AMC 8 · 2016 · #7

Grade 8 number-theory

Archetype Small-Domain Constrained Search └ Sub-archetype Sum-Target Search

perfect-squaresexponentsprime-factorizationparity systematic-enumerationeasier-related-problem ↑ Prerequisites: exponentsperfect-squares

📏 Medium solution 💡 3 insights

Problem

Five numbers are given: 1²⁰¹⁶, 2²⁰¹⁷, 3²⁰¹⁸, 4²⁰¹⁹, 5²⁰²⁰. Exactly one of them is not a perfect square. Which one?

Pick an answer.

(A)

$1^{2016}$

(B)

$2^{2017}$

(C)

$3^{2018}$

(D)

$4^{2019}$

(E)

$5^{2020}$

AMC 8 2016 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Eliminate Possibilities

Five answer choices and a yes/no test on each — Tool #3 (Eliminate) is the natural lead. The test itself comes from Tool #5: by checking small cases like 2², 2³, 2⁴, 2⁵, the pattern is clear — a power aⁿ (with a prime) is a perfect square exactly when n is even, because a²k = (a^k)². Tool #9 (Easier Problem) is what lets us trust that pattern: we verify it on tiny exponents before applying it to 2016–2020. The one trap is choice (D), where the base 4 is not prime; we rewrite 4 = 2² first so the exponent rule applies to a prime base.

1STEP 1

Build the test from small cases: 2² and 2⁴ are squares but 2³, 2⁵ aren't — so a prime power pⁿ is a perfect square exactly when n is even.

pⁿ is a perfect square ⇔ n is even (p prime)

2STEP 2

(A) 1²⁰¹⁶ = 1 = 1², a perfect square — cross off (A).

1²⁰¹⁶ = 1 = 1² ✓

3STEP 3

(B) base 2 is prime and exponent 2017 is odd, so 2²⁰¹⁷ is not a perfect square. The candidate answer; still check the other three.

2²⁰¹⁷: 2 prime, 2017 odd → not a perfect square

4STEP 4

3²⁰¹⁸ = (3¹⁰⁰⁹)² ✓

5STEP 5

(D) trap: base 4 = 2² isn't prime, so 4²⁰¹⁹ = (2²)²⁰¹⁹ = 2⁴⁰³⁸ — 4038 is even, so it's (2²⁰¹⁹)². A square; cross off (D).

4²⁰¹⁹ = (2²)²⁰¹⁹ = 2⁴⁰³⁸ = (2²⁰¹⁹)² ✓

6STEP 6

(E) base 5 is prime and exponent 2020 is even, so 5²⁰²⁰ = (5¹⁰¹⁰)² — a perfect square. Only (B) survives.

5²⁰²⁰ = (5¹⁰¹⁰)² ✓ → Answer: (B)

Answer

2²⁰¹⁷

Cross-check by prime factorization. (A) 1 has exponent 0 on every prime (even). (C) 3²⁰¹⁸: prime 3 to even exponent 2018. (D) 4²⁰¹⁹ = 2⁴⁰³⁸: prime 2 to even exponent 4038. (E) 5²⁰²⁰: prime 5 to even exponent 2020. All four pass the "every prime exponent is even" test. Only (B) 2²⁰¹⁷ has an odd exponent on the prime 2, so it cannot be written as k². The answer (B) is consistent.

💡Key takeaway

A prime raised to an odd power can never be a perfect square — and that single Grade 8 exponent rule is enough to spot the odd-one-out among (A)–(E).

More like this