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

(In reply to re: Last two digits by Richard)

"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."

Thanks Richard, you're right, I was a little hasty there! I meant to say that (a + b) divides (a^n + b^n), for *odd* n. This does follow from the remainder theorem, and I've edited my post accordingly.
Edited on January 31, 2004, 7:59 am

Posted by Nick Hobson on 2004-01-31 07:57:31