Difference between revisions of "1991 AIME Problems/Problem 3"
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Cross multiplying gives: | Cross multiplying gives: | ||
− | <math>5k+ | + | <math>5k+5>1000-k \Rightarrow k>165.8</math> |
− | Therefore, since this identity holds for all values of <math>k> | + | Therefore, since this identity holds for all values of <math>k>165.8</math>, the largest possible value of <math>k</math> is <math>\boxed{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 22:07, 9 December 2011
Contents
[hide]Problem
Expanding by the binomial theorem and doing no further manipulation gives
where for . For which is the largest?
Solution
Solution 1
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 , where denotes the greatest integer less than or equal to .
In summary, substituting and we finally find that .
Solution 2
We know that once we have found the largest value of , all values after are less than . Therefore, we are looking for the largest possible value such that:
Dividing by gives:
.
We can express these binomial coefficients as factorials.
We note that the can cancel. Also, . Similarly, .
Canceling these terms yields,
Cross multiplying gives:
Therefore, since this identity holds for all values of , the largest possible value of is .
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 |