Find a magic square of order 3 with a sum of 18

Find a magic square of order 3 with a sum of 18

The so-called magic square, also known as the vertical and horizontal graph, is to put N and 178 natural numbers starting from 1 in the n × n square matrix. Under a certain layout, the sum of the numbers in each row, column and two diagonal lines is exactly equal
The relationship between magic square and s and the order n of square matrix is as follows:
S=n（n²+1）/2
Sum of magic squares of order 3 = 3 × (3 & # 178; + 1) / 2 = 15
Therefore, it is impossible to construct a magic square with a sum of 18

How to write a magic square with 16 numbers from 11 to 26!

11 25 24 14
22 16 17 19
18 20 21 15
23 13 12 26

Use the nine numbers 6, 8, 10, 12, 14, 16, 18, 20 and 22 to form a magic square. Which number should be filled in a, B, C and D respectively
8 22 A
B 14 C
16 D 20

Imagine the Republic of magic squares: 8 + 14 + 20 = 42 (sum of three numbers of diagonal)
So it is easy to calculate a = 12
b=18
c=10
d=6
Verification, no error

For the third-order magic cube formula, I want the way of digital expression, such as 1 on 3!

Is it a beginner or an advanced one