# a six-by-six KenKen

The rules of KenKen are as follows:

• You must fill in the numbers 1 through 6, one number per box, in the grid, so each row and each column contains each number exactly once.
• Each heavily outlined section of the grid is marked for you with a number and a symbol of an arithmetic operation. It indicates that the numbers to be placed in the boxes in that section, when combined via that operation, yield the given number. For example, if it says "60×", then the numbers in the boxes in that section must multiply to sixty. For division ("/") and subtraction ("−") there will be precisely two boxes in the section: one of them divided by (or subtracted from) the other must yield the given number.

The following puzzle is taken directly from Simon Tatham's collection (where he calls the puzzle type "Keen"). It's one that I found unusually challenging, which is why I'm sharing it with you. Thanks for sharing the puzzle. It was indeed challenging and fun.

There might be a simpler initial break-in, but here is my attempt.

Green: (3,6) or (4,5)
Blue: (3,5,6)
Red: (2,5,6) or (3,4,5)
Yellow/Purple: (2,6/3,4) or (3,5/1,6)
Let's assume Green is (3,6). Then R1-3C4 form a (3,5,6) triple, forcing R6C4 to be 4.
R6C6 cannot be 3 (contradiction with Yellow/Purple).
Putting 5 on R6C6 forces R1C6 to 1.
At this point, both choices for Brown (1,3 or 2,6) are eliminated, contradiction.
Therefore, Green is (4,5). Looking at the top row, 6 must appear on the left side, making the cage (3,6) and the opposite one (1,2).
Also, the six green cells have the product of 720, which means R1C1 = R6C2.
This semi-decides C2 as well. Similar logic applies to the green cells here, giving R5C1 = R6C3.
Given that R6C2 = 3 or 6, R5C1 is limited to 1 or 2 or 4.
Red cells cannot be (3,6) due to R2,5C3, so it is either (1,2) or (2,4).
Either way, it forms a triple with R6C3, deciding the 9+ cage. Green must contain 1 since the cell below it cannot (and therefore be (1,2)).
On R5, the number 2 cannot go to Orange (due to 3/ above) or Purple, so it goes to R5C4.
Fill in the candidates for R6C6 by sum logic on C6. Look at the blue cells. The 3/ cage can be either (1,3) or (2,6).
Either case, it forms a triple with (1,2) above, banning 2 at R6C5.
Therefore, the 2 on R6 can only appear at R6C1. Now we can fill in a lot of cells. R4C6 cannot be 6, putting it at R4C5. Straightforward logic finishes the puzzle.

Finished solution: • Nice. Eventhough you start with an assumption, the right box in the top row and cell in the bottom right corner can be filled without guessing. Hint: consider blue and yellow(also red) regions in the first image. Also you may need to think about each cage in the first row. (odd/even) : )
– ACB
Sep 6, 2022 at 9:17
• Many ways to approach this, I guess. For green I used: in column 3: green+blue = 2 mod 3. This fixates green while only looking at green+blue. Sep 6, 2022 at 14:04

635421
423516
514632
352164
146253
261345

Explanation (click to enlarge):

• But how'd you get that? Sep 6, 2022 at 4:37
• There are plenty of such grid puzzles where answers get away with the lack of elaboration. Sorry but I can't stand double standards. Sep 6, 2022 at 6:01
• They don't get away with it from me. I downvote such answers when I see them. Sep 6, 2022 at 6:02
• I know it's too late, but I edited my answer. Sep 6, 2022 at 8:25
• Thanks. I was just wary of embedding such a huge image. I thought it might have been frowned upon like on most interactive boards and forums. Sep 7, 2022 at 5:58