AMC 10 · 2015 · #3
Grade 4 arithmeticPick an answer.
AMC 10 2015 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
We are not told whether 28 is the number repeated twice or the number repeated three times, so there are only two possible setups. Tool #3 (Eliminate Possibilities) tests each setup and throws out the one that cannot produce a whole number, leaving exactly one valid answer. Tool #4 (Introduce a Variable) names the unknown integer x so we can track what each case forces it to be. The decisive clue is that the other integer must come out whole — only one of the two cases survives that test.
Name the unknown, list the two cases
Let the other integer be x. Either Case A: 28 three times with x twice, or Case B: 28 twice with x three times — both must total 100.
There are only two ways to share the five slots between 28 and x, so checking both covers everything.
4.OA.A.3Use Matrix LogicTest Case A: 28 three times
If 28 appears three times, that's 3·28 = 84, leaving 100 - 84 = 16 shared equally so x = 8 — a whole number, so this case works.
If three of the numbers are pinned down, the leftover total has to be split evenly between the last two.
4.NBT.B.5Eliminate PossibilitiesTest Case B, eliminate it, read the answer
If 28 appears twice, 100 - 56 = 44 splits into three equal x, but 44 ÷ 3 isn't whole, so Case B fails and the other integer is 8 → (A).
If splitting the leftover into equal whole pieces is impossible, that case can't happen.
4.NBT.B.6Eliminate PossibilitiesWhen you don't know which slot a number fills, try every option and keep only the one that comes out as a whole number.
- Name the unknown, list the two cases
- Test Case A: 28 three times
- Test Case B, eliminate it, read the answer