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sunday31st july

 
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shadders



Joined: 03 Aug 2005
Posts: 1
Location: Spain
PostPosted: Wed Aug 03, 2005 12:12 am    Post subject: sunday31st july Reply with quote
I got this far with Sunday 31st July puzzle

--- 2-- --8
42------5
-6-5-492-
7X36---84
----4--69
64---5371
--4--6-5-
--6----92
1----8--6

I asked for a draw and got 9 on r4 c1
I can't understand why this should be 9 - as I calculate that it could be 159 and that 9 could be in r6 c3 of the same square.

Can anyone help

shadders
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pierceedd



Joined: 22 Jul 2005
Posts: 7
Location: New Jersey, USA
PostPosted: Wed Aug 03, 2005 2:27 pm    Post subject: Re: sunday31st july Reply with quote
shadders wrote:
I got this far with Sunday 31st July puzzle

--- 2-- --8
42------5
-6-5-492-
7X36---84
----4--69
64---5371
--4--6-5-
--6----92
1----8--6

I asked for a draw and got 9 on r4 c1
I can't understand why this should be 9 - as I calculate that it could be 159 and that 9 could be in r6 c3 of the same square.

Can anyone help

shadders


Did you mean r4 c2 is a 9? In your grid you gave r4 c1 as a 7.

In any case, I'm not sure why that was the hint it gave, but here's my logic:

The bottom middle cell (rows 7, 8, 9 and columns 4, 5, 6) can only have a 2 in two places: row 7, column 5 or row 9, column 5. Since those two options are both in the same column, it means that the 2 for the fifth column MUST be in that cell and cannot be anywhere else in the fifth column.

Consequently, row 6, column 5 can only have 8 or 9. Row 6, column 4 also can only have 8 or 9, which means that row 6 column 3 must be a 2.

As a result of this, the first three boxes in row 5 contain a triplet with the numbers 1, 5, and 8, which means that row 4, column 2 must be a 9.

I'd be curious to know if there is a solution that leads more directly to that 9.

John
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