perplexus.info :: Numbers : 99 for odd powers

99 for odd powers (Posted on 2012-07-14)

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

No Solution Yet Submitted by Ady TZIDON

Rating: 3.6667 (3 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)

Primary-school proof.

| Comment 2 of 3 |

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

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