2017 AMC 8

Complete problem set with solutions and individual problem pages

Problem 17 Hard

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?

  • A.

    9

  • B.

    27

  • C.

    45

  • D.

    63

  • E.

    81

Answer:C

Solution 1 

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}.

 

Solution 2

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}.