## Fiendish #237

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### Fiendish #237

So, it's the next fiendish one, and I'm stuck again.

Code: Select all
`  3|5  | 821  |3  |   84 |   |  ---+---+--- 5 1| 43|87  84|1 9|523 3 | 5 |1 4---+---+---3  |  5| 4 4  |  6| 5825 |4 7|6  `

I've noticed that the 8th column can be narrowed to the following possibles (using the 1/3 pair):

8
(6/9)
(1/3)
-
7
2
(6/9)
-
4
5
(1/3)

Where the numbers in brackets are the possibles for those squares.

I can't find any other pairs or triples, or any mistakes but I'm totally stuck - I've spent over an hour checking and searching. Any help is much appreciated.
p_court99

Posts: 9
Joined: 18 July 2005

Look for a triple of 6,7 and 9 in row 1 before eliminating them from other cells in row 1. You need to pay careful attention to box 8 in doing this.

Paul
Enigma

Posts: 53
Joined: 14 June 2005

I have this in row one:

(6/7/9) (6/7/9) 3 | 5 (1/6/7/9) (1/4) | (4/7/9) 8 2

So surely the (4/7/9) possibility stops there being a 6/7/9 triple on that row?
p_court99

Posts: 9
Joined: 18 July 2005

In box 3 the 1 has to be in row 3. This has implications for row 3 in box 2.
possum

Posts: 86
Joined: 05 April 2005

As things stand, you can fill a number in box 2.

Look carefully at the cell where you had (1/4) again.
su_doku

Posts: 30
Joined: 19 March 2005

I was able to complete the puzzle using possum's tip, thanks.
p_court99

Posts: 9
Joined: 18 July 2005

Although the (4/7/9) is a possibility, 1 cannot be in r1c5 as the 1 has to be in c5 in box 8. So you could have eliminated the 7/9 from c7 leaving just the 4, which then yields 1 in c6...
Enigma

Posts: 53
Joined: 14 June 2005