2000 AIME II Problems/Problem 15
Problem
Find the least positive integer such that
Solution 1
We apply the identity
The motivation for this identity arises from the need to decompose those fractions, possibly into telescoping.
Thus our summation becomes
Since , the summation simply reduces to . Therefore, the answer is .
Solution 2
We can make an approximation by observing the following points:
The average term is around the 60's which gives .
There are 45 terms, so the approximate sum is 60.
Therefore, the entire thing equals approximately .
Recall that the approximation of in radians is x if x is close to zero. In this case x is close to zero. Converting to radians we see that 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. Basically, it boils down to the approximation of in degrees, convert to radians and use the small angle approximation .
Solution 3 (Alternate Finish)
Let S be the sum of the sequence. We begin the same as in Solution 1 to get . Observe that this "almost telescopes," if only we had some extra terms. Consider adding the sequence . By the identity , this sequence is equal to the original one, simply written backwards. By the same logic as before, we may rewrite this sequence as , and when we add the two sequences, they telescope to give . Hence, , and our angle is .
~keeper1098
See also
2000 AIME II (Problems • Answer Key • Resources) | ||
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