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 12232011, 09:25 PM #1Senior Member
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 12232011, 09:39 PM #2
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Re: How would you solve this? Java coding for multiplication.
It cannot be solved because the righrmost digit in the fourth row has to be a zero.
kind regards,
Joscenosillicaphobia: the fear for an empty beer glass
 12232011, 09:41 PM #3Senior Member
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Re: How would you solve this? Java coding for multiplication.
What makes you think that?
I am sure it is a "can be solved" question.
 12232011, 09:59 PM #4Moderator
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Re: How would you solve this? Java coding for multiplication.
 12232011, 10:02 PM #5Senior Member
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 12232011, 10:08 PM #6
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Re: How would you solve this? Java coding for multiplication.
The last digit of the top two numbers have to be a 3 and a 5 (...3x..5 = ....5 (or ..5x.3 = ....5 is the only possibility), so the last digit of the result is 5 as well. Maybe consistent reasoning like this can help you to solve this puzzle. If there are too many alternatives I'd take the lazy approach (brute force)
kind regards,
Joscenosillicaphobia: the fear for an empty beer glass
 12232011, 10:11 PM #7Senior Member
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Re: How would you solve this? Java coding for multiplication.
I am trying to solve it with Java :)
Thanks though... :)
I have this idea of looping for all numbers in the first 2 rows and multiplying them and to check if they are all prime numbers(2,3,5,7)..
But can there be a better approach ?
 12232011, 10:15 PM #8Moderator
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Re: How would you solve this? Java coding for multiplication.
Totally slow answer adding nothing removed.
@OP: That's more or less what I would do.Last edited by pbrockway2; 12232011 at 10:17 PM.
 12232011, 10:30 PM #9
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Re: How would you solve this? Java coding for multiplication.
There are four different digits (2, 3, 5 and 7) so you should count in base 4. There are 4x4x4 possible first numbers and 4x4 possible second numbers. You can try them all after mapping 0 > 2, 1 > 3, 2 > 5 and 3 > 7 and see which numbers have the correct result.
kind regards,
Joscenosillicaphobia: the fear for an empty beer glass
 12232011, 10:34 PM #10Senior Member
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