1998 AIME Problem 13

Attempt Problem 13 of the 1998 AIME below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1998 AIME solutions, or check the answer key.

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13.

If {a1,a2,a3,,an}\{a_1, a_2, a_3, \ldots, a_n\} is a set of real numbers, indexed so that a1<a2<a3<<an,a_1 \lt a_2 \lt a_3 \lt \cdots \lt a_n, its complex power sum is defined to be a1i+a2i2+a3i3++anin,a_1 i + a_2 i^2 + a_3 i^3 + \cdots + a_n i^n, where i2=1.i^2 = -1. Let SnS_n be the sum of the complex power sums of all nonempty subsets of {1,2,,n}.\{1, 2, \ldots, n\}. Given that S8=17664iS_8 = -176 - 64i and S9=p+qi,S_9 = p + qi, where pp and qq are integers, find p+q.|p| + |q|.

Answer: 368
Concepts:complex numberrecursionbinomial theorem
Difficulty rating: 2920
Solution:

Split the nonempty subsets of {1,,9}\{1, \ldots, 9\} by whether they contain 9.9. Those without 99 contribute S8.S_8. A subset containing 99 is T{9}T \cup \{9\} for a (possibly empty) T{1,,8},T \subseteq \{1, \ldots, 8\}, and since 99 is its largest element, its complex power sum is the complex power sum of TT plus 9iT+1.9i^{|T| + 1}. Summing over all TT gives another S8S_8 plus k=08(8k)9ik+1=9i(1+i)8.\sum_{k=0}^{8} \binom{8}{k}\, 9\, i^{k+1} = 9i\,(1 + i)^8.

Since (1+i)2=2i,(1 + i)^2 = 2i, we get (1+i)8=(2i)4=16,(1 + i)^8 = (2i)^4 = 16, so S9=2S8+144i=2(17664i)+144i=352+16i. \begin{aligned} S_9 &= 2S_8 + 144i \\ &= 2(-176 - 64i) + 144i \\ &= -352 + 16i. \end{aligned}

Therefore p+q=352+16=368.|p| + |q| = 352 + 16 = 368.

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