Difference between revisions of "2000 AIME I Problems/Problem 3"
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== Solution == | == Solution == | ||
| − | {{ | + | From the [[binomial theorem]], |
| + | |||
| + | <math>\binom{2000}{2}*b^{1998}a=\binom{2000}{3}b^{1997}a^2</math> | ||
| + | |||
| + | <math>b=666a</math> | ||
| + | |||
| + | Since a and b are positive relatively prime integers, a=1 and b=666. | ||
| + | |||
| + | <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}} | ||
Revision as of 08:52, 13 November 2007
Problem
In the expansion of 
where 
and 
are relatively prime positive integers, the coefficients of 
and 
are equal. Find 
.
Solution
From the binomial theorem,
Since a and b are positive relatively prime integers, a=1 and b=666.
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 | ||
