Competition · AMC preparation · step 4 of 4
AMC 8 · 2014 · #2
Grade 3 arithmeticPick an answer.
AMC 8 2014 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
"Largest minus smallest" is really two separate optimization problems wrapped in one subtraction, so Tool #7 (Identify Subproblems) splits it cleanly: maximize the coin count, then minimize it, then subtract. For each sub-question the move is opposite — to maximize the count, use the smallest-value coin as often as possible; to minimize, use the largest-value coin first (a greedy step). Tool #17 (Pursue Parity / divisibility) confirms both extremes are actually reachable: 35 is a multiple of 5, so an all-nickel payment works, and 35 = 25 + 10 gives a clean two-coin payment.
Use the fewest coins
Fewest coins: take the biggest first — one 25-cent plus one 10-cent makes 35, so the minimum is 2 coins.
Adding two values to hit a target total is a Grade 2 word-problem skill — once you spot 25 + 10 = 35 you are done.
2.OA.A.1Identify SubproblemsUse the most coins
Most coins: use only the smallest coin — 35 is a multiple of 5, so 35 ÷ 5 = 7 coins is the maximum.
Asking "how many 5s fit into 35?" is a Grade 3 division-as-unknown-factor question — the divisibility of 35 by 5 is exactly the parity / multiple check from Tool #17.
The greatest number of coins that can total 35 cents is seven, reached by paying entirely with 5-cent coins.
▸ Why?
You can pay the whole 35 cents using only 5-cent coins, and doing so uses exactly seven coins because seven equal groups of 5 cents build up to 35.
▸ Why?
No payment can use more than seven coins, because every coin is worth at least 5 cents, so eight coins would already stack up to at least eight groups of 5, which is 40 cents and overshoots 35.
Subtract the two counts
Subtract the two counts for the requested difference: 7 - 2 = 5.
Subtracting the smaller count from the larger is the final "how many more?" step, a Grade 2 subtraction within 100.
2.OA.A.1Identify SubproblemsThis AMC 8 problem only needs Grade 3 division (35 ÷ 5) plus a tiny bit of Grade 2 adding and subtracting — split it into "fewest coins" and "most coins" and the rest is one quick subtraction.
- Use the fewest coins
- Use the most coins
- Subtract the two counts
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