop probability models and use them to find probabilities of events (Treating both stages as uniform random picks and computing each stage's probability ($\tfrac{9}{100}$ then $\tfrac{1}{2}$).)
7.SP.C.8 Find probabilities of compound events using organized lists, tables, and simulation (Combining the two-stage probabilities via $P(A) = \sum_n P(A \mid n) P(n)$ to land on $\tfrac{9}{200}$.)

⭐ 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 $\tfrac{9}{100} \cdot \tfrac{1}{2} = \tfrac{9}{200}$.