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    
99^(2n+1)=99*(99*99)*(99*99)*(99*99)...=99^(1+2+2+2...) 
It is a given that 99^2=9801      
Since the last 2 digits of 99^2 are 01, and 01*99=99, the last 2 digits under these successive multiplications by 99^2 remain unchanged i.e. 99.

Posted by broll on 2012-07-15 01:12:05