Problem 9 · 2016 AMC 8
Easy
Number Theory
factorizationprimes
What is the sum of the distinct prime integer divisors of 2016?
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Answer: B — 12.
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Hint 1 of 2
The word that matters is DISTINCT — you only care WHICH primes appear, not how many times. So you don't need the full exponents, just the list of different prime factors.
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Hint 2 of 2
Peel small primes off 2016 one at a time: it's even (keep halving), then the digits sum to a multiple of 3, and what's left will reveal the rest. Each new prime you meet goes on your list once.
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Approach: strip out each prime once; only the distinct ones count
- Halve repeatedly: 2016 → 1008 → 504 → 252 → 126 → 63, pulling out 2 (five times). The prime 2 is now on the list.
- 63 = 9 × 7 = 32 · 7, adding the primes 3 and 7. So 2016 = 25 · 32 · 7.
- Distinct primes are 2, 3, 7 (ignore the exponents — "distinct" means count each prime once). Sum = 2 + 3 + 7 = 12.
- Watch the trap: answer E = 63 is what you'd get by adding leftover factors, and 49 = 7² baits you into squaring — the question wants the PRIMES themselves, added once each.
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