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99 for odd powers (
Posted on 20120714
)
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: (
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Primaryschool 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 20120715 01:12:05
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