Competition · AMC preparation · step 4 of 4
AMC 8 · 2007 · #3
Grade 4 number-theoryPick an answer.
AMC 8 2007 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
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.
Divide out the smallest prime
250 ends in 0, so it is even — peel off the smallest prime 2 to get 125.
The smallest prime is always 2, and any even number is divisible by 2. This is the Tool #7 "peel off one prime" move.
4.OA.B.4Identify SubproblemsDivide 125 by 5
125 is odd and not a multiple of 3, but it ends in 5 — so its smallest prime is 5.
Checking divisibility in order (2, 3, 5, 7, …) guarantees we catch the smallest prime first.
Once the factor 2 is taken out of 250, the smallest prime that divides the leftover 125 is 5.
▸ Why?
5 divides 125 evenly because 125 ends in the digit 5.
▸ Why?
Whether 5 divides a number depends only on its last digit, because the tens, hundreds, and every larger place are whole numbers of tens, and every ten is already a multiple of 5.
▸ Why?
A ten is two fives and the last digit 5 is one more five, so a number ending in 0 or 5 is an exact count of fives with none left over.
▸ Why?
The smaller prime 2 does not divide 125 because 125 is odd: the tens and hundreds are all even bundles, so only the last digit decides, and 5 leaves one over when split into pairs.
▸ Why?
The smaller prime 3 does not divide 125 because 125's digits add up to 1+2+5 = 8, and 8 is not a multiple of 3.
▸ Why?
A whole number and the sum of its digits leave the same remainder when divided by 3, because each place value (10, 100, and so on) is one more than a multiple of 3.
Add the two primes
The two smallest distinct primes of 250 are 2 and 5; adding them gives 7.
7 matches choice (C). The other choices are quickly ruled out — (A) 2 and (B) 5 each forget one of the primes; (D) 10 and (E) 12 pull in composite numbers, but the problem asks for prime factors.
4.NBT.B.4Eliminate PossibilitiesTo 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.
- Divide out the smallest prime
- Divide 125 by 5
- Add the two primes
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