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

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==Problem==
 
==Problem==
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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?
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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> treasure chests empty.  So instead I put <math>6</math> gold coins in each treasure chest, but then I had <math>3</math> gold coins left over.  How many gold coins did I have?
  
 
<math>\textbf{(A) }9\qquad\textbf{(B) }27\qquad\textbf{(C) }45\qquad\textbf{(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>
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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>.
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~CHECKMATE2021
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==Solution 2 (Using Answer Choices)==
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With <math>9</math> coins, there are <math>\frac{9}{9}+2=1+2=3</math> chests, by the first condition. These don't fit in with the second condition, so we move onto <math>27</math> coins. By the same first condition, there are <math>5</math> chests(<math>\frac{27}{9}+2</math>). This also doesn't fit with the second condition. So, onto <math>45</math> coins. The first condition implies that there are <math>\frac{45}{9}+2=7</math> chests, which DOES fit with the second condition, since <math>6\cdot7+3=42+3=45</math>. Thus, the desired value is <math>\boxed{\textbf{(C)}\ 45}</math>.
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~vadava_lx
  
 
==Video Solution (CREATIVE THINKING + ANALYSIS!!!)==
 
==Video Solution (CREATIVE THINKING + ANALYSIS!!!)==
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~savannahsolver
 
~savannahsolver
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==Video Solution by SpreadTheMathLove==
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https://www.youtube.com/watch?v=TpsuRedYOiM&t=250s
  
 
==See Also==
 
==See Also==

Latest revision as of 19:38, 7 October 2024

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 $\boxed{\textbf{(C)}\ 45}$. ~CHECKMATE2021

Solution 2 (Using Answer Choices)

With $9$ coins, there are $\frac{9}{9}+2=1+2=3$ chests, by the first condition. These don't fit in with the second condition, so we move onto $27$ coins. By the same first condition, there are $5$ chests($\frac{27}{9}+2$). This also doesn't fit with the second condition. So, onto $45$ coins. The first condition implies that there are $\frac{45}{9}+2=7$ chests, which DOES fit with the second condition, since $6\cdot7+3=42+3=45$. Thus, the desired value is $\boxed{\textbf{(C)}\ 45}$.

~vadava_lx

Video Solution (CREATIVE THINKING + ANALYSIS!!!)

https://youtu.be/JvgeBrx9Q0U

~Education, the Study of Everything

Video Solution

https://youtu.be/PxO6VxSHD9A

https://youtu.be/vmg51kO7LKg

https://youtu.be/DkVbXdBAYeg

~savannahsolver

Video Solution by SpreadTheMathLove

https://www.youtube.com/watch?v=TpsuRedYOiM&t=250s

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. AMC logo.png

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