AMC 8 · 2007 · #3

Grade 4 number-theory

Archetype Arithmetic Expression Decomposition └ Sub-archetype Linear Rate Accumulation

prime-factorizationprime-numbersfactors identify-subproblems ↑ Prerequisites: prime-numbersdivisibility-rules

📏 Short solution 💡 1 insight

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Problem

Find the two smallest prime numbers that divide 250 and add them.

Pick an answer.

(A)

$2$

(B)

$5$

(C)

$7$

(D)

$10$

(E)

$12$

AMC 8 2007 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Identify Subproblems

Tool #7 (Identify Subproblems) turns "find the two smallest prime factors" into two tiny questions: (1) what is the smallest prime that divides 250? (2) once we divide that out, what is the smallest prime that divides what is left? Each step is a single divisibility check, which is much easier than factoring 250 all at once. Tool #3 (Eliminate Possibilities) is the multiple-choice safety net — once we have a candidate sum, we can verify it against the listed choices.

1STEP 1

250 ends in 0, so it is even — peel off the smallest prime 2 to get 125.

250 ÷ 2 = 125

2STEP 2

125 is odd and not a multiple of 3, but it ends in 5 — so its smallest prime is 5.

125 ÷ 5 = 25

3STEP 3

The two smallest distinct primes of 250 are 2 and 5; adding them gives 7.

2 + 5 = 7 → (C)

Answer

Continue the factorization to double-check: 25 ÷ 5 = 5 and 5 ÷ 5 = 1, so 250 = 2 × 5 × 5 × 5 = 2 × 5³. The complete list of distinct primes is just {2, 5}, and those are also the two smallest, so the sum 2 + 5 = 7 is correct. The answer (C) sits between (B) 5 (only one prime) and (D) 10 = 2 × 5 (a composite distractor), exactly where a small-prime-sum should land.

💡Key takeaway

To find the smallest prime factors, peel them off one at a time starting with 2 — a Grade 4 divisibility-rules skill is all this AMC 8 problem asks for.

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