AMC 8 · 2010 · #14
Grade 4 number-theoryPick an answer.
AMC 8 2010 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Factoring 2010 is too big to do in one shot, so Tool #7 (Identify Subproblems) breaks it into a chain of smaller divisions: peel off one small prime at a time (2, then 3, then 5, …) until what is left is itself prime. Each step is a tiny problem we can do mentally. Tool #3 (Eliminate Possibilities) is the multiple-choice safety net — once we have a candidate sum, we can check it against the listed choices and rule the others out.
2010 ends in 0, so it is even — divide by 2 to get 1005.
Splitting off one prime at a time turns a big factoring task into a sequence of easy divisions — the Tool #7 move.
4.OA.B.4Identify SubproblemsDigit sum 1+0+0+5 = 6 is divisible by 3, so divide 1005 by 3 to get 335.
The digit-sum divisibility rule is the fastest way to test for 3 without long division.
4.OA.B.4Identify Subproblems335 ends in 5, so it is divisible by 5 — divide to get 67.
Numbers ending in 0 or 5 are always divisible by 5.
4.OA.B.4Identify SubproblemsNo prime up to √67 ≈ 8.2 divides 67, so 67 is prime and 2010 = 2 × 3 × 5 × 67.
Once the leftover quotient is prime, the chain of subproblems is done.
4.OA.B.4Identify SubproblemsAdd the distinct primes: 2 + 3 + 5 + 67 = 77, which is choice (C).
77 matches choice (C); the other choices are quickly ruled out — (A) 67 forgets to add 2+3+5, (E) 210 is just the four digits of 2010 rearranged, and (D) 201 is 2010 ÷ 10, a distractor.
4.NBT.B.4Eliminate PossibilitiesBig numbers like 2010 become easy once you peel off small primes one at a time — a Grade 4 factor-finding skill is all you need!
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