by **ravel** » Thu May 11, 2006 10:11 pm

Unfortunately the 9 alone will not help you much.

How i would try to proceed (hope i dont say too much) ?

What can be seen easily:

In box 9 (right bottom 3x3 group) 5 and 9 (both in columns 8 and 9) only can be in r89c7 (row 8, column 7 and row 9,column 7).

In column 3 (both in box 4) 1 and 3 only can go to r19c3.

78 in box 4 (both in column 3) can only go to r5c1 and r6c2, which gives a 26 pair in r45c3.

Nothing helps to find a number.

So i have to count and start with box 3, where only 3 numbers are left:

r1c7 can only be 36, r1c8 either 36, r2c8 only 1, next number.

No other 1's forced by that, but the 36-pair will help to solve the puzzle.