Competition · AMC preparation · step 4 of 4
AMC 8 · 2016 · #9
Grade 5 number-theoryPick an answer.
AMC 8 2016 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
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.
Divide out the 2s
Subproblem 1: factor 2016 by dividing out 2 repeatedly until the odd quotient 63 appears.
Repeated division by 2 is a Grade 5 multi-digit division skill — no special technique needed.
5.NBT.B.6Guess And CheckDivide out the 3s
63 is odd, so switch to 3 (digit sum 6+3=9 is a multiple of 3); dividing by 3 twice leaves 7.
Using the digit-sum rule for 3 and recognizing prime factors is exactly the Grade 4 "factors and multiples" standard.
4.OA.B.4Guess And CheckStop at the prime 7
7 is itself prime, so the factorization is complete: 2⁵ × 3² × 7.
Recognizing that 7 is prime ends the factorization — a Grade 4 prime/composite check.
Dividing 2016 by primes as far as it will go gives 2016 = 2⁵ × 3² × 7, so the primes that divide 2016 are exactly 2, 3, and 7.
▸ Why?
Each exact division records one prime factor and shrinks the number: 2016 ÷ 2 = 1008 says 2016 = 2 × 1008, and repeating this on the quotient rebuilds 2016 as the product 2⁵ × 3² × 7 of the primes we divided out.
▸ Why?
An exact division can be read backward as a multiplication — 2016 ÷ 2 = 1008 with no remainder means 2 × 1008 = 2016 — so every step trades a division for a matching factor without changing the value.
▸ Why?
Because a whole number can be built out of primes in only one way, any prime that divides 2016 must already appear in this single factorization 2⁵ × 3² × 7 — so 2, 3, and 7 are exactly the primes that divide 2016, with none left out and no other prime hidden inside.
List the distinct primes
Subproblem 2: ignore the exponents and keep only the distinct prime bases 2, 3, and 7.
"Distinct" means we list each prime base once — a careful reading of the problem, not a calculation.
4.OA.B.4Identify SubproblemsAdd the three primes
Add the three distinct primes 2 + 3 + 7 and match the result to a choice.
Adding three small whole numbers is a Grade 2 fluency skill.
2.NBT.B.5Identify SubproblemsThis AMC 8 problem only needs Grade 5 division and the Grade 4 idea of prime factors that you already know!
- Divide out the 2s
- Divide out the 3s
- Stop at the prime 7
- List the distinct primes
- Add the three primes
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