AMC 8 · 2016 · #9

Grade 5 number-theory
prime-factorizationdivisibility-rulesprime-numbers identify-subproblemsguess-and-check ↑ Prerequisites: divisibility-rulesprime-numbers
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Problem
Find the sum of the distinct prime numbers that divide 2016. "Distinct" means each prime is counted once, no matter how many times it appears in the prime factorization.

Pick an answer.

(A)
9
(B)
12
(C)
16
(D)
49
(E)
63

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

How to solve
Strategy Identify Subproblems

Tool #7 (Identify Subproblems) splits the question into two clean jobs: (a) factor 2016 into primes, then (b) add the distinct prime bases. Tool #6 (Guess and Check) handles step (a) — trial-divide by the small primes 2, 3, 5, 7, … in order until the quotient becomes 1. We deliberately avoid heavier tools like #13 (Algebra) because plain trial division is the most direct path for a four-digit number.

1STEP 1

Subproblem 1: factor 2016 by dividing out 2 repeatedly until the odd quotient 63 appears.

2016 ÷ 2 = 1008, 1008 ÷ 2 = 504, 504 ÷ 2 = 252, 252 ÷ 2 = 126, 126 ÷ 2 = 63
2STEP 2

63 is odd, so switch to 3 (digit sum 6+3=9 is a multiple of 3); dividing by 3 twice leaves 7.

63 ÷ 3 = 21, 21 ÷ 3 = 7
3STEP 3

7 is itself prime, so the factorization is complete: 2⁵ × 3² × 7.

2016 = 2⁵ × 3² × 7
4STEP 4

Subproblem 2: ignore the exponents and keep only the distinct prime bases 2, 3, and 7.

Distinct primes = {2, 3, 7}
5STEP 5

Add the three distinct primes 2 + 3 + 7 and match the result to a choice.

2 + 3 + 7 = 12 → (B)
Answer
12
Multiply the factorization back together to confirm: 2⁵ = 32, 3² = 9, and 32 × 9 = 288, then 288 × 7 = 2016. The factorization is correct, so the distinct primes are truly {2, 3, 7} and their sum is 12. Choice (B) is also a small, sensible number — the trap answers (D) 49 and (E) 63 correspond to adding the prime powers (2⁵ + 3² + 7 = 32 + 9 + 7 = 48, close to 49) or to the last factor before 7 (63), both common misreadings.
💡Key takeaway

This AMC 8 problem only needs Grade 5 division and the Grade 4 idea of prime factors that you already know!