In a 12 digit number, each digit contributes its value to the sod;

and also it contributes to the mod 37 value of the whole number which depends only on its value and its position in the number; what power of 10 it is multiplied by.

If you consider the contribution to the mod 37 value of a 12 digit repunit, 111111111111, the contribution to mod 37 is:

    1,10,26,1,10,26,1,10,26,1,10,26

So for the 12 digit number to be 0 mod 37, the number of occurrences of each of 1, 5, and 9 would have to be a multiple of 3 in positions contributing 1, 10, and 26 to the mod value.

For example 111555999111 should be divisible by 37 (and it is).  Another example would be 159591915999.

Next:  count how many ways you can have 12 integers sum to 76 where each is either 1, 5, or 9.

1s 5s 9s

0 8 4

1 6 5

2 4 6

3 2 7

4 0 8

In none of these cases is each divisible by 3.

Therefore there are no such 12 digit numbers.

--------------------------------

for n1 in range(13):

    for n5 in range(13):

        for n9 in range(13):

            if n1 + 5*n5 + 9*n9 == 76  and  n1+n5+n9 == 12:

                print(n1,n5,n9)

for expo in range(12):

    print((10**expo)%37)