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pete Guest
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Posted: Sun Dec 25, 2005 5:51 pm Post subject: Dec. 25 - very hard |
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I have the following filled in:
x15 349 6x7
x97 65x 1xx
xxx x1x xxx
xx1 265 xx4
524 xxx x61
6xx 1x4 2xx
x4x 52x xxx
xx2 48x 59x
358 976 412
I got stuck here, and asked for a hint. I got r6c9 = 5. Everything else fell into place after this. Could someone tell me why this cell had to be 5? |
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Guest
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Posted: Sun Dec 25, 2005 8:50 pm Post subject: |
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I got stuck right there also, what does r6c9=5 mean? |
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Charlie
Joined: 25 Dec 2005 Posts: 6 Location: Holyoke, MA
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Posted: Sun Dec 25, 2005 10:16 pm Post subject: |
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R6c9 = row #6, column #9
I think it gave the clue as r6c9=5 because r6c3 & r6r5 are narrowed down to being either 3 or 9 -- that being the case, you can eliminate any other 3's and 9's in the row. Eliminating 9 as a possibility in r6c9 leaves the only possible choice for 9 in column 9 as r3c9 which, in turn, leaves the only possible choice for 5 as being R6c9. Hmmmmm.... then why wasn't the clue provided as being 9 in row 3c9? Now I,m confused....
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volcano vaporizers
Last edited by Charlie on Mon Jan 31, 2011 9:36 pm; edited 1 time in total |
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Harold Guest
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Posted: Sun Jan 08, 2006 9:26 am Post subject: December 25 Very hard |
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Thanks for this. I got stuck at the same point as Pete, and also couldn't understand the hint for r6c9.
At least with Charlie's help I've completed the puzzle and can stop leaving it unsolved around in the bathroom!
H |
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eddieg
Joined: 12 Jan 2006 Posts: 47 Location: San Diego, CA USA
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Posted: Fri Jan 13, 2006 12:36 am Post subject: |
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Thanks for the tip. Not only did you help me complete the puzzle, you taught me a concept that I just wasn't grasping. |
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xyberc Guest
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Posted: Fri Jan 13, 2006 1:24 am Post subject: December 25 Very hard |
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5 must be in either r6c8 or r6c9
if 5 in r6c8
put 7 in r6c2 and then 8 in r6c9
put 5 in r3c9 and then 9 in r3c7
but then 9 cannot be in r4c7, r4c8, r5c7
therefore contradiction
hence 5 must be in r6c9 |
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