AoPSWiki
Want to learn how to tackle those tough AMC/AIME/Olympiad counting and probability problems? Check out Art of Problem Solving's Intermediate Counting & Probability by David Patrick.
Personal tools

1998 AIME Problems/Problem 13

From AoPSWiki

Problem

If \{a_1,a_2,a_3,\ldots,a_n\} 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 i^2 = - 1. Let S_n be the sum of the complex power sums of all nonempty subsets of \displaystyle \{1,2,\ldots,n\}. Given that S_8 = - 176 - 64i and \displaystyle  S_9 = p + qi, were p and q are integers, find |p| + |q|.

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 9 in it is equal to the number of subsets without a 9. To easily see this, take all possible subsets of \displaystyle \{1,2,\ldots,8\}. Since the sets are ordered, a 9 must go at the end; hence we can just append a 9 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 9 to a subset doesn't change anything about that subset's complex power sum besides adding an additional term, we have that S_9 = 2S_8 + T_9, where T_9 refers to the sum of all of the 9i^x.

It a subset of size 1 has a 9, then its power sum must be 9i, and there is only 1 of these such subsets. There are {8\choose1} with 9\cdot i^2, {8\choose2} with 9\cdot i^3, 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 9i \cdot (1+i)^8. We can use De Moivre's Theorem to calculate the power: (\sqrt{2})^8\cos{8\cdot45} = 16. Hence T_9 = 16\cdot9i = 144i, 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
Want to learn how to tackle those tough AMC/AIME/Olympiad counting and probability problems? Check out Art of Problem Solving's Intermediate Counting & Probability by David Patrick.
© Copyright 2008 AoPS Incorporated. All Rights Reserved. • FoundationPrivacyContact Us