dailysudoku.com

[igolf4birdies]

I'm stuck. I need some logic for my next step. I had already found the 5,6 pair in the third row as well as the 2,7 pair in the sixth row. But now I'm stuck.

3 8 _ _ 9 4 _ 7 6
_ _ _ _ 3 8 _ 9 5
_ _ 9 _ 7 2 8 _ _
_ 5 1 7 _ 3 _ 2 4
_ 3 6 4 2 1 7 5 _
4 _ _ 9 _ 5 _ 1 _
_ _ 3 8 1 _ _ _ _
7 6 _ 2 4 9 _ _ _
1 9 _ 3 5 _ _ _ _

Thanks in advance.
Rick

[David Bryant]

Hi, Rick!

You need to spot the "hidden pair" {2, 7} in r7c9 & r9c9 ... then everything will fall into place.

-- Because of the "2" at r3c6 you can see that the "2" in column 7 must fit in the top right 3x3 box.

-- Now concentrate on the bottom right 3x3 box. Neither "2" nor "7" can fit at r7c7, r8c7, or r9c7 because of the "hidden 2" we just identified and the "7" at r5c7. Both "2" and "7" already appear in column 8 ... that rules out 3 more cells in the bottom right box. And both "2" and "7" already appear in row 8, so neither one can fit at r8c9. As there are only two cells left in the bottom right 3x3 box, we see that {2, 7} must fall in r7c7 & r9c9.

-- Now there's only one place left for a "9" in column 9 -- at r5c9. dcb

PS You can also see this by spotting the {2, 7} in row 3, the {2, 7} in the middle right 3x3 box, and the {2, 7} in row 8, all of which constrain the places in column 9 at which {2, 7} can appear.

[jeeperndbb]

How did you guys get the 3/8 in r1c1 and r1c2 and the 2 in r4c8? I understand eerything before and everything after that? Thanks

[sunpat]

Hi jeeperndbb:
r3 we have 2,7,8,9.
we need 1,3,4,5,6.
and in r3c2 only (1,4), r3c8 (3,4) and r3c9 (1,3) .
so, r3c1 and r3c4 only can put pair of (5,6),
then 3 only can put in r1c1.

8 only can put in r1c2, ( can not r1c3 because 8 also in r3c8 or r3c9 )