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In this puzzle, each square must contain a positive integer. All rows, columns and boxes must contain an arithmetic progression of length 9.

Good luck!

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7
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Is it the following?

enter image description here

Very broadly, other than standard sudoku solving techniques:

for each row,col,square find the largest common divisor of all the numbers known so far, the step size has to divide that number.

in some cases also considering the spread, if, as for example in column 4, there are two numbers more than eight apart, the step size must be larger than one. As the lcd is 2 there, we know this column must hold 1,3,5,...,17.

We can use this kind of info in much the same way as with standard sudoku to rule out some possibilities, for example from what we know about column 4 we now can tell that the step in the bottom center square must be one, not seven. From this it follows that column 5 also cannot have step 7 but must have step 1.

intersection of ranges is another motive: now that we know bottom center square has range 1,..,9 and the middle column has either range (a) 7,...,15 or (b) 8,...16, we can tell that the intersection of these two shapes must contain 9,8, and since there are three slots 7, ruling out b.

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    $\begingroup$ Perhaps the steps you took in solving it? $\endgroup$ – justhalf Aug 2 at 9:12
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    $\begingroup$ @justhalf for a sudoku? Wouldn't that be almost as boring for you to read as it would be for me to write down? That said, I could perhaps explain a bit the non-standard-sudoku steps I used. That what you were thinking about? $\endgroup$ – Paul Panzer Aug 2 at 9:22
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    $\begingroup$ @PaulPanzer Yes, you can explain the non-standard steps, that's how it's customarily done here, especially since this is a non-standard Sudoku puzzle. See these answers 1, 2, 3, for example. $\endgroup$ – justhalf Aug 2 at 9:28
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    $\begingroup$ @TobyMak Anything wrong with that? I think programming a custom solver would be a feat in itself. And there is no no computers tag. Or is that implied with sudoku? $\endgroup$ – Paul Panzer Aug 2 at 10:25
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    $\begingroup$ This puzzle is relatively easy, but the main point of interest was that such a (non-trivial) construction is actually possible i.e. the deltas between all arithmetic progressions are not always 1. I don't even know if this can be extended to a 16x16 sudoku. This is left as a proverbial exercise for the reader :) $\endgroup$ – happystar Aug 2 at 12:24

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