2010 AMC 8 Problems/Problem 7

Revision as of 13:54, 21 October 2017 by Ancientwarrior (talk | contribs) (Solution)

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 $\boxed{B(10)}$

See Also

2010 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 6
Followed by
Problem 8
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AJHSME/AMC 8 Problems and Solutions

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