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 Post subject: A Kakuro Problem
PostPosted: Wed Mar 19, 2014 4:41 am 
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Joined: Tue Apr 22, 2008 2:07 am
Posts: 104
Sorry to bother with this one, but after a long while looking, I cannot see the next step. Some elimination of combinations have been made, but still no real breakthrough. I must be missing something straightforward, but I cannot find it.

Thanks,
George

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 Post subject: Re: A Kakuro Problem
PostPosted: Sat Mar 22, 2014 5:43 am 
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Joined: Mon Aug 20, 2012 5:03 am
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In the 6-digit vertical 28 in column 8, you will note (I am assuming your pencilmarks are valid so far) that three of the digits must be 4 or less. There cannot be a fourth digit that is also 4 or less, since the four would then add up to 10, forcing the remaining two to add up to 18, which is impossible.

Thus r4c8, r6c8, r7c8 must all be 5 or more. This, in turn, prevents r7c6 from being 5, which in turn prevents r4c6 from being 3 (since the two must add up to 8).

This means that r4c6, r4c8 must be 5 and 6 (not necessarily respectively), which forces r4c7 to be 3 rather than 4. That lets you solve the vertical 10 in column 7, which in turn solves at least one digit of the vertical 28. And you're on your way.

Bill Smythe


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 Post subject: Re: A Kakuro Problem
PostPosted: Mon Apr 14, 2014 3:25 am 
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Hi Bill,

My apologies for failing to say thank you for the help with the puzzle. I looked at your message again this evening and saw that I had not replied.

Regards,
George


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