2008 AIME II Problems/Problem 8
Let . Find the smallest positive integer such that is an integer.
Thus, we need to be an integer; this can be only , which occur when is an integer. Thus . It easily follows that is the smallest such integer.
We proceed with complex trigonometry. We know that for all , we have and for some complex number . Similarly, we have and . Thus, we have
which clearly telescopes! Since the outside the brackets cancels with the inside, we see that the sum up to terms is
This expression takes on an integer value iff is an integer; that is, . Clearly, , implying that . Since we want the smallest possible value of , we see that we must have . If , then we have , which is clearly not divisible by . However, if , then , so our answer is .
It should be noted that the product-to-sum rules follow directly from complex trigonometry, so this solution is essentially equivalent to the solution above.
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