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 Difficult 7x7 Mar 23 2020 
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Posted on: Mon Mar 23, 2020 4:54 pm




Posts: 80
Joined: Sun Mar 04, 2012 3:43 am
Post Difficult 7x7 Mar 23 2020
Please tell me what my faulty assumption / math is with this puzzle.

Starting with the rightmost column, the following cages are forced:
42x (6,7)
81^ (3,4)

This leaves 1,2,5 for that column. The only way to make those work with the 0 mod cage is with the 1; (2,5) won't do.

This conflicts with the 3^ cage in the leftmost column, which must be (1,1,3) with 3 in the corner.

It's been a while since I solved these here regularly, so maybe I'm misunderstanding how the ^ and/or mod functions work?


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Posted on: Mon Mar 23, 2020 5:00 pm




Posts: 148
Joined: Sun Nov 03, 2013 10:28 pm
Post Re: Difficult 7x7 Mar 23 2020
jake4 wrote:
This conflicts with the 3^ cage in the leftmost column, which must be (1,1,3) with 3 in the corner.\


Here is where you go wrong. Since exponentiation is done from right to left, that cage can also be something like (1,3,7), since 3^(1^7)=3^1=3. In fact, any combination with both a 1 and a 3 is necessary and sufficient.


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Posted on: Mon Mar 23, 2020 5:21 pm




Posts: 80
Joined: Sun Mar 04, 2012 3:43 am
Post Re: Difficult 7x7 Mar 23 2020
Thanks!


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