2017 AMC 8 Problem 17

Attempt Problem 17 of the 2017 AMC 8 below, then check your answer against the video solution and professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2017 AMC 8 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

17.

Starting with some gold coins and some empty treasure chests, I tried to put 99 gold coins in each treasure chest, but that left 22 treasure chests empty. So instead I put 66 gold coins in each treasure chest, but then I had 33 gold coins left over. How many gold coins did I have?
9 9
27 27
45 45
63 63
81 81
Answer: C
Concepts:system of equations
Small hint:
Let nn be the number of treasure chests.
Big hint:
Write one equation for the 99-coins attempt and another for the 66-coins attempt.
Video solution:
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Written solution:
Let nn be the number of treasure chests and gg be the number of gold coins. Then 9(n−2)=g9(n - 2) = g and 6n+3=g.6n + 3 = g. Solving this system yields n=7,n = 7, so the number of gold coins is 6⋅7+3=45.6 \cdot 7 + 3 = 45.
Thus, C is the correct answer.
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