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

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

We add:

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

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


Solution 4

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

See also

1999 AHSME (ProblemsAnswer KeyResources)
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. AMC logo.png

Invalid username
Login to AoPS