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Difference in Numbers of Solutions

Source: AIME 2009II Problem 9

April 2, 2009
AMCAIME

Problem Statement

Let m m be the number of solutions in positive integers to the equation 4x\plus{}3y\plus{}2z\equal{}2009, and let n n be the number of solutions in positive integers to the equation 4x\plus{}3y\plus{}2z\equal{}2000. Find the remainder when m\minus{}n is divided by 1000 1000.
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