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!

3 posts
• Page **1** of **1**

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!

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!

- intermilan
**Posts:**5**Joined:**02 May 2009

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.

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.

- storm_norm
**Posts:**85**Joined:**27 February 2008

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.

- intermilan
**Posts:**5**Joined:**02 May 2009

3 posts
• Page **1** of **1**