-1
$\begingroup$

Note: The reason that I picked this one to self-answer was because it was a few weeks (months?) before the Wisconsin State Journal stopped printing Kenken puzzles. This was also the newest Kenken from the Wisconsin State Journal that I was able to access on the internet. There also doesn't seem to be any descriptions online on how to solve it.


Is there a logical way to deduce the solution to the "Challenging" Kenken in the Wisconsin State Journal from April 26, 2021?

enter image description here

+----+----+----+----+----+----+
|1-  |5-  |4   |10x           |
+    +    +----+----+----+----+
|    |    |7+  |5-  |13+      |
+----+----+    +    +----+    +
|80x |3   |    |    |1-  |    |
+    +----+----+----+    +----+
|         |3÷  |2-  |    |3+  |
+----+----+    +    +----+    +
|5-  |13+ |    |    |4   |    |
+    +    +----+----+----+----+
|    |         |2÷       |3   |
+----+----+----+----+----+----+
$\endgroup$
7
  • 5
    $\begingroup$ Why didn't you at least give the community a chance to answer the puzzle? I don't see any need for you to have answered it immediately after you posted it. The rules in the bottom left of the original image are, in my opinion, well written and easy to understand. $\endgroup$
    – ibanezplyr
    Nov 30, 2023 at 21:41
  • $\begingroup$ @ibanezplyr I think my thought process was that it was a question I already knew the answer to and to document the knowledge that I had (there are no other online resources for this). I do see the argument being that P.SE is mainly a site about people posting puzzles that they already know the answer to for other people to solve, however. $\endgroup$
    – CrSb0001
    Nov 30, 2023 at 21:56
  • 4
    $\begingroup$ This one is not particularly difficult or noteworthy. Why should such a common puzzle have any special attention? $\endgroup$ Nov 30, 2023 at 23:02
  • 2
    $\begingroup$ The solution is direct, there are no dead ends. This is just a random puzzle distributed to hundreds of newspapers. The fact that one particular newspaper stopped including KenKen in its puzzle page shortly after this one was published is irrelevant. I understand your reasons for posting, I just do not agree with them. $\endgroup$ Dec 1, 2023 at 0:17
  • 2
    $\begingroup$ Is it possible that you tried to solve the version of the diagram that you posted above with a '5' instead of a '4' in R1C3? That leads to an unresolvable dead end rather quickly. $\endgroup$ Dec 2, 2023 at 23:58

1 Answer 1

-1
$\begingroup$

Yes, it is solvable!

  1. Fill in the numbers that do not have a +, -, x, or ÷:

enter image description here

  1. Note that there is only one way to get 5-, and using this information, we can put 1/6 in the 3 5- boxes and then logically deduce the (mostly because I kept hitting dead ends later on if I put a 1 in that box) (note that the 5 in R1C3 should be a 4 - I fixed that in this screenshot):

enter image description here

  1. Now take the ways we can multiply to 80 - this will help us solve the 80x box: We have
+---------+---------------------------+
|80x using|Ways to do it (no using 1*)|
+---------+---------------------------+
|1 number |             80            |
+---------+---------------------------+
|2 numbers|    2*40,4*20,5*16,8*10    |
+---------+---------------------------+
|3 numbers|     2*20*2,...,4*5*4      |
+---------+---------------------------+

The way to multiply to 80 with 3 numbers that doesn't cause any contradictions later on is 4*5*4 (which is legal because there are no boxes that confine numbers to a specific area, except for the boxes. It's only required that the 1-6 are unique in each column and row).

This forces a 4 in R2C6 and then (which I forgot to do earlier) we can deduce that the 3+ must be 1 and 2 in some order (only two unique numbers that add up to 3), so now our grid is

enter image description here

  1. To get to 10x with 3 numbers, we must have 5*2*1 in some order. We know that the 2 must be in R1C4 due to Kenken rules, forcing 1 in R1C5 and 5 in R1C6. Then, 2 is forced in R2C1, and so on and so on. Our breakthrough allows us to fill in a bunch of new numbers.

enter image description here

  1. Finally, note that R4C5 and R3C5 must be 6 and 5 respectively. Only then we can fill in R4C4 and R5C4 without running into a contradiction, and then the rest of the grid is trivial, with a final unique solution:

enter image description here

+----+----+----+----+----+----+    +---+---+---+---+---+---+
|1-  |5-  |4   |10x           |    | 3 | 6 | 4 | 2 | 1 | 5 |
+    +    +----+----+----+----+    +---+---+---+---+---+---+
|    |    |7+  |5-  |13+      |    | 2 | 1 | 5 | 6 | 3 | 4 |
+----+----+    +    +----+    +    +---+---+---+---+---+---+
|80x |3   |    |    |1-  |    |    | 4 | 3 | 2 | 1 | 5 | 6 |
+    +----+----+----+    +----+    +---+---+---+---+---+---+
|         |3÷  |2-  |    |3+  |    | 5 | 4 | 1 | 3 | 6 | 2 |
+----+----+    +    +----+    +    +---+---+---+---+---+---+
|5-  |13+ |    |    |4   |    |    | 6 | 2 | 3 | 5 | 4 | 1 |
+    +    +----+----+----+----+    +---+---+---+---+---+---+
|    |         |2÷       |3   |    | 1 | 5 | 6 | 4 | 2 | 3 |
+----+----+----+----+----+----+    +---+---+---+---+---+---+
$\endgroup$

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