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

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==Problem 17==
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==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{(D) }63\qquad\textbf{(E) }81</math>
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<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}.</math>
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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==
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https://youtu.be/PxO6VxSHD9A
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https://youtu.be/vmg51kO7LKg
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https://youtu.be/DkVbXdBAYeg
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~savannahsolver
  
 
==See Also==
 
==See Also==

Latest revision as of 22:05, 7 March 2022

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?

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

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: \[9c-18 = g\] \[6c+3 = g\]

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

Video Solution

https://youtu.be/PxO6VxSHD9A

https://youtu.be/vmg51kO7LKg

https://youtu.be/DkVbXdBAYeg

~savannahsolver

See Also

2017 AMC 8 (ProblemsAnswer KeyResources)
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

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