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(2): solution --- Gosh!!!.. by Charlie)

right you are CHARLIE
Actually it is only the 5 that counts within a decade since 5+t and 5-t complement each other e.g. 2^3=8 8^3=2 8+2=0 (all taken mod 10) so go 1&9 3&7 4&6.... only 5 and 0 left.
ady

Posted by Ady TZIDON on 2004-01-23 10:38:04