AIME 1998 · 第 13 题

AIME 1998 — Problem 13

专题: Contest Math
难度: L4
来源: AIME

题目详情

Problem

If $\{a_1,a_2,a_3,\ldots,a_n\}$ is a set of real numbers, indexed so that $a_1 < a_2 < a_3 < \cdots < a_n,$ its complex power sum is defined to be $a_1i + a_2i^2+ a_3i^3 + \cdots + a_ni^n,$ where $i^2 = - 1.$ Let $S_n$ be the sum of the complex power sums of all nonempty subsets of $\{1,2,\ldots,n\}.$ Given that $S_8 = - 176 - 64i$ and $S_9 = p + qi,$ where $p$ and $q$ are integers, find $\left\lvert p\right\rvert + \left\lvert q\right\rvert.$

解析

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 $\{1,2,\ldots,8\}$ . 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 $9i^x$ .

If 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 $T_9 =\sum_{k=0}^{8} 9{8\choose{k}}i^{k+1}$ . 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, $\left\lvert p\right\rvert + \left\lvert q\right\rvert = \left\lvert -352\right\rvert + \left\lvert 16\right\rvert = \boxed{368}$ .

Video Solution

https://youtu.be/cpzXYDvkZIg?si=6pv_9cwfpW0bhBId

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