AMC 8 · 2003 · #2
Grade 4 number-theoryPick an answer.
AMC 8 2003 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
This is a multiple-choice problem with only five candidates, so Tool #3 (Eliminate Possibilities) is the natural fit: test each choice against the smallest primes in order. Tool #5 (Look for a Pattern) sharpens the test — the smallest prime is 2, and the pattern "divisible by 2 = even" lets us scan the list at a glance. If any candidate is even, it must be the winner, because no number can have a prime factor below 2.
List the primes in order: the smallest one any number can have is 2, then 3, then 5 — so begin the hunt at the bottom, with 2.
Knowing the order of primes turns the problem into a quick checklist: try 2 first, then 3, then 5.
4.OA.B.4Look For A PatternA number is divisible by 2 exactly when it is even, so scan the ones digits (5, 7, 8, 9, 1): only 58 ends in an even digit.
The even/odd pattern is the fastest divisibility test there is — one glance at the ones digit decides it.
3.OA.D.9Eliminate Possibilities58 is even, so its smallest prime factor is 2; the other four are odd (smallest prime at least 3), and nothing beats 2 — so (C).
Once one candidate hits the smallest possible prime, the search is over — no later check can produce a smaller answer.
4.OA.B.4Eliminate PossibilitiesThe smallest prime is 2, so on a "smallest prime factor" question, first ask: is any choice even? If yes, that choice wins immediately — here 58 is the only even number, so the answer is (C).