Problem 17 · 2017 AMC 8
Easy
Algebra & Patterns
substitutionsum-constraint
Starting with some gold coins and some empty treasure chests, I tried to put 9 gold coins in each treasure chest, but that left 2 treasure chests empty. So instead I put 6 gold coins in each treasure chest, but then I had 3 gold coins left over. How many gold coins did I have?
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Answer: C — 45 coins.
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Hint 1 of 2
The coin pile never changes — only how it's distributed. Write the same number of coins two ways, one for each filling, and set those expressions equal.
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Hint 2 of 2
"2 chests empty" means only n − 2 chests get the 9 coins. Equate g = 9(n − 2) with g = 6n + 3 and solve.
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Approach: count the same coins two ways
- Let n = chests and g = coins. Filling 9 each but leaving 2 chests empty uses only n − 2 chests: g = 9(n − 2). Filling 6 each with 3 left over: g = 6n + 3.
- Same g, so 9(n − 2) = 6n + 3 ⇒ 9n − 18 = 6n + 3 ⇒ 3n = 21 ⇒ n = 7.
- g = 6(7) + 3 = 45.
- Why this transfers: any "distribute the same total two different ways" puzzle becomes one equation by writing the unchanging total twice and matching them.
Another way — track the difference per chest:
- Going from 6-per-chest to 9-per-chest adds 3 coins to each of the 7 chests... but instead reason forward: the two schemes differ. Switching from 9s to 6s frees 3 coins from every filled chest, plus the 2 chests that were empty now also get 6 each.
- Quick check the answer: with 7 chests, 9×(7−2) = 45 and 6×7 + 3 = 45 — both give 45, confirming the count is consistent.
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