Question

1. put the numbers 1, 2, 3, 4, 5, 6, 7, 8 and 9 in the circles below so that the sum of each row, column and diagonal is the same. working copy here is one solution

Explanation:

Step1: Recall magic - square concept

This is a 3x3 magic - square problem. The sum of all numbers from 1 to 9 is given by the formula $\sum_{n = 1}^{9}n=\frac{9\times(9 + 1)}{2}=45$. In a 3x3 magic - square, since the sum of each row, column, and diagonal is the same, and there are 3 rows (or columns), the common sum $S$ is $\frac{45}{3}=15$.

Step2: Identify the middle number

The middle number of the 3x3 magic - square is always 5. This is because when considering the sums of rows, columns, and diagonals passing through the middle, the middle number is used in 4 different sums.

Step3: Pair numbers

We need to pair the remaining numbers to get a sum of 10 (since $15-5 = 10$). The pairs are (1,9), (2,8), (3,7), and (4,6).

Step4: Arrange the numbers

By trial - and - error and symmetry considerations, we can arrange the numbers as follows:

4	9	2
3	5	7
8	1	6

Answer:

4	9	2
3	5	7
8	1	6