## Kakuro help

For fans of Kakuro

### Kakuro help

Can someone tell the solution for this puzzle I've been working on it for ever and no luck.

mario133

Posts: 28
Joined: 22 December 2008

This was how I filled in the first two cells:

Code: Select all
`+-----+-----+-----+-----+-----+-----+-----+-----+-----+|#####| c1  | c2  | c3  | c4  | c5  | c6  | c7  | c8  |+-----+-----+-----+-----+-----+-----+-----+-----+-----+| r1  |xxxxx|xxxxx|xxxxx|14\  |28\  |xxxxx|27\  |13\  |+-----+-----+-----+-----+-----+-----+-----+-----+-----+| r2  |xxxxx|24\  |28\ 8|     |     |  \10|     |     |+-----+-----+-----+-----+-----+-----+-----+-----+-----+| r3  |  \30|     |     |     |     | 4\ 8|     |     |+-----+-----+-----+-----+-----+-----+-----+-----+-----+| r4  |  \16|     |     | 5\ 8|     |     |     |10\  |+-----+-----+-----+-----+-----+-----+-----+-----+-----+| r5  |  \30|     |     |     |     |     |     |     |+-----+-----+-----+-----+-----+-----+-----+-----+-----+| r6  |xxxxx|14\16|     |     |     | 9\ 4|     |     |+-----+-----+-----+-----+-----+-----+-----+-----+-----+| r7  |  \12|     |     |  \14|     |     |     |     |+-----+-----+-----+-----+-----+-----+-----+-----+-----+| r8  |  \ 7|     |     |  \13|     |     |xxxxx|xxxxx|+-----+-----+-----+-----+-----+-----+-----+-----+-----+`

8/2 @ r3c78={17|26|35}
=> 13/2 @ r23c8=[67|76|85]
=> 10/2 @ r2c78=[28|37|46]
=> 8/2 @ r3c78=[17|26|35]
8/3 @ r4c567={125|134}
4/2 @ r6c78={13}

In summary:
r2c7 from {234}
r3c7 from {123}
r4c7 from {12345}
r6c7 from {13}

Now consider 27/6 @ r234567c7
Any 5 cells in there has a min value of 27-9=18
i.e. can't be {12345}
=> r57c7 can't be from {12345}, must be from {6789}

But 14/4 @ r7c5678 can't have 9
=> r7c7 must be from {678}

Also, 24/3 @ r345c2={789}

Now 30/7 @ r5c2345678={1234569|1234578}
The 2 cells from {6789} must be @ r5c27, and must sum to 15
=> r5c27=[78|87|96]
=> r5c7 must be from {678}

Now r57c7 are both from {678}
=> max r57c7=7+8=15
=> min r2346c7=27-15=12, i.e. can't be {1234}
=> r4c7 can't be from {1234}, must be 5
=> min r236c7=12-5=7, i.e. can't be {123}
=> r2c7 can't be from {123}, must be 4

The rest is easy.
udosuk

Posts: 2698
Joined: 17 July 2005

Thanks for the help the answer up above.
mario133

Posts: 28
Joined: 22 December 2008