# Difference between revisions of "2017 AMC 8 Problems/Problem 17"

## Problem 17

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?

$\textbf{(A) }9\qquad\textbf{(B) }27\qquad\textbf{(C) }45{(D) }63\qquad\textbf{(E) }81$

## Solution

We can represent the amount of gold with $g$ and the amount of chests with $c$. We can use the problem to make the following equations: $$9c-18 = g$$ $$6c+3 = g$$

Therefore, $6c+3 = 9c-18.$ This implies that $c = 7.$ We therefore have $g = 45.$ So, our answer is $\boxed{\textbf{(C)}\ 45}.$

 2017 AMC 8 (Problems • Answer Key • Resources) Preceded byProblem 16 Followed byProblem 18 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AJHSME/AMC 8 Problems and Solutions

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