Problem:
1β2β3+4+5β6β7+8+9β10β11+12+13ββ―+1992+1993β1994β1995+1996=
Answer Choices:
A. β998
B. β1
C. 0
D. 1
E. 998
Solution:
Combining in groups of four yields
1β2β3+4=0,5β6β7+8=0,9β10β11+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: Cβ.
The problems on this page are the property of the MAA's American Mathematics Competitions