1999 AHSME Problems/Problem 1

Problem

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

$\mathrm{\textbf{(A)} \ -50 } \qquad \mathrm{\textbf{(B)} \ -49 } \qquad \mathrm{\textbf{(C)} \ 0 } \qquad \mathrm{\textbf{(D)} \ 49 } \qquad \mathrm{\textbf{(E)} \ 50 }$

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 $-49 + 99 = 50 \Rightarrow \mathrm{\textbf{(E)}}$ .

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 $1+49 = 50 \Rightarrow \mathrm{\textbf(E)}$ .

Solution 3

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

Therefore, $S=99-98+97-\cdots -4+3-2+1$

We add:

$2S=100-100+100-100\cdots +100=100$

$S=50\Rightarrow \mathrm{\textbf{(E)}}$

Solution 4

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

1999 AHSME (Problems • Answer Key • Resources)
Preceded by First question	Followed by Problem 2
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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

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