Construct a 5 × 5 magic square using all even numbers from 1 to 50.

First, let us understand what magic square is.

A magic square is an arrangement of numbers in a square in such a way that the sum of each row, column, and diagonal is one constant number, the so-called “magic sum” or sometimes “magic constant”.

Collect all even numbers from 1 to 50.

They are: 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36, 38, 40, 42, 44, 46, 48, 50.

Now, let us try to create a magic square using these even numbers from 1 to 50 stepwise.

Step 1: Place 2 (the smallest number is given) in the centre box in the top row.

This is where you always begin when your magic square has odd-numbered sides, regardless of how large or small that number is.

For example, in 5×5 square, you place number 2 in box 3. Similarly, if you had 11×11 square, you would have placed your number in box 6.

Step 2: Fill the next number, that is, number 3 to the one-upper row and to the one-right column.

The question arises, as to why did we place number 4 in the bottom-most row?

Notice, there is no row above the row where number 2 is placed.

This is a case while placing numbers in a magic square having odd-numbered sides:

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

Step 3: Similarly, fill the next number 6 in the one-upper row and one-right column.

Since there is a row above the row where number 4 is placed, and a column to number 4 column’s right. We have placed the number 6 to the one-upper and one-right cell.

Step 4: Similarly, place the next number 8 in the one-upper row and one-right column.

Since there is a row above the row where number 6 is placed, but no column to number 6 column’s right. We have placed the number 8 to the left-most column of the row, above the row of number 6.

This is another case while placing numbers in a magic square having odd-numbered sides:

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

Step 5: Again, place the next number 10 in the one-upper row and one-right column.

Since there is a row above the row where number 8 is placed, and a column to number 8 column’s right. We have placed the number 10 to the one-upper and one-right cell.

Step 6: Again, place the next number 12 in the one-upper row and one-right column.

Here, we have placed 12 below 10, because there was no space for the decided movement. It was already occupied by number 2.

This is the third case while placing numbers in the magic square having odd-numbered sides:

If the movement takes you to a box that is already occupied, go back to the last box that has been filled in, and place the next number directly below it.

We have covered all the possible cases while filling odd-numbered side magic square. Now, repeat the process.

We get,

Check:

If magic sum or magic constant can be calculated using the constructed magic square, then the magic square is correct.

So, let us calculate the magic sum.

Sum of first row = 34 + 48 + 2 + 16 + 30 = 130

Sum of second row = 46 + 10 + 14 + 28 + 32 = 130

Sum of third row = 8 + 12 + 26 + 40 + 44 = 130

Sum of fourth row = 20 + 24 + 38 + 42 + 6 = 130

Sum of fifth row = 22 + 36 + 50 + 4 + 18 = 130

Sum of first column = 34 + 46 + 8 + 20 + 22 = 130

Sum of second column = 48 + 10 + 12 + 24 + 36 = 130

Sum of third column = 2 + 14 + 26 + 38 + 50 = 130

Sum of fourth column = 16 + 28 + 40 + 42 + 4 = 130

Sum of fifth column = 30 + 32 + 44 + 6 + 18 = 130

Sum of diagonal = 34 + 10 + 26 + 42 + 18 = 30 + 28 + 26 + 24 + 22 = 130

⇒ Magic sum = 130

Thus, the magic square so constructed is correct.