7
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partially completed grid

There are lots of pairs and I don't want to assume any values.

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  • 2
    $\begingroup$ The cell with 125689 should be 125 because there is an 8 and 9 in the column along with a 36 36 pair. The cell with 124689 should be 12 for the same reasons plus there is a 4 in the column. $\endgroup$ – JS1 Sep 13 at 4:10
  • $\begingroup$ If this were a bit more difficult, it could be the peak of pranking offices for their workers' time! $\endgroup$ – ti7 Sep 13 at 15:28
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The bottom right hand corner, top row has to contain a 4 and a 5, which means that the bottom left hand corner top row cannot contain a 4; meaning bottom left corner top middle row must be a 1

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Bottom left (A9) can't be a $6$ due to D9=$6$, and so it is a $3$. In fact, A7=$6$ by basic Sudoku logic ($6$ in row 9, $6$ in column B).

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One step I can see is that in the bottom right block you've got two boxes which are either 4 or 5. This means that one of them must be a 4, which in turn means that the 1/4 on the far left on the same row can't be a 4 and must be a 1. This is known as a "pointing pair".

Obviously that doesn't complete the sudoku, but should at least get you unstuck.

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Your candidates are wrong. There is no 6 as a candidate in A9 leaving the 3 as a naked single. The whole bottom row can be solved as it has a hidden single for 4 in B9 and the last cell (F9) will be a 1. Another 1 will go on B7 and a 3 on A7, finishing the bottom left sub-grid.

You must also fix the candidates in the middle-right sub-grid. Especially on row H, there are no candidates for 8 or 9 as they are already filled.

With all that in mind, you could pay more attention to removing the candidates from places where they can't be anymore because they were already filled in the row, column or sub-grid.

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Bottom most row (assuming your marking are correct, I didn't verify you): there's only one place where a 4 can go, once that's done, there's only one place where the 1 can go, which will solve the 3.

With the bottom row completely done, you should be able to move on quite swiftly.

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  • $\begingroup$ Just took a loot at your markings anyway... you should pay more attention to your own marks: in square A9 you marked both 3 and 6 as possible... I'll leave it up to you to tell me why a 6 is not possible, which makes my answer here redundant. $\endgroup$ – O.F. Sep 13 at 14:31
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Once you resolve the problems with your markings, which will solve the bottom left sub-grid and reduce the markings in the right middle subgrid, you're left with a puzzle that has multiple solutions. That means it isn't a valid Sudoku at all -- at least as I've been taught, a Sudoku must have one unique solution.

An automated solver I use says that there are 25 solutions to this puzzle. To pick one, you have no choice but to assume values.

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