You can't do it with 9 different integers. Either you repeat integers or you use fractions.
Repeating integers gives the trivial solution 4-4-4; 4-4-4; 4-4-4 so we head into fraction country.
Top row: 1.6 - 7.2 - 3.2; middle row: 5.6 - 4 - 2.4; bottom row: 4.8 - 0.8 - 6.4
Sum = 3 x centre
A 3x3 magic square has the property that the sum of the numbers in each row, column, and diagonal is the same. For a 3x3 magic square using the numbers 1 to 9, the magic constant is 15, not 18. If you're referring to a different set of numbers or a modified version of a magic square, please specify the numbers used to achieve a magic constant of 18.
Yes, it is not that difficult.
use the inverse square method, it works the fastest
3*3=9 9 is the answer.
A normal 3x3 magic square has a sum of 15. So you subtract 3 from each number in the square.
Sum = 3 x centre
To solve a 3x3 magic square with decimals, you need to ensure that the sum of numbers in each row, column, and diagonal is equal. Start by placing the decimal numbers in a way that each row, column, and diagonal sums up to the same value. Adjust the numbers carefully to achieve a valid solution.
A 3x3 magic square means that each row, each column, and both diagonals all have the same sum.
Yes, it is not that difficult.
use the inverse square method, it works the fastest
3*3=9 9 is the answer.
[ -8 ] [ -1 ] [ -6 ][ -3 ] [ -5 ] [ -7 ][ -4 ] [ -9 ] [ -2 ]The sum of each row, column, and diagonal is -15.
34
There are 9 numbers. Assuming the question refers to a 3x3 "magic" square, the answer is no. The sum of all nine numbers is 36 so each of the 3 rows must sum to 12.
the magic sum is 15
45