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 | + | 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 |
<math> | <math> | ||
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</math> | </math> | ||
− | Now, <math>\log(A_{k}^{})</math> keeps increasing with <math>k_{}^{}</math> as long as the arguments <math>\frac{(N-j+1)x}{j}>1</math> in each of the terms (recall that <math>\log y_{}^{} <0</math> if <math>0<y_{}^{}<1</math>). Therefore, the integer <math>k_{}^{}</math> that we are looking for must satisfy <math>k<\frac{(N+1)x}{1+x}</math>. | + | Now, <math>\log(A_{k}^{})</math> keeps increasing with <math>k_{}^{}</math> as long as the arguments <math>\frac{(N-j+1)x}{j}>1</math> in each of the terms (recall that <math>\log y_{}^{} <0</math> if <math>0<y_{}^{}<1</math>). Therefore, the integer <math>k_{}^{}</math> that we are looking for must satisfy <math>k<\frac{(N+1)x}{1+x}</math>. |
+ | |||
+ | In summary, substituting <math>N_{}^{}=1000</math> and <math>x_{}^{}=0.2</math> we finally find that <math>k_{}^{}=166</math>. | ||
== See also == | == See also == | ||
{{AIME box|year=1991|num-b=2|num-a=4}} | {{AIME box|year=1991|num-b=2|num-a=4}} |
Revision as of 21:26, 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
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
.
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 |