The question on the screen asks how to determine the sum of all numbers in a 3x3 magic square. Based on the provided image, the magic square shown contains numbers arranged such that the sum of every row, column, and diagonal is the same.

In a 3x3 magic square, the sum of the numbers can be found using a simple method:

• The magic constant (the sum of each row, column, and diagonal) is given by the formula:

$M = \frac{n(n^2 + 1)}{2}$

Where $n$ is the size of the square, which in this case is $n = 3$.

So for a 3x3 magic square:

$M = \frac{3(3^2 + 1)}{2} = \frac{3(9 + 1)}{2} = \frac{30}{2} = 15$

The total sum of all numbers in the square is $3 \times 15 = 45$.

Given the provided choices, let's analyze the options:

• Option A: Multiply the center number by 9. This would work because in a 3x3 magic square, the center number is always 5 (for squares with consecutive numbers 1 to 9), and $5 \times 9 = 45$, which is the correct sum of the magic square.
• Option B: Multiply the largest number by 5. The largest number in the square is 9, and $9 \times 5 = 45$, so this also works.
• Option C: Multiply the bottom middle number by 6. This would depend on the specific value of that number, but it generally doesn’t work for all cases.
• Option D: Multiply the top left number by 10. This is also unlikely to work universally for magic squares.

Therefore, the best answer here is Option B: "Multiply the largest number by 5."

Would you like further details, or do you have any questions?

Here are 5 related questions:

1. How is a magic square defined?
2. What is the general formula for the magic constant in an $n \times n$ magic square?
3. What are the properties of a 3x3 magic square?
4. Why does multiplying the center number by 9 also work for finding the sum in a 3x3 magic square?
5. How can magic squares be extended to larger sizes?

Tip: In magic squares, the sum of any row, column, or diagonal always equals the magic constant.