AMC 10 · 2018 · #11

Grade 6 number-theory
Archetype Small-Domain Constrained Search └ Sub-archetype Answer-Choice Plug-In
prime-numbersmodular-arithmeticperfect-squares caseworkpattern-recognition ↑ Prerequisites: prime-numbersmodular-arithmetic
📏 Medium solution 💡 3 insights
Problem
For each of the five expressions p²+16, p²+24, p²+26, p²+46, p²+96, decide which one fails to be prime for every prime number p. Four of them produce a prime for at least one prime p; exactly one is composite no matter which prime p you plug in. Find that one.

Pick an answer.

(A)
$p^2+16$
(B)
$p^2+24$
(C)
$p^2+26$
(D)
$p^2+46$
(E)
$p^2+96$

AMC 10 2018 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Eliminate Possibilities

Tool #3 (Eliminate Possibilities): the question is lopsided. To knock out a choice I only need one prime p that turns it into a prime number; to crown the winner I need a reason it is forced composite for all p. Tool #5 (Look for a Pattern): the forcing reason comes from a clean fact about primes and the number 3 — every prime except 3 sits next to a multiple of 3. Tool #4 (Introduce a Variable): keeping p as a symbol and rewriting p²+26 exposes a hidden factor of 3 that holds for every prime at once, which a few numerical tests could never prove.

1STEP 1

Split the job two ways

"Never prime" is an all-p claim: four choices fall to one prime example each, but the winner needs a fixed divisor for every prime p.

2STEP 2

Primes hug a multiple of 3

Among p-1, p, p+1 exactly one is a multiple of 3; if p is prime and p ≠ 3 it can't be p, so (p-1)(p+1) is a multiple of 3.

one of p-1, p, p+1 is a multiple of 3 → 3 ∣ (p-1)(p+1) when p ≠ 3
3STEP 3

Rewrite (C) to expose a 3

Rewrite (C): p²+26 = (p-1)(p+1) + 27. Both pieces are multiples of 3 (27 = 3·9), so 3 ∣ p²+26 for every prime p ≠ 3.

p²+26 = (p-1)(p+1) + 27, 3 ∣ (p-1)(p+1), 3 ∣ 27 → 3 ∣ p²+26
4STEP 4

Always a multiple of 3, always composite

For every prime p, p²+26 ＞ 3 and is a multiple of 3 (p=3 gives 35 = 5·7), so it is composite for every prime p.

p=3: 3²+26 = 35 = 5· 7; p ≠ 3: 3 ∣ p²+26 and p²+26 ＞ 3
5STEP 5

The other four can be prime

The other four each hit a prime: (A) p=5→41, (B) p=7→73, (D) p=5→71, (E) p=19→457. Only p²+26 resists every prime — answer (C).

41, 73, 71, 457 are all prime → only p²+26 is never prime → (C)
Answer
p²+26
Spot-check (C) on more primes: p=5 → 51 = 3· 17, p=7 → 75 = 3· 25, p=11 → 147 = 3· 49 — every value is a multiple of 3, matching the proof. Meanwhile (A),(B),(D),(E) each hit a genuine prime (41,73,71,457), so they are correctly eliminated. Exactly one choice survives, as the problem promises, and it is (C).
💡Key takeaway

A prime that isn't 3 always sits next to a multiple of 3, and p²+26=(p-1)(p+1)+27 is built from two pieces divisible by 3 — so it's always a multiple of 3, hence never prime.

  • Split the job two ways
  • Primes hug a multiple of 3
  • Rewrite (C) to expose a 3
  • Always a multiple of 3, always composite
  • The other four can be prime

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