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 :
© Rishabh Bidya
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
It is interesting to note that in the previous solution, the additions were numbers 0-8. In which some were divisible by 3 and the rest not.
ReplyDeleteBut in this generalised solution, all numbers 'n' in (S+n)/3 or (S-n)/3 are all divisible by 3. Hence this makes all of the squares integers, given that S/3 is integer. Interesting I must say :)
Interesting indeed !! Hence the derived corollary too :)
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