Maths Olympiad Prep

Track / Stage 4 / 232 of 340 #492 of 1964

Problem 492

AMC 12 late, AIME early
Combinatorics Difficulty 4.9 Find the answer

Let the set M={1,99,−1,0,25,−36,−91,19,−2,11}M=\{1,99,-1,0,25,-36,-91,19,-2,11\}, and denote all non-empty subsets of MM as Mi,i=1,2,⋯M_{i}, i=1,2, \cdots, 2013. The product of all elements in each MiM_{i} is mim_{i}. Then ∑i=12013mi=\sum_{i=1}^{2013} m_{i}= \qquad .

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

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Official solution

∑i=12013mi=(1+1)(1+99)(1−1)(1+0)⋯(1+11)−1=−1.  \sum_{i=1}^{2013} m_{i}=(1+1)(1+99)(1-1)(1+0) \cdots(1+11)-1=-1 \text {. }

Solve:
∑i=12013mi=(1+1)(1+99)(1−1)(1+0)⋯(1+11)−1=−1.  \sum_{i=1}^{2013} m_{i}=(1+1)(1+99)(1-1)(1+0) \cdots(1+11)-1=-1 \text {. }

Source: NuminaMath-1.5, licensed Apache-2.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.