2015 AMC 10A Problem 13

Say Claudia has $n$ dimes and $12 - n$ nickels. Measure money in nickels: a dime is worth $2$ units, so her total is $(12 - n) + 2n = 12 + n$ units.

Claim: if she has at least one nickel, every amount from $1$ unit up to the total is obtainable. To make $m$ units, use $\min(n, \lfloor m/2 \rfloor)$ dimes and the rest in nickels; the number of nickels needed is at most $1$ when $m \le 2n$, and at most $12 - n$ when $m > 2n$ (because $m \le 12 + n$). So the obtainable values are exactly $1, 2, \ldots, 12 + n$, which is $12 + n$ distinct values.

Setting $12 + n = 17$ gives $n = 5$. (If she had no nickels, only $12$ values would be possible, not $17$.)

The answer is $\boxed{\textbf{(C)}\ 5}$.

With coins of $5$ and $10$ cents and at least one $5$, the reachable amounts have no gaps: dimes handle the even part, a single nickel fixes parity, and leftover nickels extend the range to the total. So "number of distinct values" is just "total divided by $5$," a one-line equation in $n$.

Check the choices directly. With $5$ dimes and $7$ nickels the total is $85$ cents, giving values $5, 10, \ldots, 85$: that is $17$ values. With $4$ dimes the total is $80$ ($16$ values) and with $6$ dimes it is $90$ ($18$ values), so only choice (C) fits.

• Trying to count subsets or assuming the number of values depends on the coin arrangement, rather than realizing every multiple of $5$ up to the total appears.
• Forgetting that the total counts as a value, or including $0$ (the empty combination), which shifts the count by one.