Difference between revisions of "2017 AMC 8 Problems/Problem 17"
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Therefore, <math>6c+3 = 9c-18.</math> This implies that <math>c = 7.</math> We therefore have <math>g = 45.</math> So, our answer is <math>\boxed{\textbf{(C)}\ 45}</math>. | Therefore, <math>6c+3 = 9c-18.</math> This implies that <math>c = 7.</math> We therefore have <math>g = 45.</math> So, our answer is <math>\boxed{\textbf{(C)}\ 45}</math>. | ||
| + | |||
| + | ==Video Solution== | ||
| + | https://youtu.be/vmg51kO7LKg | ||
==See Also== | ==See Also== | ||
Revision as of 13:04, 18 January 2021
Contents
Problem
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?
Solution
We can represent the amount of gold with 
and the amount of chests with 
. We can use the problem to make the following equations:
![\[9c-18 = g\]](http://latex.artofproblemsolving.com/2/2/5/225c8355639347fb2649ec641ef2715c6d7064c7.png)
![\[6c+3 = g\]](http://latex.artofproblemsolving.com/e/e/4/ee449126e0ac01646778ea0d6f12fbdce6e711d7.png)
We do this because for 9 chests there are 2 empty and if 9 were in each 9 multiplied by 2 is 18 left.
Therefore, 
This implies that 
We therefore have 
So, our answer is 
.
Video Solution
See Also
| 2017 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 16 |
Followed by Problem 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
