Problem:
For all positive integers n, let f(n)=log2002n2. Let
N=f(11)+f(13)+f(14)
Which of the following relations is true?
Answer Choices:
A. N>1
B. N=1
C. 1<N<2
D. N=2
E. N>2
Solution:
We have
N=log2002112+log2002132+log2002142=log2002112⋅132⋅142=log2002(11⋅13⋅1
Simplifying gives
N=log2002(11⋅13⋅14)2=log200220022=2
The problems on this page are the property of the MAA's American Mathematics Competitions