# Difference between revisions of "2015 AMC 10A Problems/Problem 13"

## Problem 13

Ashraful 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 Ashraful have?

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

## Solution 1

Let Ashraful have $x$ 5-cent coins and $\left( 12 - x \right)$ 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 $12 - 7 = \boxed{\textbf{(C) } 5}$

## Solution 2

Since the coins are 5-cent and 10-cent, all possible values that can be made will be multiples of $5.$ To have exactly $17$ different multiples of $5,$ we will need to make up to $85$ cents. If all twelve coins were 5-cent coins, we will have $60$ cents possible. Each trade of a 5-cent coin for a 10-cent coin will gain $5$ cents, and as we need to gain $25$ cents, the answer is $\boxed{\textbf{(C) } 5}$

For small scale substitutions of 5 cent for 10 cent, like in this problem, it is quite fast, however, if the problem is a little bit more complex, knowing that you need 85 cents, explained above, it is also possible to use a system of equations. It would be $$5x+10y=85$$ $$x+y=12$$ Solving this $x$ (5-cent coins) $= 7$ and $y$ (10-cent coins) $= 5$, so again the answer is $\boxed{\textbf{(C) } 5}$

~savannahsolver