Difference between revisions of "2017 AMC 8 Problems/Problem 17"
Nukelauncher (talk | contribs) m (→Solution) |
Nukelauncher (talk | contribs) |
||
Line 11: | Line 11: | ||
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> | ||
+ | |||
+ | ==See Also== | ||
+ | {{AMC8 box|year=2017|num-b=16|num-a=18}} | ||
+ | |||
+ | {{MAA Notice}} |
Revision as of 14:39, 22 November 2017
Problem 17
Starting with some gold coins and some empty treasure chests, I tried to put gold coins in each treasure chest, but that left treasure chests empty. So instead I put gold coins in each treasure chest, but then I had 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:
Therefore, This implies that We therefore have So, our answer is
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.