Difference between revisions of "1991 AIME Problems/Problem 3"
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== Solution == | == Solution == | ||
− | Let <math>0<x_{}^{}<1</math>. Then we may write <math>A_{k}^{}=\frac{N!}{k!(N-k)!}x^{k}=\frac{(N-k+1)!}{k!}x^{k}</math>. Taking logarithms in both sides of this last equation, we have | + | Let <math>0<x_{}^{}<1</math>. Then we may write <math>A_{k}^{}=\frac{N!}{k!(N-k)!}x^{k}=\frac{(N-k+1)!}{k!}x^{k}</math>. Taking logarithms in both sides of this last equation, and recalling that <math>\log(a_{}^{}b)=\log a + \log b</math> (valid if <math>a_{}^{},b_{}^{}>0</math>), we have |
<math> | <math> |
Revision as of 20:02, 20 April 2007
Problem
Expanding by the binomial theorem and doing no further manipulation gives
where for . For which is the largest?
Solution
Let . Then we may write . Taking logarithms in both sides of this last equation, and recalling that (valid if ), we have
See also
1991 AIME (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 |