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First of all here is the spoiler free Kenken-Puzzle for everyone who likes to solve this on her/his own.

spoiler free version

Now here is how far I come with this, but now I struggle a lot. What I am looking for is only one more field solved with a logical explanation why this number belongs there (and can’t be anything else)

my solution so far

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    $\begingroup$ Welcome to Puzzling! Great question - we don't see many questions about puzzles around here. c: $\endgroup$ – Deusovi Jul 18 '16 at 14:51
  • $\begingroup$ I thought numbers couldn't repeat in a Kenken cage. $\endgroup$ – Joe Z. Jul 19 '16 at 5:33
  • $\begingroup$ How did you know that it was 2,1 in the first row, and not 1,2? Also, how do you know it is 4,5 in the third column, and not 6,5? $\endgroup$ – shoover Jul 20 '16 at 15:46
  • $\begingroup$ @shoover. The first row of the 10+ is built by a 1-2 pair or a 4-2 pair, because of the 4x next to it. If it is a 4-2 pair this leaves you with a 3-1 pair to complete the 10+ field. If it is the 1-2 pair it leaves you with either 6-1 or 4-3 since the 5 is already blocked off. If you put a 4 in the third column in the 18+ field this leaves you with a 3-5-6 solution only. This 3-5-6 solution “kills” all possible solutions of the 10+ field so it can’t be a 4 there. $\endgroup$ – solid Jul 21 '16 at 7:33
  • $\begingroup$ @shoover. If you take a look at the 5 th column than there must be a 1 on the top. Since the 144x and 19+ would need more 6’s to complete the fields than you can fit there. Therefor it is a 2 1 not a 1 2 $\endgroup$ – solid Jul 21 '16 at 7:38
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How can you get four numbers that multiply to 144?

  • 6,6,4,1
  • 6,6,2,2
  • 6,4,3,2
  • 4,4,3,3

There's already a 1 in the last two columns, so it can't be 6,6,4,1. We can also eliminate 6,6,2,2 and 4,4,3,3 because those won't fit in the cage. So the 144x cage has 6,4,3,2.

There's already a 6 in the 4th row, so the 6 in that cage must be in the 2nd or 3rd row. The 6 in the 1st column must also be in the 2nd or 3rd row. That's two 6s in the 2nd and 3rd rows, so the 10+ cage can't have a 6. It also can't have a 5, so the last two numbers in that cage must be 4 and 3.

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  • $\begingroup$ You are absolutely right. Thank you. I don’t know how I missed this. Answer accepted. $\endgroup$ – solid Jul 18 '16 at 14:52
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If you dislike multiplication, the other way is:

  • The sum of the multiples of the 144x cage is 15 (total of 2 columns (21*2) - (19+2+4+1+1)) = (42-27) = 15.

The multiples that for 144 with sum 15 = 2,3,4,6

[Explanation]

Take the last two columns. Each column total would be (6+5+4+3+2+1=21).
So, for the last two columns, its 42.
Subtract all the filled parts from 42 i.e 42 - (19+2+4+1+1) = 15.
So, the total of 144x cage is 15.

And from hereon you can easily proceed forward.

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  • $\begingroup$ Can you explain this a bit more? I don't really follow but it seems like it would be a good tool for ones that I get stuck on. $\endgroup$ – Ed T Jul 18 '16 at 17:40
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    $\begingroup$ The sum of any given row is 21 since it contains the digits 1-6. The sum of two adjacent columns is therefore 42. From the rightmost two columns, you have filled in 1, 4, 1, 2 and there is a cage that adds to 19. 42 - (19 + 1 + 4 + 1 + 2) = 15, so the sum of the 144 cage must be 15. $\endgroup$ – Mad Physicist Jul 18 '16 at 17:56
  • $\begingroup$ Thank you, I never realized this method even though I've completed hundreds of these puzzles. This might explain the ones I got stuck on! $\endgroup$ – Ed T Jul 20 '16 at 16:45

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