2000 AIME I Problems/Problem 3

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Problem

In the expansion of $(ax + b)^{2000},$ where $a$ and $b$ are relatively prime positive integers, the coefficients of $x^{2}$ and $x^{3}$ are equal. Find $a + b$.

Solution

From the binomial theorem,

$\binom{2000}{2}*b^{1998}a=\binom{2000}{3}b^{1997}a^2$

$b=666a$

Since a and b are positive relatively prime integers, a=1 and b=666.

$a+b=\boxed{667}$

See also

2000 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 2
Followed by
Problem 4
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All AIME Problems and Solutions