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Lulu
Joined: 21 Sep 2005 Posts: 11 Location: Manchester, England
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Posted: Wed Sep 28, 2005 9:49 pm Post subject: Sept 28th hard puzzle |
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I am stuck and neglecting husband and children, please can anyone help me get further with this one?
So far I have:-
8 1 9/7 3 2 9/7/5 9/6/5 4/5/6/9 5/4
9/7/3 9/7/4/3 6 9/7/5/1 4 8 9/5/1 1/5/9 2
5 2 4 9/1 6 9/5/1 8 7 3
7/6/3 8 2 7/5/1 5/3/1 4 6/5/1 1/3/5/6 9
9/7/6/3 9/7/6/3 5 2 8 9/7/1 4 3/1/6 7
4 9/7/3 1 6 5/3/1 9/7/5/1 2 3/5 8
1 5 8 4 9 3 7 2 6
2 9/7/4 9/7 8 1/5 6 3 9/5/4/1 5/4/1
9/6 9/6/4 3 5/1 7 2 9/5/1 8 5/4/1
Based on what I've read would I be correct in assuming in box 9 that I have a triplet of 541 so r9c7 is nr9?
Lulu |
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chobans
Joined: 21 Aug 2005 Posts: 39
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Posted: Wed Sep 28, 2005 11:39 pm Post subject: |
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Sudokus are fun... but don't neglect your family to play it! Just get them hooked on it as you are. Then you all can neglect one another. :D
r3c6 should be {1,9}. You have a 5 in there but that's not possible since you have a 5 at r3c1.
In box 5(r4c4-r6c6), 9 can ONLY go in column 6. So you can eliminate 9 from other cells in column 6. So r3c6 becomes 1.
But to answer your question about naked triple... no, that's not a naked triple because that's only true if 3 cells have ONLY numbers {1,4,5} combination in them. r8c8 also can be a 9 so the naked triple possibility is out the window. And it's not a hidden triple since 1 and 5 are also possible at r9c7. |
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Lulu
Joined: 21 Sep 2005 Posts: 11 Location: Manchester, England
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Posted: Fri Sep 30, 2005 11:40 am Post subject: |
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Thank you chobans your hints really helped and I've got it finished now.
Lulu |
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