Francesca put the integers $1,2,3,4,5,6,7,8,9$ in the nine squares in the grid. She put one integer in each square and used no integer twice. She calculated the product of the three integers in each row and wrote the products to the right of the corresponding rows. She calculated the product of the integers in each column and wrote the products below the corresponding columns. Finally, she erased the integers from the nine squares. Which integer

![](https://cdn.mathpix.com/cropped/2024_04_20_ac36362783317e0251fdg-033.jpg?height=350&width=371&top_left_y=1820&top_left_x=1343)
was in the square marked $N$ ?
(A) 3
(B) 8
(C) 9
(D) 6
(E) 4
Each entry in the grid must be a divisor of the product of the numbers in its row and the product of the numbers in its column.

![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-066.jpg?height=349&width=377&top_left_y=758&top_left_x=928)

Only two of the products are multiples of 5, namely 160 and 135.

This means that the 5 must go in the second row and third column.

From this, we can see that the product of the other two numbers in the second row is $\frac{135}{5}=27$. Since all of the entries are between 1 and 9, then the remaining two numbers in this row must be 3 and 9 .

Since 9 is not a divisor of 21 , then 9 must be in the middle column.

This means that the product of the remaining numbers in the middle column is $\frac{108}{9}=12$.

This means that the remaining digits in the middle column are 3 and 4 , or 2 and 6 . (These are the only factor pairs of 12 from the list of possible entries.)

Since 3 already occurs in the second row, then the entries in the second column must be 2 and 6 . Since 6 is not a divisor of 56 , then 6 cannot go in the first row.

This means that 6 goes in the third row and so $N=6$.

We can complete the grid as follows:

| 7 | 2 | 4 |
| :---: | :---: | :---: |
| 3 | 9 | 5 |
| 1 | 6 | 8 |

ANSWER: (D)