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 <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>. | ||
+ | ~CHECKMATE2021 | ||
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+ | ==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>. | ||
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+ | ~vadava_lx | ||
==Video Solution (CREATIVE THINKING + ANALYSIS!!!)== | ==Video Solution (CREATIVE THINKING + ANALYSIS!!!)== |
Latest revision as of 19:38, 7 October 2024
Contents
Problem
Starting with some gold coins and some empty treasure chests, I tried to put gold coins in each treasure chest, but that left treasure chests empty. So instead I put gold coins in each treasure chest, but then I had 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 . ~CHECKMATE2021
Solution 2 (Using 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.