Difference between revisions of "2000 AIME I Problems/Problem 3"

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== Solution ==
 
== Solution ==
Using the [[binomial theorem]], <math>\binom{2000}{2} b^{1998}a^2 = \binom{2000}{3}b^{1997}a^3 \Longrightarrow b=666a</math>.
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Using the [[binomial theorem]], <math>\binom{2000}{1998} b^{1998}a^2 = \binom{2000}{1997}b^{1997}a^3 \Longrightarrow b=666a</math>.
  
 
Since <math>a</math> and <math>b</math> are positive relatively prime integers, <math>a=1</math> and <math>b=666</math>, and <math>a+b=\boxed{667}</math>.
 
Since <math>a</math> and <math>b</math> are positive relatively prime integers, <math>a=1</math> and <math>b=666</math>, and <math>a+b=\boxed{667}</math>.

Revision as of 18:41, 11 June 2016

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

Using the binomial theorem, $\binom{2000}{1998} b^{1998}a^2 = \binom{2000}{1997}b^{1997}a^3 \Longrightarrow b=666a$.

Since $a$ and $b$ are positive relatively prime integers, $a=1$ and $b=666$, and $a+b=\boxed{667}$.

See also

2000 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 2
Followed by
Problem 4
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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