use all digits from 1 - 9 inclusively to make a 3 by 3 magic square.each row ,column and diagonal has the same total

8 1 6

3 5 7
4 9 2

total of each row, column and diagonal is 15

thanks ,but how u work it out?

my method works for any odd-numbered square

draw the square, add an imaginary row to the top and an imaginary column at the right
start with 1 in the middle of the top row and and place consecutive numbers following a righ diagonal pattern.
If you reach the imaginary top row, drop down to the bottom of that column.
If you reach the imaginary right column, place your number in the left-most open space of that row
If you are blocked, drop down one space and continue the pattern.
If you reach the top - right corner, drop down one
(notice the 7 after the 6 for the 3by3 , and the 16 following the 15 in the one below)
Your last number should end up in the middle of the bottom row.
Practise with some larger squares

here is my 5 by 5

17 24 1 8 15
23 5 7 14 16
4 6 13 20 22
10 12 19 21 3
11 18 25 2 9

everything add up to 65

To create a 3 by 3 magic square using the digits from 1 to 9, we need to arrange these digits in a way that each row, each column, and each diagonal has the same total.

Here's an explanation of how you can approach this problem step by step:

1. Begin by placing the number 5 in the center of the square. In a 3 by 3 magic square, the number in the center is always half the sum of all the digits, which in this case is (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9) / 2 = 45 / 2 = 22.5. We can't use fractions, so we round it to the closest integer value, which is 22. Since 5 is the only integer between 1 and 9, it goes in the center.

The magic square now looks like this:
- - -
- 5 -
- - -

2. Next, place the remaining numbers around the center in a clockwise pattern starting from the top-left corner. Place the number 1 in the top-left corner, 9 in the top-right corner, and 3 in the bottom-right corner of the square.

The magic square now looks like this:
1 - 9
- 5 -
- - 3

3. Now, we need to fill in the remaining numbers. Since the sum of each row, column, and diagonal should be 15 (since the total sum of all digits is 45 and 45 divided by 3 is 15), we can calculate the missing values based on the values already placed.

Looking at the square, we can observe that the sum of the top row is 1 + x + 9 = 15, where x is the missing number. Solving this equation gives us x = 5.

The magic square now looks like this:
1 5 9
- 5 -
- - 3

4. Continuing with the same approach, we can determine the missing numbers for the second and third rows. The sum of the second row is 5 + 5 + y = 15, where y is the missing number. Solving this equation gives us y = 5. Similarly, the sum of the third row is z + 5 + 3 = 15, where z is the missing number. Solving this equation gives us z = 7.

The magic square now looks like this:
1 5 9
7 5 3
- - 3

5. Finally, we fill in the remaining cells based on the sum of each column. The sum of the first column is 1 + 7 + a = 15, where a is the missing number. Solving this equation gives us a = 7. Similarly, the sum of the second column is 5 + 5 + b = 15, where b is the missing number. Solving this equation gives us b = 5.

The completed magic square is:
1 5 9
7 5 3
7 5 3

And that's how you can create a 3 by 3 magic square using the digits from 1 to 9 inclusively.