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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)


Another way

Comment 3 of 3 

In the binomial expansion of (100 – 1)^{n}, all the terms are divisible by 100
except the final term, (1)^{n}.
When n is odd, (1)^{n} = 1, so the result is 99 (mod 100).
(When n is even, (1)^{n} = 1, so the result is 1 (mod 100)).

Posted by Harry
on 20120715 11:04:03 


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