Problem: The sum of the digits in base ten of (104n2+8+1)2, where n is a positive integer, is
Answer Choices:
A. 4
B. 4n
C. 2+2n
D. 4n2
E. n2+n+2
Solution:
Let k be any positive integer. Then 10k+1=100⋯01, and
100⋯0100⋯0100⋯01100⋯01100⋯011200⋯01
The sum of the digits is therefore 1+2+1=4.