Competition · AMC preparation · step 4 of 4
AMC 8 · 2025 · #6
Grade 4 number-theoryPick an answer.
AMC 8 2025 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
There are only five candidates — the five answer choices themselves. That is a textbook setup for Tool #3 (Eliminate Possibilities): test each choice against the rule "sum of the remaining four is a multiple of 4" and keep the one that survives. Tool #2 (Systematic List) keeps the bookkeeping tidy: first compute the total once, then list "85 minus each candidate" in order so no case is missed or double-counted. We deliberately avoid Tool #13 (Algebra) or modular-arithmetic shortcuts — they work, but a 4th-grader can solve this with addition and a divisibility check, which is the whole point.
Add all five numbers
Add all five numbers once — pairing the ends keeps it easy — and the total comes to 85.
Adding five two-digit numbers is exactly the Grade 4 "fluently add multi-digit whole numbers" skill — no shortcut needed.
4.NBT.B.4Make A Systematic ListSubtract each candidate
Erase x and the leftover sum is 85 - x; list that value in order for all five choices so none is missed.
A clean ordered list of five subtractions makes sure no candidate is skipped — same Grade 4 add/subtract fluency.
4.NBT.B.4Make A Systematic ListKeep only multiples of 4
Check each leftover sum for divisibility by 4: only 68 passes, so every choice but (C) is eliminated.
Checking whether a number is a multiple of 4 is exactly the Grade 4 "factors and multiples" skill — count by 4s or divide and look for remainder 0.
Of the five possible leftover sums — each one the total 85 minus a different candidate — exactly one is a multiple of 4, so only one erased number can satisfy the rule.
▸ Why?
The sum of the four kept numbers is just 85 minus the single erased number, because the five numbers split with no gaps or overlaps into the four that stay and the one that goes.
▸ Why?
A number is a multiple of 4 exactly when dividing it into groups of 4 leaves remainder 0, so we run each of the five leftover sums through that remainder test: 68 = 4 × 17 + 0 passes, while 70, 69, 67, 66 leave remainders 2, 1, 3, 2 and fail — exactly one of the five passes.
Pick the erased number
The lone survivor is 17, so that is the number Sekou erased and the answer is (C).
Tool #3 says: when exactly one choice survives every test, that is the answer.
4.OA.B.4Eliminate PossibilitiesThis AMC 8 problem only needs Grade 4 addition and multiples-of-4 checking you already know!
- Add all five numbers
- Subtract each candidate
- Keep only multiples of 4
- Pick the erased number
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