The digits form three 3-digit numbers reading left-to-right, and three 3-digit numbers reading top-to-bottom. Also consider these six 3-digit numbers reversed, that is, reading right-to-left and bottom-to-top.

For each of those nine digits, d, you count how many of the twelve 3-digit numbers are divisible by the number d. In the case of each d, that number, the count, is itself divisible by d.

How are the numbers arranged, with the understanding that rotations and reflections of that arrangement are possible alternatives?

Im not all the way there yet, but I'm close..

The divisibility condition for 5 and 2 depend only on the last digit of the 3 digit numbers (=3dns). Therefore, the number of occurences of numbers divisible by 2 (5) is the number of occurences of an even number (multiple of 5) at an edge midpoint (=emp) plus twice the number of occurences at a corner. It follows that {5} must be located at the centre and that there are an even number of emps occupied by evens.

The divisibility condition for 9 depends on the sum of the digits of the 3dns and, as such, is independent of the arrangement of digits (!) The number of 3dns divisible by 9 is divisible by 9 if 4 times the sum of the digits 1..9 is divisible by 9. This condition is met (which also implies the the condition for 3 is met as well), so we are still in the game.

Let’s take a closer look at the evens. If we locate all the evens at the emps, there will be exactly 1 3dns divisible by 8. Namely, the 3dn ending in 56 No good. If we locate all the evens in the corners, then we have exactly 2 3dns divisible by 8, namely those ending in 6. Again, no good. So we are left splitting the evens between corners and emps, and have to take care to place 6 in a corner diagonally opposite one of {48} so as to avoid the 3dn 256 (divisible by by 8!). This also assures six 3dns divisible by 6, as there will be exactly six even 3dns.

We are left to consider arrangement that produce either seven 3dns divisible by 7, or none. This choice is decisive for deciding which even to put in the corner opposite {6}, as 658 is a multiple of 7.

Finding an arrangement that gives exactly sevens multiples of 7 sounds like a tall order, as this would uniquely constrains a lot of locations. Let’s try:

XX8

X5X

6XX

of the emp combinations (A5B), 952 is uniquely prodcutive, as it gives multiples of 7 in both directions. The only possibility to get a multiple of 7 ending or beginning with {4} is 154. This leaves {37} to fill the empty corners, in the best case producing two further multiples of 7. This leads to

798

154

623

which comes close with six multiples of 7, but still has to be rejected.

It seems is easier to avoid multiples of 7. For instance,

391

752

684

does the trick. Or, proceeding more systematically, we start with

ab4

c5d

6ef

2a + b <> 3,10,17,24

a + b <> 6,13

2a+c <> 8,15,22

a+c <> 9,16

2c+d <> 6,13,20

c+2d <> 6,13,20

e+f <> 9,16

e+2f <> 8,15,22

2b+e <> 6,13,20

b+2e <> 6,13,20

d+f <> 6,13

d+2f <> 10,17,24

this leads to a set of possibilities. Maybe if i have energy and interest in the next days i will post it.