It's A Factor of 24

The only way we can factor $28$ using 3 distinct digits is some order of $1 \times 4 \times 7$ , so these 3 digits lie in the first row. Now $7$ is not a factor of either $48$ or $180$ , but is one of $42$ , so $7$ must go in the upper left box. Now for the first column, since $42 = 7 \times 6$ and since $1$ has already been used in the first row, the lower two boxes in the first column must be $2$ and $3$ in some order.

Now we cannot have $1$ in the upper right box since there are no two digits whose product is $180$ , thus $1$ must go in the middle of the upper row and thus $4$ must be placed in the upper right box. The lower two boxes of the middle column must then be $6$ and $8$ in some order, as this is the only way to have a product of two digits be $48$ . Next, since $180 = 4 \times 45$ and the only way to factor $45$ using two digits is $5 \times 9$ , the lower two boxes of the rightmost column must be $5$ and $9$ in some order.

Next, as $9$ is a factor of $108$ but not $120$ , we must place $9$ in the lower right box. Then since $108 = 9 \times 12$ and $4$ has already been used, we must have $2$ and $6$ in the first two boxes of the bottom row, which in turn then forces $3$ and $8$ , in that order, into the first two boxes of the middle row. Thus the digit $\boxed{8}$ must be placed in the center box.

The completed grid looks like

$\Large 7 \space \space \space \space 1 \space \space \space \space 4 \space \space 28$

$\Large 3 \space \space \space \space 8 \space \space \space \space 5 \space \space 120$

$\Large 2 \space \space \space \space 6 \space \space \space \space 9 \space \space 108$

$\Large 42 \space 48 \space 180$

[This solution is written based on knowing that the answer is 8, and then finding a fast way to justify it is 8. It also doesn't guarantee that we can always find the answer using this way, but I felt that this approach was nice and should be presented.
But otherwise, Brian's logical deductive approach was how I first solved the problem.]

For the primes of 7 and 5, the product of each row and column will uniquely determine where they are placed. For example, the first row and first column has products that are multiples of 7, so the top left square must be a 7. Similarly, the second row and third column has products that are multiples of 5, so the middle right square must be a 5.

At first glance, it is hard to extend this argument to the other primes, because there are different numbers that contribute to the product, especially in the case of 2. However, let's see what happens when we focus on just the contribution of 2. These number are 2, 4, 6, 8, and the total number of 2's is $1 + 2 + 1 + 3 = 7$ . By looking at just the power of 2 in each term, our goal is to fill out the squares with the number 1, 2, 1, 3 (and several 0's), such that the sum of each row/column gives us the indicated number

If we consider the columns, only the center column can contain 3.
If we consider the rows, only the center row can contain 3.
Hence, the center square must be 3, which means that in the original problem, the center square is equal to $2 ^ 3 = 8$ .

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Note: It doesn't always yield a solution, because we are not able to hunt down the exact placement of the contributions of 2. There are 2 ways of completing the 2-square

Extension: In a similar manner, can you conclude where 9 must be?