2004 AMC 10A Problem 14

With $n$ coins the total value is $20n$ cents. Adding a quarter gives $$20n + 25 = 21(n + 1),$$ so $n = 4.$ If there were at most two quarters, the other coins would be worth at most $10$ cents each, so four coins would total at most $2(25)+2(10)=70$ cents. Thus there must be three quarters, leaving $5$ cents for the fourth coin. The purse contains three quarters and one nickel, so it has $0$ dimes. Thus, the correct answer is A.