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Ken Ken


Go to solution Solved by kukupai,

Question

I've gotten hooked on Ken Ken. It's like Sudoku with extra constraints.

Numbers 1-9 must occur in every row and column.

Clusters of numbers must also obey rules.

Within a cluster,

  • 15+ means they must sum to 15. e.g. 6 and 9.
  • 5- means the (two) numbers must differ by 5. e.g. 3 and 8.
  • 72x means their product must be 72, e.g. 2 4 9.
  • 4/ means the (two) numbers must have 4 as a quotient. e.g. 2 and 8.

Clusters for + and x can have any number of members.

For - and / there obviously can be only two members in a cluster.

I find the "expert" 9x9 puzzles can provide 30-60 minutes of fun.

Here's one for you:

post-1048-0-25764600-1378921331_thumb.jp

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i think i found a logical contradiction.

_ _ _ _ 3?7?_ _ _
_ _ 7 _ _ _ _ _ _
_ _ _ _ _ _ _ _ _
_ _ _ _ _ _ _ _ _
_ _ _ _ _ _ _ _ _
_ _ _ _ _ _ _ _ _
_ _ _ 5 _ _ _ _ _
_ 5 _ _ _ _ _ _ _
_ _ _ _ _ _ _ _ _

336 = 7*8*6. 15 has to be 9+6 since 7 is already on that row. therefore...

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

84 = 3*4*7; therefore...

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

now, what can possibly go in the 3 divide?

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i think i found a logical contradiction.

_ _ _ _ 3?7?_ _ _
_ _ 7 _ _ _ _ _ _
_ _ _ _ _ _ _ _ _
_ _ _ _ _ _ _ _ _
_ _ _ _ _ _ _ _ _
_ _ _ _ _ _ _ _ _
_ _ _ 5 _ _ _ _ _
_ 5 _ _ _ _ _ _ _
_ _ _ _ _ _ _ _ _

336 = 7*8*6. 15 has to be 9+6 since 7 is already on that row. therefore...

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

84 = 3*4*7; therefore...

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

now, what can possibly go in the 3 divide?

84 also is 2x6x7

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i think i found a logical contradiction.

_ _ _ _ 3?7?_ _ _
_ _ 7 _ _ _ _ _ _
_ _ _ _ _ _ _ _ _
_ _ _ _ _ _ _ _ _
_ _ _ _ _ _ _ _ _
_ _ _ _ _ _ _ _ _
_ _ _ 5 _ _ _ _ _
_ 5 _ _ _ _ _ _ _
_ _ _ _ _ _ _ _ _

336 = 7*8*6. 15 has to be 9+6 since 7 is already on that row. therefore...

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

84 = 3*4*7; therefore...

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

now, what can possibly go in the 3 divide?

In your diagram, I would insert 6 above 2. Cells (5,5) and (5,6) respectively.

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On row 1

x x x x 3 7 x x x

yields

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

Yet column 7or 8 must contain a 5 for 280x

Something wrong here

What you have found is that the 5's in columns 7 and 8 go, one each, in the 3- and the 280x clusters.

Possibly in cells (7,1) and (8,5). If not, then in cells (8,1) and (7,4) or (7,5).

And you have found that the 7's in columns 5 and 6 go, one each, in the 21x and the 84x clusters.

Possibly in cells (6,1) and (5,7). If not, then in cells (5,1) and (6,6) or (6,7).

You're making progress.

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