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Karen's Solution


Karen's solution for part a is correct, but for part b I don't agree. There was nothing that said he threw in the lowest he could, although that would be smart. He could throw in either a nickel or a dime for anything 5 cents and under, thus there are two ways to pay. Then for anything 10 and under there are three, he could throw nickel nickel, nickel dime, or just a dime. Other than that I agree with Karen. ---Kristen 21:39, 4 March 2009 (UTC)

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Abstract algebra continues the conceptual developments of linear algebra, on an even grander scale.

Dr. Paul Garrett