Competition · AMC preparation · step 4 of 4
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.
Divide out the factor 2
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 SubproblemsDivide out the factor 3
Digit 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 SubproblemsDivide out the factor 5
335 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 SubproblemsCheck whether 67 is prime
No 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.
The prime factorization of 2010 is 2 × 3 × 5 × 67, so its distinct prime factors are exactly 2, 3, 5, and 67.
▸ Why?
Multiplying the four factors back together returns 2010, so 2 × 3 × 5 × 67 names the very number the problem starts from.
▸ Why?
Each peeling step was an exact division, and an exact division just undoes a multiplication, so 5 × 67 = 335, 3 × 335 = 1005, and 2 × 1005 = 2010 each hold and rebuild 2010 step by step.
▸ Why?
The way the factors are grouped never changes their product, so the nested 2 × (3 × (5 × 67)) is the same number as the flat product 2 × 3 × 5 × 67.
▸ Why?
These four are exactly the prime factors, because each one is prime and the peeling stopped only once the leftover was itself prime, so nothing was skipped and no factor can be split further.
▸ Why?
2, 3, and 5 are among the first primes, and 67 can be laid out only as one row of 67 — no equal rows of 2 up to 8 fill it evenly — so each factor is prime and hides no smaller factors inside.
▸ Why?
Since 2, 3, 5, and 67 are each prime, 2 × 3 × 5 × 67 is a way of building 2010 out of primes — and a whole number above 1 can be built from primes in only one way apart from order, so this is the complete list: no other prime divides 2010, and none of these four can be dropped.
Add the distinct primes
Add 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!
- Divide out the factor 2
- Divide out the factor 3
- Divide out the factor 5
- Check whether 67 is prime
- Add the distinct primes
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