2000 AIME II Problems/Problem 15
Problem
Find the least positive integer such that
Solution
We apply the identity
The motivation for this identity arises from the need to decompose those fractions, possibly into telescoping series.
Thus our summation becomes
Since , the summation simply reduces to . Therefore, the answer is .
Solution 2(Not Rigorous)
We can make an approximation by observing the following points: Each of the terms is greater than 1, so the overall sum will be greater than 1. The average term is around the 60's which gives 4/3. There are 45 terms so the approximate sum is 60. Therefore the entire thing equals approximately . Recall that the approximation of sinx in radians is x if x is close to zero. In this case x is close to zero. Converting to radians we see that sin1 in degrees is about sin in radians, or is about because of the approximation. What we want is apparently close to that so we make the guess that n is equal to 1 degree. -jasonhu4
See also
2000 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Last Question | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.