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Well, the fact that you found only one solution doesn't mean there aren't more. That doesn't mean I don't agree with you though. There's only one solution. It rests mostly on the fact that P must equal 8. That established, 8 plus 2D must equal one of the other remaining digits excluding the carry it creates. All that remain at that point are 0, 1, 5, 7, and 9. Even trial-and-error is fast enough to show that D then must equal 1 and H equal 0. The next line up does the same thing only with J = 9 and Y = 5, since those are the only tho remaining digits that can be made to work around the other numbers (which sum up to 6 [3 + 2 + 1carry]).
..ok, so maybe this isn't perfectly clear after all, but there's no need to go overboard explaining it, so ill stop with a simple: Yes, Berto, there's only one solution. :p
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