Counting Problem

here is one

1  2  3

5  4  6

7  8  9

11 10 12       there are many more....see Alan's answer below:

Think of each number in terms of their remainder when divided by 3.

1 has rem 1

2 has rem 2

3 has rem 0

4 has rem 1

5 has rem 2

6 has rem 0

7 has rem 1

8 has rem 2

9 has rem 0

10 has rem 1

11 has rem 2

12 has rem 0

In order that each column sums to a number divisible by three we must have combinations such that each number with remainder 0 is combined with one of the numbers with remainder 1 and one with remainder 2.

There are 4*4*4 = 64 ways of doing this. This is the result if the columns are distinguishable. If they aren't distinguishable this number should be divided by 4 (I think!).

I started with Alan's technique.  

Thanks Alan I would not have thought of it myself.

I am thinking of the numbers as four distinct zeros, 4 distinct ones and 4 distinct twos.

I am going to say it is 3 columns and 4 rows with each row adding to a multiple of 3 because that will be much easier for me to work with.

Here is one possibility

0,0,0,0     4! = 24

1,1,2,2     4*3*4*3  * 4!/(2!2!) = 144*6 = 864

1,1,2,2     4!                     =24

So far I get 24*864*24 = 497664 ways

But the rows can be in any order. I think that is 3 ways =3 ways

So that is  497664*3 = 1 492 992 ways.

Another possibility is

2,2,2,0    4*3*2*4*4 = 384 ways

1,2,0,0    4*1*4*3*(4!/2!) = 48*12 = 576 ways

1,1,1,0    3*2*4 = 24 ways

384*576*24 = 5308416

Times by 6 = 31 850 496 ways

Another possibility is 

1,2,0,0    4*4*4*3*(4!/2!) =   193*12 =  2304

1,2,0,0    3*3*2*1*12      =    18*   12=  216

1,2,1,2     2*1*2*1*(4!/4) =     4*    6 =     24

2304*216*24= 11 943 936

Times by 3 = 35 831808

1 492 992+31 850 496+35 831808 = 69 175 296 ways

This does seem like a lot and maybe I have double counted

but if there were no restrictions at all then there would be 12! ways = 479 001 600.

69175296/497001600 = 0.1391852581561106

So my answer is only 14% of all possible answers with no conditions at all.

So it may be correct. It passes my reasonableness test.

I found that a few other combinations of remainders are three 0 rems and two 2 rems and one 1 rem. I hope this will help...