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Topic: 1967 And A Digital Rearrangement Puzzle (Read 356 times) |
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K Sengupta
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1967 And A Digital Rearrangement Puzzle
« on: Jul 27th, 2007, 7:50am » |
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Let us consider a positive decimal number D. The number F is constituted by rearranging the digits of D.
Analytically determine whether it is possible that :
D + F = 99….9, where the digit 9 is repeated precisely 1967 times
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Barukh
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Re: 1967 And A Digital Rearrangement Puzzle
« Reply #1 on: Jul 27th, 2007, 11:04am » |
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No because of parity?
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jollytall
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Re: 1967 And A Digital Rearrangement Puzzle
« Reply #2 on: Jul 28th, 2007, 6:05am » |
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No. Barukh's solution a bit more in details.
If you take the last digits d+f=x9. Even if d=f=9 then x9 (x>0) cannot be reached, i.e. no carry on digit. So look at the next one, etc. and we can see that for every digit of the sum it must be true that d+f=9. Because 9 is odd, d and f must be different. But for every d+f=9 there is a d'+f'=f+d=9 rearranged. So the total number of 9-s in the total is always even, i.e. 1967 is not possible (netiher would be 2007 just to actualise the 40 years old question).
Any even number of digits is easily possible though 333666+666333=999999.
An odd number of even digits is also possible: 123+321=444.
Also odd number of odd digits can be done if <9, using carry digits, e.g. 27827+27728=55555.
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