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1998 AIME Problems/Problem 13

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Problem

If is a set of real numbers, indexed so that \displaystyle a_1 < a_2 < a_3 < \displaystyle  \cdots < a_n, its complex power sum is defined to be \displaystyle a_1i + a_2i^2 \displaystyle + a_3i^3 + \cdots + a_ni^n, where Let be the sum of the complex power sums of all nonempty subsets of \displaystyle \{1,2,\ldots,n\}. Given that and were and are integers, find

Solution

We note that the number of subsets (for now, including the empty subset, which we will just define to have a power sum of zero) with in it is equal to the number of subsets without a . To easily see this, take all possible subsets of . Since the sets are ordered, a must go at the end; hence we can just append a to any of those subsets to get a new one.

Now that we have drawn that bijection, we can calculate the complex power sum recursively. Since appending a to a subset doesn't change anything about that subset's complex power sum besides adding an additional term, we have that , where refers to the sum of all of the .

It a subset of size 1 has a 9, then its power sum must be , and there is only of these such subsets. There are with , with , and so forth. So T_9 = \displaystyle\sum_{k=0}^{8} 9{8\choose{k}}i^{k+1}. This is exactly the binomial expansion of . We can use De Moivre's Theorem to calculate the power: (\sqrt{2})^8\cos{8\cdot45} = 16. Hence , and S_9 = 2S_8 + 144i = 2(-176 -64i) + 144i = -352 + 16i. Thus, |p| + |q| = |-352| + |16| = 368.

See also

1998 AIME (ProblemsResources)
Preceded by
Problem 12
Followed by
Problem 14
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
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