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From the position you posted I would look at 'column' 8. There are only 2 places for a 4.

If the 4 is in row 5 then r5c7=1

If the 4 is in row 6 then r6c7=1, r6c6=5

Either way, 1 can be eliminated from other cells in column 7. This leaves r4c7 as being 2 or 5, so the 8 cage (is that the right term in a kakuro?) in column 7 must be 1,5,2.

Can you progress from that point?

If the 4 is in row 5 then r5c7=1

If the 4 is in row 6 then r6c7=1, r6c6=5

Either way, 1 can be eliminated from other cells in column 7. This leaves r4c7 as being 2 or 5, so the 8 cage (is that the right term in a kakuro?) in column 7 must be 1,5,2.

Can you progress from that point?

- jimbob
**Posts:**47**Joined:**07 March 2006

Okay, since r4c7 has to be 2|5, r4c8 cannot be 6.

And if r6c8=6, r6c6=3 => r6c7=1 => r5c7=2 => r5c8=3, and we must have 9s in both r28c8... Therefore r6c8 cannot be 6.

Which leaves r8c8=6, r8c5=1 and r7c5=7, and also r2c8=9...

And the rest would flow from there...

And if r6c8=6, r6c6=3 => r6c7=1 => r5c7=2 => r5c8=3, and we must have 9s in both r28c8... Therefore r6c8 cannot be 6.

Which leaves r8c8=6, r8c5=1 and r7c5=7, and also r2c8=9...

And the rest would flow from there...

- udosuk
**Posts:**2698**Joined:**17 July 2005

Another way:

Since R7C45 <> {15} -> R7C6 <> 2

-> R6C6 <> 5

Since R6C6 <> {15} -> R6C8 <> 4

-> R5C8 = 4 (hidden single in C8)

R5C7 = 1, R46C7 = [25], R4C68 = [71]

...

Since R7C45 <> {15} -> R7C6 <> 2

-> R6C6 <> 5

Since R6C6 <> {15} -> R6C8 <> 4

-> R5C8 = 4 (hidden single in C8)

R5C7 = 1, R46C7 = [25], R4C68 = [71]

...

- Jean-Christophe
**Posts:**149**Joined:**22 January 2006

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