2015 AMC 10A Problem 13

Let the number of $5$-cent coins be $x$ and the number of $10$-cent coins be $12 - x.$ If $x=0,$ the only possible values are $10,20,\ldots,120$ cents, giving just $12$ values. Hence $x>0.$ Having at least one $5$-cent coin then makes every multiple of $5$ from $5$ through the total value $$5x+10(12-x)=120-5x$$ obtainable. There are $24 - x$ such multiples of $5,$ which means that $x = 7$ to get $17$ possible different values. The number of $10$-cent coins is therefore $12 - 7 = 5.$ Thus, C is the correct answer.