The New Sudoku Players' Forum

[davidlawton@talk21.com]

help please, can normally do hard ones


original

6-- -84 ---
-2- --- ---
8-3 -1- ---

--- --8 6-7
7-- 1-9 --8
9-4 5-- ---

--- -5- 8-1
--- --- -6-
--- 42- --3

me


6 1579 1579 379 8 4 123579 12357 59
14 2 179 3679 79 5 13479 8 469
8 4579 3 679 1 2 4579 457 4569

135 135 15 2 4 8 6 9 7
7 6 2 1 3 9 45 45 8
9 8 4 5 6 7 13 13 2

24 479 6 79 5 3 8 247 1
2345 34579 579 8 79 1 24579 6 459
15 1579 8 4 2 6 579 57 3


answera probably staring at me , cant see it cannot see any unique 3 variables

[scrose]

Make some eliminations based on a pair in column 8. You will then be able to continue filling cells.

Update: Fixed the hyperlink.

[simes]

Look at column 8, and see where 1 and 3 can go. Think about the effects on the candidates for those cells. Then see what candidates are remaining in the column.

[davidlawton@talk21.com]

Sorry still cannot see where your leading me . I cannot follow the logic of a pair in column 8, tried 1 and 3 seems im in the world of trial and error

Dave

[scrose]

In column 8, there are only two cells (r1c8 and r6c8) where the candidates 1 and 3 can be. That means one of these cells is a 1, and the other is a 3. The important thing to note is that no other candidates can be placed in these two cells. So the candidates 2, 5, and 7 can be removed from the cell r1c8. These eliminations will let you fill the cell r1c7.

[davidlawton@talk21.com]

to scrose

thanks - penny finally dropped, once you pointed out 1-3 was a tied pair.

Dave