The lines of the first given table are all arithmetic progressions of the same ratio, and the columns of this table are all arithmetic progressions of the same ratio. In the second given table, the same formation rule was used, but someone erased some numbers, leaving only three. What is the number that was in the position indicated with

| 5 | 8 | 11 | 14 | 17 |
| :---: | :---: | :---: | :---: | :---: |
| 12 | 15 | 18 | 21 | 24 |
| 19 | 22 | 25 | 28 | 31 |
| 26 | 29 | 32 | 35 | 38 |
| 33 | 36 | 39 | 42 | 45 |


|  |  | 39 |  |  |
| :--- | :--- | :--- | :--- | :--- |
|  |  |  |  |  |
|  |  |  |  | 87 |
| 56 |  |  |  |  |
|  |  |  | $\star$ |  |
We see that in the first table, each row is an arithmetic progression with a common difference of 3, and each column is an arithmetic progression with a common difference of 7.

| 5  | 8  | 11 | 14 | 17 |
| :---: | :---: | :---: | :---: | :---: |
| 12 | 15 | 18 | 21 | 24 |
| 19 | 22 | 25 | 28 | 31 |
| 26 | 29 | 32 | 35 | 38 |
| 33 | 36 | 39 | 42 | 45 |

|  |  | 39 |  |  |
| :--- | :--- | :--- | :--- | :--- |
|  |  |  |  |  |
|  |  |  |  | 87 |
| 56 |  |  |  |  |
|  |  |  | $\star$ |  |

Let's say that in the second table, the common difference of the arithmetic progressions in the rows is $a$ and in the columns is $b$. Thus, we get $39 + 2a + 2b = 87$ and $39 - 2a + 3b = 56$.

| $39 - 2a$ | $39 - a$ | 39 | $39 + a$ | $39 + 2a$ |
| :---: | :---: | :---: | :---: | :---: |
| $39 - 2a + b$ |  |  |  | $39 + 2a + b$ |
| $39 - 2a + 2b$ |  |  |  | 87 |
| 56 |  |  |  |  |
|  |  |  | $\star$ |  |

Adding these two equations, we get $78 + 5b = 143$, so $b = 13$. Subtracting the two equations that were previously added, we get $4a - b = 31$. Therefore, $a = 11$. Thus, the number that was in the position $\star$ was $39 + a + 4b = 39 + 11 + 4 \times 13 = 102$.