2015 AMC 10A Problem 13

Measure all money in nickels, so a $5$-cent coin is $1$ unit and a $10$-cent coin is $2$ units.

1. Try one choice to see the pattern. Say Claudia has $5$ dimes and $7$ nickels. Her total is $5 \times 2 + 7 = 17$ units ($85$ cents).
2. Which amounts can she make? Here is how to make each one:
- $1$ unit: one nickel. $2$ units: one dime. $3$ units: a dime and a nickel. $4$ units: two dimes.
- Keep going: an even amount up to $10$ uses only dimes, and an odd amount up to $11$ uses dimes plus one nickel.
- Past $11$ units all $5$ dimes are used, and the extra nickels fill in $12, 13, \ldots, 17$.

So she can make every amount from $1$ to $17$ units: $17$ different values. That matches.
3. The same reasoning works for any number of dimes $n$, as long as there is at least one nickel: every amount from $1$ unit up to the total can be made. The total is $2n + (12 - n) = 12 + n$ units, so she can make exactly $12 + n$ values.
4. Set $12 + n = 17$, so $n = 5$.
5. (With no nickels at all, the $12$ dimes make only $12$ values, so that case does not give $17$.)

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

With coins worth $1$ and $2$ units and at least one $1$, there are no gaps: the $2$'s build even amounts, a single $1$ makes the odd ones, and the rest of the $1$'s reach the top. So "how many values" just means "how big is the total". Spotting that there are no gaps turns a counting problem into one short equation.

Check nearby choices. With $4$ dimes and $8$ nickels the total is $80$ cents, giving $5, 10, \ldots, 80$: $16$ values. With $6$ dimes and $6$ nickels the total is $90$ cents: $18$ values. Only $5$ dimes gives $17$.