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)

... I left out 3:

A number ending in 3 will have powers with ending digits that cycle 3, 9, 7, 1 in a cycle of 4. As 99 is congruent to 3 mod 4, 3^99 ends in 7, the third in the cycle.

That makes one decade add up to 5 mod 10.

Posted by Charlie on 2004-01-23 10:22:38