# 1999 AHSME Problems/Problem 1

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## Problem $1 - 2 + 3 -4 + \cdots - 98 + 99 =$ $\mathrm{(A) \ -50 } \qquad \mathrm{(B) \ -49 } \qquad \mathrm{(C) \ 0 } \qquad \mathrm{(D) \ 49 } \qquad \mathrm{(E) \ 50 }$

## Solution

### Solution 1

If we group consecutive terms together, we get $(-1) + (-1) + \cdots + 99$, and since there are 49 pairs of terms the answer is $-49 + 99 = 50 \Rightarrow \mathrm{(E)}$.

### Solution 2

( Similar to Solution 1 ) If we rearranged the terms, we get $1+3-2+5-4 \cdots + 99-98$ then $1 + 1 + \cdots + 1$, and since there are 49 pairs of terms and the $1$ in the beginning the answer is $1+49 = 50 \Rightarrow \mathrm{(E)}$.

### Solution 3

Let $1 - 2 + 3 -4 + \cdots - 98 + 99 = S$.

Therefore, $S=99-98+97-\cdots -4+3-2+1$ $2S=100-100+100-100\cdots +100=100$ $S=50\Rightarrow \mathrm{(E)}$

### Solution 4

We proceed with addition, 1 -2 + 3 -4.... Once done we find $\mathrm{(E)}$

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 