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.
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 SubproblemsMost 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.
3.OA.B.6Visualize Spatial RelationshipsSubtract 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.