5
$\begingroup$ 
 .  2  . | .  7  . | .  .  9 
 .  9  8 | .  .  . | .  .  . 
 3  .  . | 9  .  . | .  .  . 
---------+---------+--------
 2  .  7 | .  .  . | .  .  . 
 .  1  5 | 7  .  . | .  .  . 
 .  .  . | 1  5  8 | .  6  . 
---------+---------+--------
 .  .  . | .  3  6 | .  .  . 
 .  .  . | .  .  . | .  .  3 
 .  .  . | 4  .  5 | .  .  8
  • Normal Sudoku rules apply
  • When divided by 11 the 4 marked numbers (see example) have the same remainder

Clarification

This is an example of the remainder constraint:


 .  .  . | .  .  . | .  .  . 
 .  .  . | 9  5  2 | .  .  . 
 .  .  . | .  .  . | .  .  . 
---------+---------+--------
 .  3  . | .  .  . | .  1  . 
 .  2  . | .  .  . | .  3  . 
 .  5  . | .  .  . | .  8  . 
---------+---------+--------
 .  .  . | .  .  . | .  .  .
 .  .  . | 2  3  7 | .  .  . 
 .  .  . | .  .  . | .  .  .

Because:

  • 952 = 86 x 11 + 6
  • 138 = 12 x 11 + 6
  • 237 = 21 x 11 + 6
  • 325 = 29 x 11 + 6

 2  6  5 | 3  1  4 | 8  7  9 
 7  1  8 | 9  5  2 | 4  6  3 
 3  9  4 | 7  6  8 | 1  5  2 
---------+---------+--------
 6  3  7 | 8  4  9 | 2  1  5 
 8  2  1 | 5  7  6 | 9  3  4 
 4  5  9 | 1  2  3 | 7  8  6 
---------+---------+--------
 5  4  2 | 6  8  1 | 3  9  7 
 9  8  6 | 2  3  7 | 5  4  1 
 1  7  3 | 4  9  5 | 6  2  8

Important

The remainder in the example Sudoku is not necessarily the same as the remainder in the actual challenge.

The location of the special numbers is the same however.

Improve this question
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3
  • $\begingroup$ That's an interesting variation: one that breaks the usual equivalence under digit/band/row/col permutations, and also one that allows puzzles with fewer than 17 clues. Here's an example: .7....2.....83.......4........7....4.2.....5..........3...5.1..4.9...........26.. $\endgroup$
    – 53x15
    Commented Dec 19, 2019 at 19:14
  • $\begingroup$ Starting from 17 clue puzzles, randomly permuting, restricting to those satisfying the remainder property, minimizing, and running this process for an hour, I find 1192 17-clue puzzles, 2160 16-clue puzzles, 217 15-clue puzzles, and 1 14-clue puzzle. Here's the one with 14 clues: .4.....6....51..................2...5.1......8..............1........8.5.26..4... $\endgroup$
    – 53x15
    Commented Dec 19, 2019 at 20:40
  • $\begingroup$ That was sort of my take on generating Sudokus. Trying to find a 'fun' constraint that will allow me to generate a Sudoku with minimal clues. I'm a software architect by profession. $\endgroup$
    – Joris Schellekens
    Commented Jan 9, 2020 at 9:18

1 Answer 1

Reset to default
2
$\begingroup$

Solution:

 6  2  1 | 5  7  3 | 4  8  9
 7  9  8 | 6  2  4 | 3  1  5
 3  5  4 | 9  8  1 | 2  7  6
 --------+---------+--------
 2  6  7 | 3  4  9 | 8  5  1
 8  1  5 | 7  6  2 | 9  3  4
 4  3  9 | 1  5  8 | 7  6  2
 --------+---------+--------
 5  4  2 | 8  3  6 | 1  9  7
 9  8  6 | 2  1  7 | 5  4  3
 1  7  3 | 4  9  5 | 6  2  8
As you can see, 624 % 11 = 613 % 11 = 536 % 11 = 217 % 11 =8

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5
  • $\begingroup$ Is that the only solution? $\endgroup$
    – Joris Schellekens
    Commented Dec 18, 2019 at 19:15
  • $\begingroup$ @JorisSchellekens As far as I can tell, it's unique. $\endgroup$
    – Avi
    Commented Dec 18, 2019 at 19:15
  • $\begingroup$ rot13(Gurer vf va snpg nabgure fbyhgvba. Jvgu nabgure erznvaqre boivbhfyl.) $\endgroup$
    – Joris Schellekens
    Commented Dec 18, 2019 at 19:19
  • $\begingroup$ @JorisSchellekens Check again, I fixed a derp $\endgroup$
    – Avi
    Commented Dec 18, 2019 at 19:39
  • $\begingroup$ Congratulations. You've found it. $\endgroup$
    – Joris Schellekens
    Commented Dec 18, 2019 at 19:40

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