Mystery Number Hunt

Find the eight three digit numbers that fill in the grid below such that the product of the digits in the rows and columns equal the numbers in the rightmost column and last row.

$$\begin{array}{|c|c|c|c|} \hline & & & 294 \\ \hline & & & 216 \\ \hline & & & 135 \\ \hline & & & 98 \\ \hline & & & 112 \\ \hline & & & 84 \\ \hline & & & 245 \\ \hline & & & 40 \\ \hline 8,890,560 & 156,800 & 55,566 & \\ \hline \end{array} $$

First, we break down each of the entries in the rightmost column and bottom row in terms of prime decompositions, noting that each has no prime factor larger than 7. That is, \begin{align*} 294 = 2 \cdot 3 \cdot 7^2, \,\,\, 216 = 2^3 \cdot 3^3, &\,\,\, 135 = 3^3 \cdot 5, \,\,\, 98 = 2 \cdot 7^2,\\ 112 = 2^4 \cdot 7, \,\,\, 84 = 2^2 \cdot 3 \cdot 7,& \,\,\, 245 = 5 \cdot 7^2, \,\,\, 40 = 2^3 \cdot 5, \\ 8,890,560 = 2^6 \cdot 3^4 \cdot 5 \cdot 7^3, \,\,\, 156,800 &= 2^7 \cdot 5^2 \cdot 7^2, \,\,\, 55,566 = 2 \cdot 3^4 \cdot 7^3\end{align*}

The third row must have a $5$ in the second column, since the second column must have no multiples of $3$ in it. Similarly, since $294$ can only be expressed as the product of three digits as $6 x 7 x 7,$ not necessarily in that order, the first row must have a $7$ in the second column. Also, the second column of the second row can only be a $4$ or an $8,$ since it cannot have a multiple of $3$ in it. Additionally, there must be a $7$ in the third column of the seventh row as the only other option is $5$ and the third column must have no multiples of $5$ in it.

In the third column there can be only one multiple of $2$ and no more than two additional copies of $7.$ The possible places for a seven in the third column are in the first, fourth, fifth and sixth rows, while a $2$ could be in the fourth, fifth, sixth and eighth rows and a $6$ could occur in the first, second and sixth rows. If a $2$ or $6$ occurs in the second, sixth or eighth rows, then the first, fourth and fifth rows must each have a $7,$ which is too many. So in particular, we see that there must be a $1$ in the third column of the eighth row, leaving the other digits in the eighth row to ba a $5$ and an $8$. Further, since there cannot be a multiple of $2$ in the third column of the second row, the third column must be either a $3$ or a $9,$ which leaves the first column of the second row to either be a $3,$ $6$ or $9.$

Similarly, if a $7$ is in the third column of the sixth row, then only one additional row of the first, fourth and fifth may have a $7$, which leaves at least two multiples of $2,$ which is too many. Therefore, the third column of the sixth row must be $3,$ which leaves the remaining digits of the sixth row to be a $4$ and a $7.$ This leaves us with the following grid of possibilities:

$$\begin{array}{|c|c|c|c|} \hline \{6,7\} & 7 & \{6,7\}& 294 \\ \hline \{3, 6, 9\} & \{4, 8\}& \{3, 9\} & 216 \\ \hline \{3, 9\} & 5 & \{3, 9\} & 135 \\ \hline \{2, 7\} & \{2, 7\} & \{2, 7\} & 98 \\ \hline \{2, 4, 7, 8\} & \{2, 4, 7, 8\} & \{2, 7\} & 112 \\ \hline \{4, 7\} & \{4, 7\} & 3 & 84 \\ \hline \{5, 7\} & \{5, 7\} & 7 & 245 \\ \hline \{5, 8\} & \{5, 8\} & 1 & 40 \\ \hline 8,890,560 & 156,800 & 55,566 & \\ \hline \end{array} $$

At this point, I sort of decided that the number of brute force combinations was roughly small enough. Playing around with the possibilities for a while by hand, yielded the solution:

$$\begin{array}{|c|c|c|c|} \hline 7 & 7 & 6 & 294 \\ \hline 9 & 8 & 3 & 216 \\ \hline 9 & 5 & 3 & 135 \\ \hline 7 & 2 & 7 & 98 \\ \hline 8 & 2 & 7 & 112 \\ \hline 7 & 4 & 3 & 84 \\ \hline 5 & 7 & 7 & 245 \\ \hline 8 & 5 & 1 & 40 \\ \hline 8,890,560 & 156,800 & 55,566 & \\ \hline \end{array} $$