Q. 34.7( 3 Votes )

# Starting with the

Answer :

Here, we are supposed to start with the middle cell in the bottom row of the square using numbers from 1 to 9 in the 3×3 magic square. This is another way of filling magic square of odd-numbered sides.

First, let us understand what a 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”.

Now, let us try to create one using the numbers from 1 to 9 stepwise.

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

This is the step according to the question.

**Step 2:** Fill the next number, that is, number 2 to the one-lower row and to the one-left column.

The question arises, as to why did we place number 2 in the uppermost row?

Notice, there is no row below the row where number 1 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” below the magic square’s bottom-most row, remain in the one left box’s column, but place the number in the upper row of the column.

**Step 3:** Similarly, fill the next number 3 in the one-lower row and one-left column.

Since there is a row below the row where number 2 is placed, but no column to number 2 column’s left. We have placed the number 3 to the right-most column of the middle row.

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 left of the magic square’s left column, remain in the one below row, but place the number in the furthest right column of that row.

**Step 4:** Similarly, place the next number 4in the one-lower row and one-left column.

Here, we have placed 4 above 3, because there was no space for the decided movement. It was already occupied by number 1.

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 above 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 = 2 + 9 + 4 = 15

Sum of second row = 7 + 5 + 3 = 15

Sum of third row = 6 + 1 + 8 = 15

Sum of first column = 2 + 7 + 6 = 15

Sum of second column = 9 + 5 + 1 = 15

Sum of third column = 4 + 3 + 8 = 15

Sum of diagonal = 2 + 5 + 8 = 4 + 5 + 6 = 15

⇒ Magic sum = 15

Thus, the magic square so constructed is correct.

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