Method for another Challenger sudoku

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Method for another Challenger sudoku

Postby JeJ » Wed Jan 12, 2011 2:40 am

I reviewed these numbers many times I hope I am not missing a simple clue.
So the original game is:
Code: Select all
. 1 .|2 . .|5 . .
5 . 2|. . 8|3 . .
. 8 .|. 5 .|. 6 1
-----------------
3 . 8|. 4 .|. . .
. . .|5 . 2|. . .
. . .|. 8 .|1 . 9
------------------
4 2 .|. 6 .|. 7 .
. . 5|7 . .|6 . 3
. . 7|. . 5|. 1 .


And what I have so far is:
Code: Select all
79    1     346  |2     379   34679|5     489   478
5     46    2    |1469  179   8    |3     49    47
79    8     34   |349   5     3479 |2     6     1
-----------------------------------------------------
3     56    8    |169   4     169  |7     25    26
1     467   9    |5     37    2    |48    348   468
2     4567  46   |36    8     367  |1     345   9
-----------------------------------------------------
4     2     1    |389   6     39   |89    7     5
8     9     5    |7     12    14   |6     24    3
6     3     7    |489   29    5    |489   1     248


Finding the (7,9) pair at B1 only helped to cross out a 7 at r2c2, but that was it. So ideas please?
Thanks.
JeJ
 
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Joined: 06 January 2011

Re: Method for another Challenger sudoku

Postby JasonLion » Wed Jan 12, 2011 2:48 am

Think about R4 and B4 and see if you can find what is hidden.
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Re: Method for another Challenger sudoku

Postby JeJ » Wed Jan 12, 2011 6:45 pm

JasonLion wrote:Think about R4 and B4 and see if you can find what is hidden.

Row 4:
Code: Select all
3     56    8    |169   4     169  |7     25    26


And Box 4:
Code: Select all
3     56    8    |
1     467   9    |
2     4567  46   |


I just don't see anything that could lead me to eliminate numbers or to define numbers at specific cells. One more hint please.
JeJ
 
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Joined: 06 January 2011

Re: Method for another Challenger sudoku

Postby JasonLion » Wed Jan 12, 2011 9:58 pm

Ah, sorry. I meant B5, not B4.

Look for a hidden pair.
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Re: Method for another Challenger sudoku

Postby JeJ » Wed Jan 12, 2011 11:27 pm

JasonLion wrote:Ah, sorry. I meant B5, not B4.

Look for a hidden pair.

A Hidden pair in B5? I don't see it. 1 is with 6 and 9, that is not a pair, 3 is with 6 and 7 again not a pair, 7 is with 3 and 6 same situation and 6 is with about everyone else. I am not as good at this as I thought or there is something missing from the theory I studied about pairs.
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Re: Method for another Challenger sudoku

Postby JasonLion » Wed Jan 12, 2011 11:35 pm

1 and 9 is a hidden pair in both row 4 and block 5. There is also a naked triple in R4 that makes the same elimination: 2,5 and 6 in R4C289
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Re: Method for another Challenger sudoku

Postby JeJ » Thu Jan 13, 2011 1:24 am

I see it now. Thanks. This means I still have to work on my pair descovering abilities.
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