Difference between revisions of "2015 AMC 10A Problems/Problem 13"
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| − | + | ==Problem 13== | |
| + | Claudia has 12 coins, each of which is a 5-cent coin or a 10-cent coin. There are exactly 17 different values that can be obtained as combinations of one or more of her coins. How many 10-cent coins does Claudia have? | ||
| − | + | <math> \textbf{(A) }3\qquad\textbf{(B) }4\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad\textbf{(E) }7 </math> | |
| − | + | ==Solution== | |
| − | + | Let Claudia have <math>x</math> 5-cent coins and <math>12-x</math> 10-cent coins. It is easily observed that any multiple of 5 between 5 and <math>5x + 10(12 - x) = 120 - 5x</math> inclusive can be obtained by a combination of coins. Thus, <math>24 - x = 17</math> combinations can be made, so <math>x = 7</math>. But the answer is not 7, because we are asked for the number of 10-cent coins, which is 12 - 7 = 5. <math>\textbf{(C)}</math> | |
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Revision as of 18:51, 4 February 2015
Problem 13
Claudia has 12 coins, each of which is a 5-cent coin or a 10-cent coin. There are exactly 17 different values that can be obtained as combinations of one or more of her coins. How many 10-cent coins does Claudia have?
Solution
Let Claudia have 
5-cent coins and 
10-cent coins. It is easily observed that any multiple of 5 between 5 and 
inclusive can be obtained by a combination of coins. Thus, 
combinations can be made, so 
. But the answer is not 7, because we are asked for the number of 10-cent coins, which is 12 - 7 = 5. 
