Difference between revisions of "2017 AMC 8 Problems/Problem 17"
(→Solution) |
m (→Solution 2(Answer Choices)) |
||
| Line 15: | Line 15: | ||
==Solution 2(Answer Choices)== | ==Solution 2(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>. | 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!!!)== | ||
Revision as of 17:18, 28 December 2023
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:
![\[9c-18 = g\]](http://latex.artofproblemsolving.com/2/2/5/225c8355639347fb2649ec641ef2715c6d7064c7.png)
![\[6c+3 = g\]](http://latex.artofproblemsolving.com/e/e/4/ee449126e0ac01646778ea0d6f12fbdce6e711d7.png)
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 
.
Solution 2(Answer Choices)
With 
coins, there are 
chests, by the first condition. These don't fit in with the second condition, so we move onto 
coins. By the same first condition, there are 
chests(
). This also doesn't fit with the second condition. So, onto 
coins. The first condition implies that there are 
chests, which DOES fit with the second condition, since 
. Thus, the desired value is .
~vadava_lx
Video Solution (CREATIVE THINKING + ANALYSIS!!!)
~Education, the Study of Everything
Video Solution
~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. 
