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.
© Rishabh Bidya
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
Rishabh, I had not thought of using a variable "a", brilliant !! :)
ReplyDeleteThanks a lot, Jaishankar !! The general solution of magic squares coming soon :D
DeleteHowever: left to right, top to bottom diagonal is 3a + 3... hm?
DeleteRight Pablo! Because if you see the question statement, I've merely considered the sum of row and column elements. The sum of right to left, top to bottom diagonal elements is a mere co-incidence.
Delete