Problem 954

First, it is investigated which of the numbers $1,2, \ldots, 30$ are suitable for the solution. For this purpose, the product 270 is factored into prime factors: $270=2 \cdot 3 \cdot 3 \cdot 3 \cdot 5=2^{1} \cdot 3^{3} \cdot 5^{1}$.

The numbers to be inserted may therefore only contain the factors $2,3,3^{2}, 3^{3}, 5$. These are the 10 numbers 2, 3, 5, 6, 9, 10, 15, 18, 27, and 30, and also the number 1. Of these 11 numbers, 2 must be eliminated. The product of all 11 numbers is $2^{5} \cdot 3^{11} \cdot 5^{4}$, while the product of the 9 numbers to be squared is $270^{3}=2^{3} \cdot 3^{9} \cdot 5^{3}$. The product of all 11 numbers thus contains the factor $2^{2} \cdot 3^{2} \cdot 5=180$ in excess compared to the product of the 9 numbers squared. The product of the two numbers to be eliminated is therefore 180.

There are two possible cases: $180=6 \cdot 30=10 \cdot 18$.

Now we check which arrangements of the prime factors are possible. Since each of the prime factors 2, 3, and 5 must occur in the same number in each row and in each column if the conditions of the problem are to be met, the following arrangements are possible:

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The individual arrangements can be transformed into each other a) by swapping columns b) by swapping rows

Due to the equivalence of these arrangements, it does not matter which field the 1 is placed in. If the upper left field is chosen, the schemes I and II are no longer applicable for placing the remaining factors.

Since only 4 schemes (III to VI) are available for the distribution of the prime factors 2, 3, 3, 3, 5, two factors are combined: $3 \cdot 3=9$. Thus, $2,3,5,9$ are distributed.

It must be noted that the factors 5 and 9 must be distributed in such a way that they do not meet in a common field; because the product $5 \cdot 9=45$ is outside the allowed numbers. Therefore, the numbers 5 and 9 are placed according to schemes III and VI (or IV and V) as a partial solution (a):

$$\begin{array}{lll} 1 & 5 & 9 \\ 5 & 9 & . \\ 9 & . & 5 \end{array}$$

Now the factors 2 and 3 are to be placed according to schemes IV and V. One obtains:

$$\left|\begin{array}{ccc} 1 & 5 \cdot 2 & 9 \cdot 3 \\ 5 \cdot 3 & 9 & \cdot \\ 9 \cdot 2 & \cdot & 5 \end{array}\right|=\left|\begin{array}{ccc} 1 & 10 & 27 \\ 15 & 9 & 2 \\ 18 & 3 & 5 \end{array}\right|$$

The numbers 6 and 30 do not appear. It is not possible to omit the numbers 10 and 18. If one tries to place the 2 in partial solution (a), it will inevitably meet a 5 or 9 at one point, resulting in 10 or 18, which is precisely the number one does not want to include in the scheme.

Finally, it should be investigated how many different arrangements of the numbers from the solution found above can be generated by swapping rows or columns. In the following scheme (b)

| 11 | 12 | 13 |
| :--- | :--- | :--- |
| 21 | 22 | 23 |
| 31 | 32 | 33 |

the first digit of each number represents the row number and the second the column number. 3 elements, here the numbers 1, 2, 3, can be arranged in 6 different ways (permutations).

The swaps can be performed both with the row numbers and with the column numbers. Thus, 36 different rearrangements of scheme (b) are found.

Another 36 new arrangements that meet the given conditions are obtained by swapping the rows and columns in each of the 36 solutions found so far. Therefore, there are 72 different arrangements as solutions to the given problem.