The table shown is to be filled in with integers so that

- the integers in each row form an arithmetic sequence that increases from left to right,
- the integers in each column form a geometric sequence that is increasing from top to bottom, and
- the three geometric sequences have the same common ratio, which is an integer.

Determine the sum of all possible values of $b$.

| 5 |  |  |
| :---: | :---: | :---: |
|  |  | $b$ |
|  |  | 900 |
Suppose the common ratio is $r$ and the common difference in the first row is some integer $d>0$. This means the entries in the first row will be $5,5+d$, and $5+2 d$ and the entries in the first column will be $5,5 r$, and $5 r^{2}$.

We know that $d$ must be an integer since all entries in the table are integers.

The entry in the top-right cell is equal to $\frac{900}{r^{2}}$ which must be an integer, so $r$ is an integer with the property that $r^{2}$ is a divisor of 900 .

The prime factorization of 900 is $2^{2} 3^{2} 5^{2}$, from which it is possible to deduce that the perfectsquare divisors of 900 are $1^{2}, 2^{2}, 3^{2}, 5^{2}, 6^{2}, 10^{2}, 15^{2}$, and $30^{2}$.

We have now limited the possibilities of $r$ to 1, 2, 3, 5, 6, 10, 15, and 30 .

However, the geometric sequences are increasing, so $r=1$ is not possible.

We also cannot have $r=30$ since this would lead to $\frac{900}{30^{2}}=1$ in the top-right cell, which is less than 5 .

Similarly, if $r=15$, then the entry in the top right cell would be $\frac{900}{15^{2}}=4$, which is less than 5 . The entry in the top-right cell is $5+2 d$, which must be odd since $d$ is an integer.

The numbers $\frac{900}{3^{2}}=100$ and $\frac{900}{5^{2}}=36$ are even, so we cannot have $r=3$ or $r=5$.

We are now down to the values $r=2, r=6$, and $r=10$ as possibilities.

Suppose $r=2$. Then $b=\frac{900}{r}=\frac{900}{2}=450$ and the entry in the top-right cell is $\frac{450}{2}=225$.

Since $5+2 d=225, d=110$. The entries in the first column are $5 \times 2=10$ and $10 \times 2=20$, so the common difference in the middle row is $\frac{450-10}{2}=220$ and the common difference in the bottom row is $\frac{900-20}{2}=440$. Thus, we get the following table with $r=2$ :

| 5 | 115 | 225 |
| :---: | :---: | :---: |
| 10 | 230 | 450 |
| 20 | 460 | 900 |

which satisfies all of the properties, so $b=450$ is a possibility.

With $r=6$ and $r=10$, we get the following tables:

| 5 | 15 | 25 | 5 | 7 | 9 |
| :---: | :---: | :---: | :---: | :---: | :---: |
| 30 | 90 | 150 | 50 | 70 | 90 |
| 180 | 540 | 900 | 500 | 700 | 900 |

which both satisfy the required properties. Therefore, the possible values of $b$ are $b=450$, $b=150$, and $b=90$, so the answer is $450+150+90=690$.

ANSWER: 690