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| Difficult 7x7 Mar 23 2020 https://www.calcudoku.org/forum/viewtopic.php?f=16&t=1229 |
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| Author: | jake4 |
| Post subject: | 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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| Author: | jpoos |
| Post subject: | 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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| Author: | jake4 |
| Post subject: | Re: Difficult 7x7 Mar 23 2020 |
Thanks! |
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