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

(1)^99+(2)^99+....+(98)^99+(99)^99
=1+(((2)^100)/2)+(((3)^100)/3)+....+(((98)^100)/98)+(((99)^100)/99)
=1+last digit(2/2)+last digit(3/3)+last digit(4/4)+....+last digit(99/99)
=1+1+1.....99 times(last digit)
=9.
so the last digit of the given series is 9.

Posted by Purna on 2004-01-24 05:08:27