Problem:
1990−1980+1970−1960+…−20+10=
Answer Choices:
A. −990
B. −10
C. 990
D. 1000
E. 1990
Solution:
By grouping as shown below, there are 2199+1=100 groups of 10 for a sum of 1000:
[1990−1980]+[1970−1960]+…+[30−20]+10.
Answer: D.
The problems on this page are the property of the MAA's American Mathematics Competitions