Difference between revisions of "2000 AIME I Problems/Problem 3"
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== Problem == | == Problem == | ||
| − | In the expansion of <math>(ax + b)^{2000},</math> where <math>a</math> and <math>b</math> are relatively prime positive integers, the | + | In the expansion of <math>(ax + b)^{2000},</math> where <math>a</math> and <math>b</math> are [[relatively prime]] positive integers, the [[coefficient]]s of <math>x^{2}</math> and <math>x^{3}</math> are equal. Find <math>a + b</math>. |
== Solution == | == Solution == | ||
| − | + | Using the [[binomial theorem]], <math>\binom{2000}{2} b^{1998}a = \binom{2000}{3}b^{1997}a^2 \Longrightarrow b=666a</math>. | |
| − | <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>. |
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| − | <math>b | ||
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| − | |||
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| − | <math>a+b=\boxed{667}</math> | ||
== See also == | == See also == | ||
{{AIME box|year=2000|n=I|num-b=2|num-a=4}} | {{AIME box|year=2000|n=I|num-b=2|num-a=4}} | ||
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| + | [[Category:Intermediate Algebra Problems]] | ||
Revision as of 16:04, 31 December 2007
Problem
In the expansion of 
where 
and 
are relatively prime positive integers, the coefficients of 
and 
are equal. Find 
.
Solution
Using the binomial theorem, 
.
Since and are positive relatively prime integers, 
and 
, and 
.
See also
| 2000 AIME I (Problems • Answer Key • Resources) | ||
| 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 | ||
