2010 AMC 8 Problem 7

After making every amount through $24$ cents, quarters can extend the range to $99$ cents.

First force the coins needed to make small amounts like $4$, $9$, and $24$ cents.

At least four pennies are necessary to pay $4$ cents. After those four pennies, the next coin must be worth at most $5$ cents so that $5$ cents can be paid; a nickel is the best choice. Those five coins total only $9$ cents. The next coin must be worth at most $10$ cents, and even a dime brings the total to only $19$ cents. One more coin worth at most $20$ cents is therefore necessary, so at least seven coins are needed before quarters can be used without leaving a gap.

With only nine coins total, at most two coins could remain after those required seven. The first seven can total at most $4+5+10+10=29$ cents, and two more quarters would bring the total to only $79$ cents, so nine coins cannot pay every amount through $99$ cents.

Ten coins do work: four pennies, one nickel, two dimes, and three quarters can make every amount from $1$ through $99$ cents. Therefore the minimum is $10$.

Therefore, the answer is B .