There is another 3x3 square of numbers, with some entries missing, such that the first number in each row or column divided by the second equals the third entry in that row or column. It so happens that we can write four numbers in the square in almost any way so that the rest of the square can be completed.

\[ \begin{array}{ccc} 2 & 3 & ? \\ 4 & 6 & ? \\ ? & ? & ? \end{array} \]

The square is completed for the first two rows with

\[2 \div 3=6\]

\[4 \div 6=2/3\]

We have then

\[ \begin{array}{ccc} 2 & 3 & 2/3 \\ 4 & 6 & 2/3 \\ ? & ? & ? \end{array} \]

Now the columns.

\[ \begin{array}{ccc} 2 & 3 & 2/3 \\ 4 & 6 & 2/3 \\ 1/2 & 1/2 & 1 \end{array} \]

.Now notice for the last row

\[1/2 \div 1/2 =1\]

.This always happens.

If we fill out the square differently, say

\[ \begin{array}{ccc} 1 & ? & 7 \\ 1 & ? & 8 \\ ? & ? & ? \end{array} \]

.Completing the columns first gives

\[ \begin{array}{ccc} 1 & ? & 7 \\ 1 & ? & 8 \\ 1 & ? & 7/8 \end{array}\]

.Completing the rows gives

\[ \begin{array}{ccc} 1 & 1/7 & 7 \\ 1 & 1/8 & 8 \\ 1 & 8/7 & 7/8 \end{array}\]

.Again the first enty in a row or column divided by the second entry in that row or column equals the third entry.