2008 AIME II Problems/Problem 8
Problem
Let . Find the smallest positive integer such that is an integer.
Solution
Solution 1
By the product-to-sum identities, we have that . Therefore, this reduces to a telescoping series:
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.
Solution 2
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.
See also
2008 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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