Jane places the integers 1 through 9 in the nine cells of a $3 \times 3$ grid. The sum of the three integers in a row is called a "row sum" and the product of the three integers in a column is called a "column product". After Jane arranges the integers, the following properties hold.

- Each integer is used exactly once.
- The integer 1 is in the leftmost cell of the second row.
- The integer 8 is in the centre cell.
- The three row sums are equal.
- The largest column product and the smallest column product differ by at most 40 .

Determine all possibilities for the largest column product.

## Question 25 is on the next page
The sum of the positive integers from 1 to 9 inclusive is 45 . Since each row sum must be the same, this means each row must be one third of 45 which is 15 .

In the second row, the first two cells contain the integers 1 and 8 , so the third must contain $15-8-1=6$.

The integers that still need to be placed are 2, 3, 4, 5, 7, and 9 .

Notice that $15-9=6$ and that 2 and 4 are the only two integers in this list that have a sum of 6 .

Therefore, the integers 9,4 , and 2 must be in one of the remaining rows and the integers 3,5 , and 7 must be in the other.

In fact, if the integers are placed in such a way that all conditions are satisfied and the top and bottom rows are exchanged, all properties will still be satisfied and the column products will not have changed.

Therefore, we can assume that 2, 4, and 9 are in the top row.

This gives six possible ways to fill in the top two rows. They are shown below.

| 2 | 4 | 9 |
| :--- | :--- | :--- |
| 1 | 8 | 6 |
|  |  |  |


| 2 | 9 | 4 |
| :--- | :--- | :--- |
| 1 | 8 | 6 |
|  |  |  |


| 4 | 2 | 9 |
| :--- | :--- | :--- |
| 1 | 8 | 6 |
|  |  |  |


| 4 | 9 | 2 |
| :--- | :--- | :--- |
| 1 | 8 | 6 |
|  |  |  |


| 9 | 2 | 4 |
| :--- | :--- | :--- |
| 1 | 8 | 6 |
|  |  |  |


| 9 | 4 | 2 |
| :--- | :--- | :--- |
| 1 | 8 | 6 |
|  |  |  |

The digits that remain to be placed are 3,5 , and 7 .

We will first focus on the first partially filled in grid in which the first row contains 2,4 , and 9 from left to right.

The third column already contains 9 and 6 . The integer to be placed in the third column in the bottom row is at least 3 , so this means the third column product is at least $9 \times 6 \times 3=162$. By similar reasoning, the first column product is at most $2 \times 1 \times 7=14$.

Therefore, no matter how the remaining integers are placed, the difference between the largest and smallest column product will be at least $162-14=148$, which is greater than 40 .

Therefore, the 2, 4, and 9 cannot be placed in that order.

In the second grid, the second column product is at least $9 \times 8 \times 3=216$ and the first column product is at most $2 \times 1 \times 7=14$, so the difference between the largest and smallest column product is at least $216-14=202$ which is greater than 40 , so the 2,4 , and 9 cannot be placed in the order 2, 9, 4, either.

By similar reasoning, the third and fourth possibilities above can be eliminated.

Therefore, the top row must contain, from left to right, either 9,2 , and 4 , or 9,4 , and 2 .

If the 5 is placed in the third column of the grid below,

| 9 | 2 | 4 |
| :--- | :--- | :--- |
| 1 | 8 | 6 |
|  |  |  |

then the third column product will be $4 \times 6 \times 5=120$.

This would mean that the largest the first column product can be is $9 \times 1 \times 7=63$.

This would lead to the largest and smallest column products having a difference greater than 40, so 5 cannot be placed in the third column.

If 7 is placed in the third column, the difference between the largest and smallest column products would be even larger, so the 3 must be placed in the third column.

By similar reasoning, the 3 must be placed in the second column of the last grid.

We have now reduced to four possible ways to place the integers, and they are shown below with the column products given below each column.

| 9 | 2 | 4 |
| :---: | :---: | :---: |
| 1 | 8 | 6 |
| 5 | 7 | 3 |
| 45 | 112 | 72 |


| 9 | 2 | 4 |
| :--- | :--- | :--- |
| 1 | 8 | 6 |
| 7 | 5 | 3 |
| 63 | 80 | 72 |


| 9 | 4 | 2 |
| :---: | :---: | :---: |
| 1 | 8 | 6 |
| 5 | 3 | 7 |
| 45 | 96 | 84 |


| 9 | 4 | 2 |
| :---: | :---: | :---: |
| 1 | 8 | 6 |
| 7 | 3 | 5 |
| 63 | 96 | 60 |

The column products in the first grid are 45, 112, and 72 . Since 45 and 112 differ by more than 40 , this grid does not satisfy the conditions.

The column products in the second grid are 63,80 , and 72 .

The difference between the largest and smallest column product is $80-63=17$ which is less than 40 , so this grid satisfies all of the conditions.

Its largest column product is 80 , so 80 is one of the possibilities.

The column products in the third grid are 45, 96, and 84 . Since 96 and 45 differ by more than 40 , this grid does not satisfy the conditions.

The column products in the fourth grid are 63, 96, and 60 .

The difference between the largest and smallest column product is $96-60=36$ which is less than 40 , so this grid satisfies all of the conditions.

Its largest column product is 96 , so 96 is another possibility.

We have shown that there are exactly two ways to arrange the integers so that the conditions are satisfied, and the largest column products in these two arrangements are 80 and 96 .

ANSWER: 80,96