2010 AMC 8 Problem 7
Attempt Problem 7 of the 2010 AMC 8 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2010 AMC 8 solutions, or check the answer key.
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7.
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?
Answer: B
Solution:
At least four pennies are necessary to pay cents. After those four pennies, the next coin must be worth at most cents so that cents can be paid; a nickel is the best choice. Those five coins total only cents. The next coin must be worth at most cents, and even a dime brings the total to only cents. One more coin worth at most 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 cents, and two more quarters would bring the total to only cents, so nine coins cannot pay every amount through cents.
Ten coins do work: four pennies, one nickel, two dimes, and three quarters can make every amount from through cents. Therefore the minimum is .
Therefore, the answer is B .
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