AMC 8 · 2016 · #9
Grade 5 number-theoryProblem
What is the sum of the distinct prime integer divisors of 2016?
Pick an answer.
AMC 8 2016 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.
Toolkit + CCSS Solution
Understand
Restated: 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.
Givens: The target number is $2016$; We want only the prime divisors (not all divisors); We want distinct primes — repeats are counted once; Answer choices: (A) $9$, (B) $12$, (C) $16$, (D) $49$, (E) $63$
Unknowns: The sum of the distinct prime divisors of $2016$
Understand
Restated: 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.
Givens: The target number is $2016$; We want only the prime divisors (not all divisors); We want distinct primes — repeats are counted once; Answer choices: (A) $9$, (B) $12$, (C) $16$, (D) $49$, (E) $63$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #6 Guess and Check
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 $2, 3, 5, 7, \ldots$ 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.
Execute — Answer: B
5.NBT.B.6 Step 1 - Subproblem 1: factor $2016$.
- Start by dividing out the smallest prime, $2$, as many times as possible.
- $2016$ is even, so we keep halving until we hit an odd number.
💡 Repeated division by $2$ is a Grade 5 multi-digit division skill — no special technique needed.
4.OA.B.4 Step 2 - $63$ is odd, so $2$ is finished.
- Try the next prime, $3$.
- The digit sum of $63$ is $6+3=9$, which is a multiple of $3$, so $3$ divides $63$.
- Keep dividing by $3$ until you cannot anymore.
💡 Using the digit-sum rule for $3$ and recognizing prime factors is exactly the Grade 4 "factors and multiples" standard.
4.OA.B.4 Step 3 - The remaining quotient $7$ is itself prime, so the factorization is complete.
- Write out the full prime factorization, collecting like primes into exponents.
💡 Recognizing that $7$ is prime ends the factorization — a Grade 4 prime/composite check.
4.OA.B.4 Step 4 - Subproblem 2: list the distinct prime bases (ignore the exponents — they only tell us how many times each prime appears, not which primes appear).
- The distinct primes are $2$, $3$, and $7$.
💡 "Distinct" means we list each prime base once — a careful reading of the problem, not a calculation.
2.NBT.B.5 Step 5 Add the three distinct primes and match the result to the answer choices.
💡 Adding three small whole numbers is a Grade 2 fluency skill.
5.NBT.B.6 Subproblem 1: factor $2016$. Start by dividing out the smallest prime, $2$, as m 4.OA.B.4 $63$ is odd, so $2$ is finished. Try the next prime, $3$. The digit sum of $63$ 4.OA.B.4 The remaining quotient $7$ is itself prime, so the factorization is complete. Wr 4.OA.B.4 Subproblem 2: list the distinct prime bases (ignore the exponents — they only te 2.NBT.B.5 Add the three distinct primes and match the result to the answer choices. Review
Reasonableness: Multiply the factorization back together to confirm: $2^5 = 32$, $3^2 = 9$, and $32 \times 9 = 288$, then $288 \times 7 = 2016$. 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 ($2^5 + 3^2 + 7 = 32 + 9 + 7 = 48$, close to $49$) or to the last factor before $7$ ($63$), both common misreadings.
Alternative: Tool #9 (Solve an Easier Related Problem): break $2016$ into familiar pieces — $2016 = 2 \times 1008 = 2 \times 2 \times 504 = \ldots$ or notice $2016 = 2000 + 16 = 16 \times 126 = 16 \times 2 \times 63 = 32 \times 63$. Then $32 = 2^5$ and $63 = 9 \times 7 = 3^2 \times 7$, so $2016 = 2^5 \times 3^2 \times 7$. Same distinct primes, same sum $2+3+7 = 12$.
CCSS standards used (min grade 5)
2.NBT.B.5Fluently add and subtract within 100 (Adding the three distinct primes $2 + 3 + 7 = 12$ to produce the final answer.)4.OA.B.4Find all factor pairs and recognize multiples; determine prime or composite (Recognizing $2$, $3$, and $7$ as prime divisors and using the digit-sum rule to test divisibility by $3$.)5.NBT.B.6Find whole-number quotients of whole numbers with up to four-digit dividends (Performing the repeated divisions $2016 \div 2 = 1008$, $1008 \div 2 = 504$, $\ldots$ to peel off the prime factors.)
⭐ This AMC 8 problem only needs Grade 5 division and the Grade 4 idea of prime factors that you already know!
⭐ This AMC 8 problem only needs Grade 5 division and the Grade 4 idea of prime factors that you already know!
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