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99 for odd powers (Posted on 2012-07-14) Difficulty: 2 of 5
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)

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Solution 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 2012-07-15 11:04:03
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