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)
• 8 years agoHelpfull: Yes(0) No(0)
• 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
• 8 years agoHelpfull: Yes(0) No(0)
• 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.
• 8 years agoHelpfull: Yes(0) No(0)