Can you find the other possible positions for 1 and 9?

Now, we have one full row or column of the magic square! Try completing it!

[Hint: First fill the row or columns containing 1 and 9]

Question

Can you find the other possible positions for 1 and 9? Now, we have one full row or column of the magic square! Try completing it! [Hint: First fill the row or columns containing 1 and 9]

Homi · Accepted Answer

A 3x3 magic square using numbers 1 to 9 has a magic constant of 15 and 5 in the center.

Step 1 — Finding other positions for 1 and 9

The center number in a 3x3 magic square (1-9) is always 5. Any line (row, column, or diagonal) passing through the center must sum to 15. If 1 and 5 are in a line, the third number must be 9. This is because $15 - 1 - 5 = 9$ . So, 1 and 9 must always be in a line with 5. The given images show 1, 5, 9 in the middle row and middle column. The other lines passing through the center are the two main diagonals. Therefore, 1 and 9 can be in the corners. If 1 is in the top-left corner, 9 must be in the bottom-right corner. If 1 is in the top-right corner, 9 must be in the bottom-left corner.

Diagram 1

Step 2 — Completing the magic square

Let us complete the magic square where 1, 5, 9 are in the middle row. The magic constant is 15. The numbers 1 to 9 are used, each exactly once. Let the square be represented as: $\begin{array}{|c|c|c|} \hline A & B & C \\ \hline 1 & 5 & 9 \\ \hline G & H & I \\ \hline \end{array}$ The numbers used are 1, 5, 9. The remaining numbers are {2, 3, 4, 6, 7, 8}. Let us write down the sum equations for rows, columns, and diagonals.

The middle row sum is already 15: $1 + 5 + 9 = 15$

For the first column: $A + 1 + G = 15$ $A + G = 15 - 1$

$\boxed{A + G = 14}$

For the middle column: $B + 5 + H = 15$ $B + H = 15 - 5$

$\boxed{B + H = 10}$

For the third column: $C + 9 + I = 15$ $C + I = 15 - 9$

$\boxed{C + I = 6}$

For the main diagonal (top-left to bottom-right): $A + 5 + I = 15$ $A + I = 15 - 5$

$\boxed{A + I = 10}$

For the anti-diagonal (top-right to bottom-left): $C + 5 + G = 15$ $C + G = 15 - 5$

$\boxed{C + G = 10}$

We have the following equations for the unknown numbers:

$A + G = 14$
$B + H = 10$
$C + I = 6$
$A + I = 10$
$C + G = 10$

Let us find pairs from the remaining numbers {2, 3, 4, 6, 7, 8}. From equation (1), $A+G=14$ . Possible pairs are (6, 8) or (8, 6). From equation (3), $C+I=6$ . Possible pairs are (2, 4) or (4, 2).

Let us try setting $A = 8$ . Using equation (1), $A + G = 14$ : $8 + G = 14$ $G = 14 - 8$ $G = 6$ Using equation (4), $A + I = 10$ : $8 + I = 10$ $I = 10 - 8$ $I = 2$ Now we have $A=8$ , $G=6$ , $I=2$ . These are distinct and from the remaining numbers. Using equation (3), $C + I = 6$ : $C + 2 = 6$ $C = 6 - 2$ $C = 4$ Let us check this with equation (5), $C + G = 10$ : $4 + 6 = 10$ This is correct.

So far, we have found: $A = 8$ $C = 4$ $G = 6$ $I = 2$ The numbers used are 1, 2, 4, 5, 6, 8, 9. The remaining numbers for $B$ and $H$ are {3, 7}. Using equation (2), $B + H = 10$ : $3 + 7 = 10$ This is correct. We can set $B=3$ and $H=7$ .

Now, let us fill the magic square with these values: $\begin{array}{|c|c|c|} \hline 8 & 3 & 4 \\ \hline 1 & 5 & 9 \\ \hline 6 & 7 & 2 \\ \hline \end{array}$ Let us verify all sums: Rows: $8+3+4=15$ , $1+5+9=15$ , $6+7+2=15$ . Columns: $8+1+6=15$ , $3+5+7=15$ , $4+9+2=15$ . Diagonals: $8+5+2=15$ , $4+5+6=15$ . All sums are 15. The magic square is complete.

Diagram 2

Answer

(i) The other possible positions for 1 and 9 are along the diagonals. (ii) If 1 is in the top-left corner, 9 is in the bottom-right corner (with 5 in the center). (iii) If 1 is in the top-right corner, 9 is in the bottom-left corner (with 5 in the center). (iv) One completed magic square is: $\begin{array}{|c|c|c|} \hline 8 & 3 & 4 \\ \hline 1 & 5 & 9 \\ \hline 6 & 7 & 2 \\ \hline \end{array}$

Question 29

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