Please help on this problem

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Please help on this problem

Postby intermilan » Mon May 11, 2009 11:02 pm

It seems that there's no hidden pair or tripple. How to solve it? (I have used simple rules to get so far).

r9c3 should reduce from 257 to 25 (7 is impossible because 7 in r9c8, r9c9), but I don't think it is very helpful ...

Thanks!

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intermilan
 
Posts: 5
Joined: 02 May 2009

Postby storm_norm » Mon May 11, 2009 11:58 pm

familiar with a skyscraper?
take a look at the 6's that i have marked.
imagine placing a 6 in either r9c6 or r8c7
what happens to the other 6's?? you will find a contradiction.
you can thus eliminate the 6's in r9c6 and r8c7.

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storm_norm
 
Posts: 85
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Postby intermilan » Wed May 13, 2009 8:40 pm

3x a lot!!
storm_norm wrote:familiar with a skyscraper?
take a look at the 6's that i have marked.
imagine placing a 6 in either r9c6 or r8c7
what happens to the other 6's?? you will find a contradiction.
you can thus eliminate the 6's in r9c6 and r8c7.

Image
intermilan
 
Posts: 5
Joined: 02 May 2009


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