by PaulIQ164 » Mon Sep 05, 2005 11:43 pm
Right. If you look at the sixth column - the one you already have 8, 9 and 2 in. You still need to put in 1, 3, 4, 5, 6 and 7. Now, the central 3x3 box on the grid already has 3, 6 and 7 in it, so the two squares in row 6 that are in this middle box can't be 3, 6 or 7 (meaning they can still be 1, 4 or 5). Now, if you look at the seventh row down, you'll see that row already has 1 3, a 6 and a 7 in it too. So again, the cell on the seventh row and the sixth column cannot be 3, 6 or 7. So, out of the six cells left to fill in row 6, there are three that cannot be 3, 6, or 7; and three that can be. So we must have 3, 6 and 7 in these cells.
Now look at box 2 (the 3x3 box in the top-center). On the face of it, there are two places where the 4 can go (hopefully you can see this from your pencilmarks). But, one of them is one of the three cells we've already identified as between themselves containing the 3, 6 and 7. So this one can't take the 4. That means the other cell must be the 4. Luckily, after this the rest of the puzzle is fairly simple.