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

m (Solution 2(Answer Choices))
 
(6 intermediate revisions by 4 users not shown)
Line 1: Line 1:
 
==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 <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>
Line 12: Line 12:
  
 
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>.
 +
~CHECKMATE2021
 +
 +
==Solution 2 (Using Answer Choices)==
 +
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>.
 +
 +
~vadava_lx
  
 
==Video Solution (CREATIVE THINKING + ANALYSIS!!!)==
 
==Video Solution (CREATIVE THINKING + ANALYSIS!!!)==

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 (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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png