Difference between revisions of "1991 AIME Problems/Problem 3"
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− | 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>\frac{(N- | + | 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>\frac{(N-k+1)x}{k}>1</math>. Hence, <math>k<\frac{(N+1)x}{1+x}</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 20:21, 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
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 . Hence, .
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 |