1consider
4k, 4k+1,4k+2 and 4k+3
now,
for a fixed k,
for any choice of a,b, and c out of these, (each can be chosen any number of times), there exists just one of the 4, which can be d, such that,
a+b+c+d= multiple of 4.
now call element in ith row and jth column as aij
what we'll do is that we fill up the ith column, then go to the ith row, then the i+1th column and so on...
to fill up the first column,
we have 4*4*4*1 ways (a11,21,31filled with any number and the last one left with just one option)
lets look at the 1st row now,
with a11 fixed, 12, 13 can be fixed in 4*4 ways for which there is only one way of a14
(total ways so far are 43*42)
now, look at column 2,
with a12 fixed, we have 4*4 ways of fixing a22,32
(total ways so far =43*42*42)
similarly now, for row 2,
progressing similarly we have 4*1 ways for this, which makes the total count 43*42*42*4
now going to column 3,
and doing it the same way, we have the final count as 43*42*42*4*4
moving to the 3rd row, we see that there is just 1 way left,
and looking at the 4th column, we see again 1 way left,
so total count= 43+2+2+2=49
check calculations again.
cheers!
