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really hard one

 
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germanboy
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PostPosted: Mon Oct 24, 2005 7:14 pm    Post subject: really hard one Reply with quote

iam desparated. I think this one is even a hard one for the freaks here who know almost every trick. Please help me and explain why the next number fits. Thank you very much Very Happy


080 695 700
300 278 600
000 314 000

000 836 000
568 147 329
000 529 000

000 752 000
054 903 001
009 401 050
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Louise56



Joined: 21 Sep 2005
Posts: 94
Location: El Cajon, California USA

PostPosted: Mon Oct 24, 2005 8:20 pm    Post subject: Reply with quote

Hello Germanboy!

Do you see a 2/8 pair in r8c7 and r9c7? If you can see that, then you can eliminate 8's from the lower right 3x3 box except for r8c7 and r9c7. Now in row 7 there is only one 8 and that is in r7c1. Try that and write back if you need more help!
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germanboy
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PostPosted: Tue Oct 25, 2005 7:01 pm    Post subject: Reply with quote

don'T understad it Question
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David Bryant



Joined: 29 Jul 2005
Posts: 559
Location: Denver, Colorado

PostPosted: Tue Oct 25, 2005 9:47 pm    Post subject: Louise is right, but maybe too succinct Reply with quote

OK, I'll try to explain it in more detail.

The only numbers missing in row 8 are {2, 6, 7, 8}. The only numbers missing in row 9 are {2, 3, 6, 7, 8}.

Now look at column 7. It contains {3, 6, 7}. If we delete the members of this set from the two sets listed above we are left with {2, 8} in each case. So the only values that can fit in r8c7 & r9c7 are {2, 8}.

Since there are two values and two cells we can be certain that the values "2" and "8" cannot appear anywhere else in the bottom right 3x3 box. In particular neither of them can appear in r6c7, r6c8, or r6c9. But row 6 has to contain an "8" somewhere, and the only place left for it to fit into that row is at r8c1.

Does it make sense now? dcb Smile
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PostPosted: Wed Nov 02, 2005 2:47 am    Post subject: Reply with quote

germanboy wrote:
don'T understad it Question
Shocked
What helps me is a format to organize the possibilities. it helps me to write 367 in a box to represent that only a 3, 6, or 7 could be an option for that square. Then go thru your series of reductions and simply erase the 7 for example if your logic warrants it. I have figured all the logic tricks I think in these puzzles as I now can solve even the very hard versions in under an hour even with some distractions. I don't speak German as I am an American but I will help if you ask again. Look for (amo loves to bowl) for a response in a few days time as I am busy as a math teacher.
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