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Topic: Multiple with all odd digits (Read 819 times) |
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Aryabhatta
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Multiple with all odd digits
« on: Mar 11th, 2010, 6:18pm » |
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For any natural number n, there is an n digit multiple of 5n such that each of the n digits is odd (working in base-10, of course).
Prove.
E.g:
54*15 = 9375, a 4-digit number with all odd digits.
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towr
wu::riddles Moderator
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Re: Multiple with all odd digits
« Reply #1 on: Mar 12th, 2010, 2:18am » |
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Suppose we have an n-1 all-odd-digit number 5n-1*k, then we can try to put an odd digit in front of it, such that 5n-1*k + 5n-1*2n-1*m = 5n*l (where m  {1,3,5,7,9} is the new front digit)
So given a k, we need to find an m that satisfies k + 2n-1*m = 5*l. Because 5 and 2 are coprime, we can always find an m = -k/2n-1 (mod 5) = -k*3n-1 (mod 5). This gives us an 0 <= m < 5, but if m is even we can just add 5 to get an odd 0 < m < 10.
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| « Last Edit: Mar 12th, 2010, 2:29am by towr » |
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Wikipedia, Google, Mathworld, Integer sequence DB
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Aryabhatta
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Re: Multiple with all odd digits
« Reply #2 on: Mar 12th, 2010, 8:54am » |
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Correct! Well done.
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