Consider the following results:
99^1 = 99
99^2 = 9801
99^3 = 970299
99^4 = 96059601
99^5 = 9509900499

Prove that 99^n ends in 99 for all odd n.

Source: mathschallenges 2003

99^1=99, so it is true for 1. Now, suppose 99^(2n-1) ends in 99. Let 99^(2n-1)=100x+99. Then, 99^2n=9900x+9801=100(99x+98)+1, so 99^2n ends in 01. That implies that 99^(2n+1)=9900(99x+98)+99=100(9801x+9702)+99, so 99^(2n+1) ends in 99. Therefore, by induction, it is true for all odd numbers.

Posted by Math Man on 2012-07-14 20:50:13