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Take all the integers below 100 that don’t contain a digit 9. There are 81 of them.

Is it possible to arrange these 81 numbers in a 9×9 grid in such a way that both the one’s places and the ten’s places form two separate Sukodu squares (using the digits 0–8 instead of 1–9)?

(I’m currently running a brute-force algorithm to discover a solution, but I have reason to believe that it’ll take longer than the universe, therefore any clever input is appreciated.)

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    Can't you just put two different sudokus together (using one to define the 10s place, and the other the 1s place)? – 2018-04-17

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You're looking at order-9 Mutually Orthogonal Latin Squares (MOLS).

Try looking at the data at Data on MOLS.

Also try a search on "orthogonal sudoku".

Here's a puzzle version by Paul Vaderlind.

And the solution:

59 23 96 | 18 42 75 | 84 61 37
88 62 35 | 57 21 94 | 16 43 79
17 41 74 | 89 63 36 | 55 22 98
---------+----------+---------
26 99 53 | 45 78 12 | 67 34 81
65 38 82 | 24 97 51 | 49 76 13
44 77 11 | 66 39 83 | 28 95 52
---------+----------+---------
93 56 29 | 72 15 48 | 31 87 64
32 85 68 | 91 54 27 | 73 19 46
71 14 47 | 33 86 69 | 92 58 25
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    Sudokus are more restrictive than Latin Squares because of the requirements on the $3 \times 3$ boxes. – 2017-01-13
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    Ross -- correct. But there are fewer order-9 MOLS than there are Sudoku, so start with the smaller set with existing data and see which ones satisfy the additional requirement. – 2017-01-13