Question

the magic square given above uses all integer from 7 to 15.the sum of the number in each row column and the two main diagonal are the same.which number would replace*in the given square once completed​

Answer 1

Answer:

It is a very easy question to solve,

So I will give some hints - how to solve it,

follow that and do it very easy,

Solving an Odd-Numbered Magic Square

1

Calculate the magic constant.[1] You can find this number by using a simple math formula, where n = the number of rows or columns in your magic square. So, for example, in a 3x3 magic square, n = 3. The magic constant = n[(n^2+1)/2]. So, in the example of the 3x3 square:

sum = 3 * [(9 + 1) / 2]

sum = 3 * (10 / 2)

sum = 3 * (5)

sum = 15

The magic constant for a 3x3 square is 15.

All rows, columns, and diagonals must add up to this number.

2

Place the number 1 in the center box on the top row. This is always where you begin when your magic square has odd-numbered sides, regardless of how large or small that number is. So, if you have a 3x3 square, place the number 1 in Box 2; in a 15x15 square, place the number 1 in Box 8.

3

Fill in the remaining numbers using an up-one, right-one pattern. You will always fill in the numbers sequentially (1, 2, 3, 4, etc.) by moving up one row, then one column to the right. You’ll notice immediately that in order to place the number 2, you’ll move above the top row, off the magic square. That’s okay — although you always work in this up-one, right-one manner, there are three exceptions that also have patterned, predictable rules:

If the movement takes you to a “box” above the magic square’s top row, remain in that box’s column, but place the number in the bottom row of that column.

If the movement takes you to a “box” to the right of the magic square’s right column, remain in that box’s row, but place,

The above are some small hints,

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