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| 99 for odd powers (Posted on 2012-07-14) |
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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
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No Solution Yet
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Submitted by Ady TZIDON
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Rating: 3.6667 (3 votes)
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Another way
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Comment 3 of 3 | 
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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)).
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Posted by Harry
on 2012-07-15 11:04:03 |
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