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 and using the well-known fact <math>\log(a_{}^{}b)=\log a + \log b</math> (valid if <math>a_{}^{},b_{}^{}>0</math>), we have | + | Let <math>0<x_{}^{}<1</math>. Then we may write <math>A_{k}^{}={N\choose k}x^{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 using the well-known fact <math>\log(a_{}^{}b)=\log a + \log b</math> (valid if <math>a_{}^{},b_{}^{}>0</math>), we have |
<math> | <math> | ||
Revision as of 20:44, 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 using the well-known fact 
(valid if 
), we have
![$\log(A_{k})=\log\left[\frac{(N-k+1)!}{k!}x^{k}\right]=\sum_{j=1}^{k}\log\left[\frac{(N-j+1)x}{j}\right]\, .$](http://latex.artofproblemsolving.com/4/c/4/4c439debe68426a1311aa27c6dda653aef3ac535.png)
Now, 
keeps increasing with as long as the arguments 
in each of the terms (recall that 
if 
). Therefore, the integer that we are looking for must satisfy 
, where 
denotes the greatest integer smaller than or equal to 
.
In summary, substituting 
and 
we finally find that 
.
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 | ||
