1999 AHSME Problems/Problem 1

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

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

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

We add:

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

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

1999 AHSME (Problems • Answer Key • Resources)
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1999 AHSME Problems/Problem 1

Contents

Problem

Solution

Solution 1

Solution 2

See also