4*4 array...

[man111]

Find the numbers of [tex]4\times 4[/tex] array whose entries are from the set [tex]\left\{0,1,2,3\right\}[/tex] and which are such

that the sum of the numbers in each of the four rows and four columns is divisible by [tex]4[/tex].

[mkmarinov]

Can a number belong more than once in the same row?
i.e. is the row (0,0,0,0) acceptable?

[man111]

Yes it s acceptable.....

[drago]

Let G be an abelian group of all 4x4 arrays mod (4) (above[tex]Z_4[/tex]) with natural sum of arrays.
Let [tex]H \subset G, h \in H[/tex] if sum of the numbers in each row and column of [tex]h[/tex] is 0 (divisible by 4). Obviously [tex]H[/tex]is a subgroup of [tex]G[/tex]. Then [tex]|G|=|H|.|G/H|[/tex] ,([tex]G/H[/tex]is the quotient group).
[tex]|G|=4^{16}.[/tex] Lets find [tex]|G/H|[/tex]. There is a bijection between elements of [tex]G/H[/tex] and possible 8 sums of 4 rows and 4 columns. If [tex]c_1,...,c_4, r_1,...,r_4[/tex]are these sums then [tex]r_4=c_1+...+c_4-r_1-r_2-r_3.[/tex] If we choose arbitrary the latter 7 numbers (mod 4), then there exists an array [tex]a_i_j[/tex] with these sums. For example: [tex]a_{1,1}:= c_1, a_{1,2}=r_1-a_{1,1},a_{2,2}=c_2- a_{1,2}, a_{2,3}=r_2-a_{2,2},...,a_{4,4}=c_4-a_{3,4}[/tex]and all of the rest elments of [tex]a_{i,j}[/tex] are 0.
So [tex]|G/H|=4^7[/tex]and [tex]|H|=4^9[/tex].

[man111]

Drago thanks for nice answer.

is there is any other method....

[allier]

Drago's solution is in fact very simple. If you arbitrarily complete the upper left 3 by 3 array, there is a unique way to choose the rest of the elements (since you got all possible remainders mod 4). Thus, the answer is simply [tex]4^9[/tex], i.e. the number of ways to choose the 9 elements of the 3 by 3 array.

[drago]

Drago's solution is in fact very simple. If you arbitrarily complete the upper left 3 by 3 array, there is a unique way to choose the rest of the elements (since you got all possible remainders mod 4). Thus, the answer is simply [tex]4^9[/tex], i.e. the number of ways to choose the 9 elements of the 3 by 3 array.

Yes, It is very simple, and no need of groups and etc...
Congratulations!
But it is Yours solution, not mine.