AMC 8 · 2016 · #9
Easy mode Grade 5Problem
Take the number 2016 and break it down into prime numbers that multiply together to give 2016.
Some of those primes may appear more than once. Look at just the different ones — count each kind only once.
What do you get when you add those different prime numbers together?
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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