AMC 10 · 2019 · #2
Easy mode Grade 4Problem
Someone makes this claim: "If is not a prime number, then is a prime number."
A counterexample is a single value of that proves the claim wrong. So we need an where is not prime, but is also not prime.
Which of the answer choices is such a counterexample?
Pick an answer.
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
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) $27$
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
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.
💡 If $n$ is prime, the 'if' part of the claim is false and there's nothing to break.
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: $15 - 2 = 13$, $21 - 2 = 19$, $27 - 2 = 25$.
💡 Subtract $2$ from each survivor and check the result.
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.
💡 Only one survivor leaves $n - 2$ composite — and that one breaks the claim.
4.OA.B.4 First filter: $n$ must NOT be prime. Check each choice. $11$ is prime (only $1, 4.NBT.B.4 Second filter: for each survivor, check whether $n - 2$ is prime. If $n - 2$ is 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.4Fluently 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.4Find 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).
⭐ 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).