2010 AMC 8 Problems/Problem 7

Problem

Using only pennies, nickels, dimes, and quarters, what is the smallest number of coins Freddie would need so he could pay any amount of money less than a dollar?

$\textbf{(A)}\ 6 \qquad\textbf{(B)}\ 10\qquad\textbf{(C)}\ 15\qquad\textbf{(D)}\ 25\qquad\textbf{(E)}\ 99$

Solution

You have to have 4 pennies so you can have 4 cents, 9 cents, 14, etc. Then you start with 1 dime, 1 nickel, and 1 quarter. You go through 1-9, 10-19, etc. until 90-99, and along the way, you add coins with the most value. You end up with 4 pennies, 1 nickel, 2 dimes, and 3 quarters. $4+1+2+3=10$ , so our answer is $\boxed{\qquad\textbf{(B)}\ 10}$

2010 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
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All AJHSME/AMC 8 Problems and Solutions

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2010 AMC 8 Problems/Problem 7

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