2009 AMC 10A Problem 2

Below is the professionally curated solution for Problem 2 of the 2009 AMC 10A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2009 AMC 10A solutions, or check the answer key.

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Concepts:money divisibility bounding to limit cases

Difficulty rating: 940

Four coins are picked out of a piggy bank that contains a collection of pennies, nickels, dimes, and quarters. Which of the following could not be the total value of the four coins, in cents?

$15$

$25$

$35$

$45$

$55$

Solution:

To get a multiple of $5$ cents, the number of pennies must be a multiple of $5.$ With only four coins, that means using no pennies, but then the four coins are each worth at least $5$ cents, for a total of at least $20$ cents.

So $15$ cents cannot be made. The others can: $25 = 10 + 3(5),$ $35 = 3(10) + 5,$ $45 = 25 + 10 + 2(5),$ and $55 = 25 + 3(10).$

Thus, the correct answer is A.

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