Magic Squares 2 : Creating a 3x3 magic square using 9 consecutive integers given the sum of a row or a column

Problem Background : A 3x3 magic square is an arrangement of 9 squares arranged in 3 rows and 3 columns. We are given a sum S which is the sum of any row or column. We need to arrange 9 consecutive integers in the square.

Solution : If the sum of a row is S, S also needs to be the sum of any other row or any other column.

Derived corollary : It can be noted that for the numbers in individual squares to be integers, S must be divisible by 3, i.e. S must be 0(mod3) and S must be greater than 12.

The above result can be derived from  Magic Squares 1 :

After this, we merely substitute for a and simplify.

Q.E.D

© Rishabh Bidya

Magic Squares 1 : To create a 3x3 magic square given 9 consecutive integers

Problem Background : A 3x3 magic square is an arrangement of 9 squares arranged in 3 rows and 3 columns. We are given 9 consecutive integers to place in each of the squares such that the sum of each of the rows and columns is equal.

Solution : If the given integers begin from a, rest if the integers are a+1, a+2, a+3, ..., a+7, a+8.

In the above figure, Sum of every row as also sum of every column is 3a+12.

Q.E.D

© Rishabh Bidya