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Topic: An impossible division (Read 778 times) |
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pcbouhid
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An impossible division
« on: Jan 19th, 2006, 4:03pm » |
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Prove that the division below where each letter represents a digit from 0-9 (different letters, different digits) has no answer (DEF divided by BC gives A, with remainder IJ, and GH = A*BC):
BC_| DEF |_ A
GH
--
IJ
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| « Last Edit: Jan 19th, 2006, 4:06pm by pcbouhid » |
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TimMann
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Re: An impossible division
« Reply #1 on: Aug 25th, 2006, 7:45pm » |
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I may be missing the "easy" solution to this one. I have a solution that's straightforward but tedious.
Basically, the way one usually does these puzzles is by accumulating inferences that either show some letter must be some number or some letter can't be some number. Eventually, by a process of elimination, you either get a solution or show there isn't one. It can be useful to make a table with the rows labelled by letters and the columns by numbers, then fill in an X each time you eliminate a possibility.
Here's the reasoning I went through until I hit a point where I'd proved there was no solution. Possibly this could be pruned down...
1) A obviously can't be 0. B, D, G, and I can't be 0 because they are leading digits. C can't be 0 because then H would also be 0. H can't be 0 because then F = J. J can't be 0 because then F = H. So either E or F must be 0.
2) Because DEF is the sum of two 2-digit numbers, DEF <= 198. So D = 1.
3) Since A * BC = GH, we know A * B < 10. With 0 and 1 eliminated as possible values for A and B, and because A != B, the only possible assignments for (A, B) are (2, 3), (3, 2), (2, 4), (4, 2). So in particular, either A or B must be 2.
4) The remainder IJ < BC, and I != B, so I < B. Since B <= 4, I < 4. We already know that I is not 0, 1, or 2, so I = 3. Then since I < B, B = 4. Then by lemma (3), A = 2.
5) Since A = 2 and B = 4, G is 8 or 9. But then if E = 0, we have a contradiction, since 10F - 8H or 10F - 9H must be less than 3J. So F = 0.
6) Since F = 0, H + J = 10. We can't have H = J = 5, so one of H, J must be less than 5 and the other greater. But we've already used up all the digits less than 5: F = 0, D = 1, A = 2, I = 3, and B = 4. So the puzzle has no solution.
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| « Last Edit: Aug 25th, 2006, 7:48pm by TimMann » |
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