Problem:
What is the greatest power of 2 that is a factor of 101002−4501?
Answer Choices:
A. 21002
B. 21003
C. 21004
D. 21005
E. 2^
Solution:
Note that
101002−4501=21002⋅51002−21002=21002(51002−1)=21002(5501−1)(5501+1)=21002(5−1)(5500+5499+⋯+5+1)(5+1)(5500−5499+⋯−5+1)=21005(3)(5500+5499+⋯+5+1)(5500−5499+⋯−5+1)
Because each of the last two factors is a sum of an odd number of odd terms, they are both odd. The greatest power of 2 is (D)21005.
The problems on this page are the property of the MAA's American Mathematics Competitions