Find the last digit of summation of the series:
(1)^99 + (2)^99 + (3)^99 + (4)^99 + ……… + (98)^99 + (99)^99

(In reply to Last two digits by Nick Hobson)

"By the remainder theorem (also known as the factor theorem), non-zero (a + b) divides (a^n + b^n)."

But if I take a=3, b=1, and n=2, a^n+b^n=10, while a+b=4 and 4 does not divide 10. It is true that a-b divides a^n-b^n.

Posted by Richard on 2004-01-30 22:09:15