Problem:
1−2−3+4+5−6−7+8+9−10−11+12+13−⋯+1992+1993−1994−1995+1996=
Answer Choices:
A. −998
B. −1
C. 0
D. 1
E. 998
Solution:
Combining in groups of four yields
1−2−3+4=0,5−6−7+8=0,9−10−11+12=0
and so on. Since there are 499 groups of four in 1996, it follows that the sum is zero.
OR
Combining in pairs yields
so the sum is 0.
Answer: C​.
The problems on this page are the property of the MAA's American Mathematics Competitions
