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 digit by Purna)

How come you get from last_digit((2^100)/2 to last_digit(2/2)?? The last digit of 2^100 is 6, so the result should be 3, not 1; the whole reasoning is wrong.

Posted by e.g. on 2004-01-27 09:16:50