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2009 AIME II Question 9

AIME · 2009 AIME II

Question 9

Let $m$ be the number of solutions in positive integers to the equation $4x + 3y + 2z = 2009,$ and let $n$ be the number of solutions in positive integers to the equation $4x + 3y + 2z = 2000.$ Find the remainder when $m - n$ is divided by $1000.$