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: solution by Charlie)

THE RESULT IS STILL ZERO AND ALL I WROTE IS CORRECT EXCEPT THE CYCLE BEING FOUR AND NOT FIVE i.e. 99=3 mod 4.
being a newcomer to the site I do not know how to edit my solution and so I post this comment

ady

Posted by Ady TZIDON on 2004-01-23 10:21:36