Latest Self Aptitude Question SOLUTION: arrange consecutive 16 numbers in a 4 X 4 mtrix such that the sum of elements in diagonal =sum of elemnets in each row =sum of elements in each column

let no will be 34
34-30=4
4/4=1
bottom no is 1
increment diagonal by 5=(16 11 6 1)
left bottom n+3=4 and increment diagonal by 3(4 7 10 13)
now 8 are remain let have n+1 =2 and fill remaining box (2 3 4 7 8 11 13 14)

we need to arrange 4*4 matrix in such a way that sum of element of each row = sum of each column = sum of each element of diagonal . so first we need to calculate the sum that we need to get after summing up the elements of a row or column or diagonal it is calculated by this formula :- n^ 2(n^2+1)/2 where n is the order i.e 4 coz we are working for 4*4 matrix it would have been 10 if we worked on 10*10 matrix.
so we have the sum as 34 for each row or column or diagonal .
make a matrix of 4*4 and write numbers from 1 to 16 starting from P(11)to P(44) .

now mark the diagonals and swap values on the basis of this logic :-
if magic sq. has order n, swap the no. that add up to (n^2 +1).

for this problem we are having order 4 and therefore we get 4^2+1 = 17 . for eg. 16 and 1 , 7and 10 etc
the matrix u receive after swapping the values of the diagonal will be ur answer as that matrix would have sum of rows elements = column elements = diagonal elements = 34 .

matrix would look like :-
16 2 3 13
5 11 10 8
9 7 6 12
4 14 15 1

remember this method is only applicable for even order matrix i.e matrix with order 2 ,4,6,8,10,12 etc

really sorry but in the answer that i submitted has one correction i wrote a formula to calculate the sum of the row or column or diagonal elements respectively is

M(n) = n(n^2+1)/2 where n = order of matrix.