Problem 14 · 2010 AMC 8
Easy
Number Theory
prime-factorization
What is the sum of the prime factors of 2010?
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Answer: C — 77.
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Hint 1 of 2
2010 ends in 0, so it's begging to be split as 201 × 10. That instantly hands you the primes 2 and 5 — now just crack 201.
Still stuck? Show hint 2 →
Hint 2 of 2
To prime-factor, peel off the easy small primes first (2, 3, 5) using divisibility shortcuts, then factor whatever's left.
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Approach: peel off small primes using divisibility tests
- The trailing 0 gives 2010 = 201 × 10 = 2 · 5 · 201. For 201, its digits sum to 3, so 3 divides it: 201 = 3 · 67.
- 67 has no small factor (not even, not a multiple of 3 or 5, and 7·7 > 67 once you've ruled out 7) — so it's prime.
- Primes are 2, 3, 5, 67; sum = 77.
- Why this transfers: divisibility tests (even → 2, digit-sum → 3, ends in 0/5 → 5) let you factor by inspection. And you only test primes up to the square root before declaring what's left prime.
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