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2017 AMC 8 Problems/Problem 17

Revision as of 14:34, 22 November 2017 by LearningMath (talk | contribs) (Created page with "==Problem 17== Starting with some gold coins and some empty treasure chests, I tried to put <math>9 </math> gold coins in each treasure chest, but that left <math>2</math> tr...")
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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?

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

Solution

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 $\text{C)}$ $45.$

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