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A normal 3x3 magic square has a sum of 15. So you subtract 3 from each number in the square.

Q: How do you solve a 3x3 magic square with the sum of 6?

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If you have three cells in a row, column, or diagonal, and you know the sum of each, you can find the fourth.

Assuming a 3x3 square, yes. If you want to know a solution where all rows, columns and diagonals sum to 15, it is: 2 9 4 7 5 3 6 1 8

It's not. Take 49 and 16 for example. The square root of the sum is the square root of 65. The sum of the square roots is 11.

sum of 14th square number and 10th square number

25 is a square number, and the sum of 9 and 16 is 25. 9 and 16 are also square numbers.

Related questions

Sum = 3 x centre

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

I recently studied a magic square. It is a square that when each row, diagonal, horizontally, or vertically is added up, it equals the same positive integer.

To make a fraction magic square, start by filling in the grid with fractions so that each row, column, and diagonal has the same sum. Use different fractions that have the same sum but different denominators to create a variety of solutions. You can also adjust the value of the fractions to make the magic square more challenging.