Problem:
What is the remainder when 30+31+32+⋯+32009 is divided by 8?
Answer Choices:
A. 0
B. 1
C. 2
D. 4
E. 6
Solution:
The sum of any four consecutive powers of 3 is divisible by 30+31+32+33=40 and hence is divisible by 8. Therefore
(32+33+34+35)+⋯+(32006+32007+32008+32009)
is divisible by 8. So the required remainder is 30+31=(D)4​.
The problems on this page are the property of the MAA's American Mathematics Competitions