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

It must be zero, since whatever result is achieved from the summation of the first ten members (actually it is 5 but that is irrelevant) is then repeated nine more times so we have summation of this digit 10 times .

Remark(NOT NEEDED TO SOLVE THE PUZZLE): the last digit of Nth power of any number is equal to the last digit of Kth power of said number iff(=if and only if) M mod 4= N mod 4 i.e. consider the 3th power instead of the 99th.

ady
Edited on January 23, 2004, 10:29 am

Posted by Ady TZIDON on 2004-01-23 09:57:59