AMC 8 · 2017 · #17

Grade 4 algebra

Archetype Small-Domain Constrained Search └ Sub-archetype Sum-Target Search

linear-equations-one-varsystems-of-equationsmulti-digit-arithmetic convert-to-algebraguess-and-check ↑ Prerequisites: multi-digit-arithmeticlinear-equations-one-var

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Problem

You have a pile of gold coins and a row of empty treasure chests. If you try to put 9 coins in every chest, you run out of coins before filling everyone — 2 chests are left empty. If instead you put 6 coins in every chest, every chest is full and you still have 3 coins left in your hand. How many gold coins do you have?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 8 2017 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Eliminate Possibilities

It is a multiple-choice AMC problem with only five candidate values for G, and each candidate is easy to plug back into the story. Tool #3 (Eliminate Possibilities) is the natural first move on any AMC multiple-choice problem — for every candidate G we can compute the number of chests two different ways (once from Plan A and once from Plan B) and keep only the candidate where the two answers agree. Tool #6 (Guess and Check) is what powers each individual test: guess a G, compute, and see if the two pictures line up. This is faster and more elementary than setting up a system of equations (Tool #13).

1STEP 1

Turn each story into arithmetic: Plan A gives C = G/9 + 2, Plan B gives C = (G-3)/6.

C_A = G/9 + 2 C_B = (G-3)/6

2STEP 2

Try (A) G = 9: Plan A wants 3 chests but Plan B wants 1 — a mismatch, so eliminate (A).

C_A = 3, C_B = 1 → mismatch

3STEP 3

Try (B) G = 27: Plan A wants 5 chests but Plan B wants 4 — still a mismatch, so eliminate (B).

C_A = 5, C_B = 4 → mismatch

4STEP 4

Try (C) G = 45: both plans want the same 7 chests, so the counts agree and (C) survives.

C_A = 7 = C_B → G = 45 ✓

5STEP 5

(D) G = 63 gives 9 vs 10 and (E) G = 81 gives 11 vs 13 — both mismatch, so only (C) 45 works.

45 = 9 × (7-2) = 9 × 5, 45 = 6 × 7 + 3 → (C)

Answer

With G = 45 and C = 7: Plan A fills 5 chests with 9 coins each (5 × 9 = 45, all coins used, 2 chests left empty — matches the problem). Plan B fills all 7 chests with 6 coins (7 × 6 = 42) and leaves 45 - 42 = 3 coins over — also matches. Both stories are satisfied by the same numbers, so 45 is consistent.

💡Key takeaway

This AMC 8 problem only needs Grade 4 multi-step word-problem skills you already know — just check each answer choice in both stories and keep the one that fits!

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