Track / Stage 4 / 220 of 340 #480 of 1964

Problem 480

AMC 12 late, AIME early

Combinatorics Difficulty 4.9 Find the answer

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

Your answer

A number or a short expression. Spacing, $ signs and \frac vs / are all fine.

Official solution

$\sum_{i=1}^{2013} m_{i}=(1+1)(1+99)(1-1)(1+0) \cdots(1+11)-1=-1 \text {. }$

Solve:
$\sum_{i=1}^{2013} m_{i}=(1+1)(1+99)(1-1)(1+0) \cdots(1+11)-1=-1 \text {. }$

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.