Difference between revisions of "2017 AMC 8 Problems/Problem 17"
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− | ==Problem | + | ==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? | 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? | ||
− | <math>\textbf{(A) }9\qquad\textbf{(B) }27\qquad\textbf{(C) }45\qquad{(D) }63\qquad\textbf{(E) }81</math> | + | <math>\textbf{(A) }9\qquad\textbf{(B) }27\qquad\textbf{(C) }45\qquad\textbf{(D) }63\qquad\textbf{(E) }81</math> |
==Solution== | ==Solution== | ||
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<cmath>6c+3 = g</cmath> | <cmath>6c+3 = g</cmath> | ||
− | 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} | + | We do this because for 9 chests there are 2 empty and if 9 were in each 9 multiplied by 2 is 18 left. |
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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>. | ||
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+ | ==Video Solution== | ||
+ | https://youtu.be/PxO6VxSHD9A | ||
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+ | https://youtu.be/vmg51kO7LKg | ||
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+ | https://youtu.be/DkVbXdBAYeg | ||
+ | |||
+ | ~savannahsolver | ||
==See Also== | ==See Also== |
Latest revision as of 21:05, 7 March 2022
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:
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
~savannahsolver
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.