2015 AMC 10A Problems/Problem 13

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?

$\textbf{(A) }3\qquad\textbf{(B) }4\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad\textbf{(E) }7$

Solution

Let Claudia have $x$ 5-cent coins and $12-x$ 10-cent coins. It is easily observed that any multiple of 5 between 5 and $5x + 10(12 - x) = 120 - 5x$ inclusive can be obtained by a combination of coins. Thus, $24 - x = 17$ combinations can be made, so $x = 7$ . But the answer is not 7, because we are asked for the number of 10-cent coins, which is 12 - 7 = 5. $\textbf{(C)}$

2015 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
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2015 AMC 10A Problems/Problem 13

Problem 13

Solution

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