There are 2 different solutions. Can you find them?

Use all of the digits 1 - 9 to make this multiplication sum correct.
There are 2 different solutions. Can you find them?
14 Comments
Darlene Pantaleo
3/1/2014 08:26:00 am
Solved it! Should I post the answer here? Once I do, no one else has the challenge.
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mehtamatics
4/1/2014 03:38:56 am
1738 and 1963
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mehtamatics
4/1/2014 03:40:11 am
Blast!, I should have read Darlene's email and done likewise!
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Paul
4/1/2014 06:53:04 am
Well done, and do not worry, Mehtamatics.
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Darlene Pantaleo
4/1/2014 09:14:16 am
No worries, mehtamatics, there is still another answer to find!
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Darlene P
5/1/2014 04:39:23 pm
Hahaha! I just compared your numbers to mine and discovered that you gave both answers! I just assumed you were giving the same numbers that I did--2 sets of numbers for one equation. 4/1/2014 05:36:31 am
Great puzzles! I've bookmarked you so I can play each month AND pass them along to students, peers, and friends! Keep them coming!
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Paul
4/1/2014 06:50:20 am
Many thanks Norma
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B R Sitaram
12/1/2014 03:41:35 am
I would be interested in knowing how the solutions are obtained. Brute force search will of course yield the solutions, is there any analysis?
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Paul
12/1/2014 06:39:40 am
There was a nice solution posted on my LinkedIn page, which used modulo arithmetic. I have asked the solver to post on this page, so watch this space!
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Darlene P
12/1/2014 08:00:11 am
I used the old-fashioned brute force search--trying different combos until the results yielded themselves. I like doing math, so even if I had a faster, more efficient way of solving this, I probably would have done it the way I did it anyhow.
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Richard Pennington
16/1/2014 05:41:16 am
For the problem: a b c d * 4 = e f g h [where a b c d and e f g h re 4-digit numbers], with {a, b, c, d, e, f, g, h, 4, 0} all distinct:
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Jason
9/4/2014 09:33:21 am
This is a really great method Richard, where did you get the idea of using mod 9(but not mod 1-8?)
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Paul
16/1/2014 09:10:20 am
Many thanks Richard
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