AMC 10 · 2019 · #2

Easy mode Grade 4

Problem

Someone makes this claim: "If $n$ is not a prime number, then $n - 2$ is a prime number."

A counterexample is a single value of $n$ that proves the claim wrong. So we need an $n$ where $n$ is not prime, but $n - 2$ is also not prime.

Which of the answer choices is such a counterexample?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 10 2019 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

Try it yourself first — the explanation is most useful after you’ve attempted it.

View mode:

Toolkit + CCSS Solution

Understand

Restated: Consider the claim: 'If $n$ is not prime, then $n - 2$ is prime.' Among the five answer choices, find a value of $n$ that breaks this claim — that is, $n$ is not prime BUT $n - 2$ is also not prime.

Givens: The claim has the form: if $n$ is composite (not prime), then $n - 2$ is prime; Answer choices: (A) $11$, (B) $15$, (C) $19$, (D) $21$, (E) $27$

Unknowns: The single value of $n$ that is a counterexample

Understand

Givens: The claim has the form: if $n$ is composite (not prime), then $n - 2$ is prime; Answer choices: (A) $11$, (B) $15$, (C) $19$, (D) $21$, (E) $27$

Plan

Primary tool: #3 Eliminate Possibilities

Secondary: #2 Make a Systematic List

Tool #3 (Eliminate): the answer must satisfy two conditions. First, $n$ itself must be composite, which kills any prime choice — that removes $11$ and $19$ immediately. Tool #2 (Systematic List): for each surviving composite $n$, compute $n - 2$ and check primality. The single survivor is the counterexample.

Execute — Answer: E

#3 Eliminate Possibilities 4.OA.B.4 Step 1

First filter: $n$ must NOT be prime.
Check each choice.
$11$ is prime (only $1, 11$ divide it) — eliminate.
$15 = 3 \cdot 5$ — composite, keep.
$19$ is prime — eliminate.
$21 = 3 \cdot 7$ — composite, keep.
$27 = 3 \cdot 9$ — composite, keep.

$11, 19$ prime $\Rightarrow$ eliminated. Survivors: $15, 21, 27$.

💡 If $n$ is prime, the 'if' part of the claim is false and there's nothing to break.

#2 Make a Systematic List 4.NBT.B.4 Step 2

Second filter: for each survivor, check whether $n - 2$ is prime.
If $n - 2$ is prime, the claim holds for that $n$ — not a counterexample.
If $n - 2$ is composite, we found one.
List them: $15 - 2 = 13$, $21 - 2 = 19$, $27 - 2 = 25$.

$15 - 2 = 13$, $21 - 2 = 19$, $27 - 2 = 25$

💡 Subtract $2$ from each survivor and check the result.

#3 Eliminate Possibilities 4.OA.B.4 Step 3

Check primality of each $n - 2$ value.
$13$ is prime (no divisors but $1, 13$) — claim holds for $n = 15$, not a counterexample.
$19$ is prime — claim holds for $n = 21$, not a counterexample.
$25 = 5 \cdot 5$ is composite — claim FAILS for $n = 27$.
This is the counterexample.

$13$ prime, $19$ prime, $25 = 5 \cdot 5$ composite $\;\Rightarrow\; n = 27 \;\Rightarrow\; \textbf{(E)}$

💡 Only one survivor leaves $n - 2$ composite — and that one breaks the claim.

[1] #3 4.OA.B.4 First filter: $n$ must NOT be prime. Check each choice. $11$ is prime (only $1,

[2] #2 4.NBT.B.4 Second filter: for each survivor, check whether $n - 2$ is prime. If $n - 2$ is

[3] #3 4.OA.B.4 Check primality of each $n - 2$ value. $13$ is prime (no divisors but $1, 13$) —

Review

Reasonableness: Verify the counterexample at $n = 27$. Is $27$ not prime? Yes — $27 = 3 \cdot 9 = 3^3$, so composite. Is $27 - 2 = 25$ prime? No — $25 = 5 \cdot 5$, also composite. So the implication 'not prime $\Rightarrow$ $n - 2$ prime' has a true premise but a false conclusion — exactly what a counterexample requires. The other composites $15$ and $21$ both give a prime $n - 2$, so they're consistent with the claim.

Alternative: Tool #2 (Systematic List): make a small table with columns 'choice $n$', 'is $n$ prime?', '$n - 2$', 'is $n - 2$ prime?', and 'counterexample?'. Walk down it row by row — the only Yes/No/Yes/No row is $27$, giving (E).

CCSS standards used (min grade 4)

4.NBT.B.4 Fluently add and subtract multi-digit whole numbers (Subtracting $2$ from each candidate to get $n - 2$: $15 - 2 = 13$, $21 - 2 = 19$, $27 - 2 = 25$.)
4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite (Classifying each value as prime or composite — the core tool for both filters.)

⭐ This AMC 10 problem only needs Grade 4 prime-vs-composite checks you already know — toss out the primes ($11, 19$), then subtract $2$ from each leftover and look for the first composite result. $27 - 2 = 25 = 5 \cdot 5$, so (E).

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: E

Review

CCSS standards used (min grade 4)

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