[Algebra]
Problem: The Sieve of Sundaram uses the following infinite table of positive integers:

| 4 | 7 | 10 | 13 | $\cdots$ |
| :---: | :---: | :---: | :---: | :---: |
| 7 | 12 | 17 | 22 | $\cdots$ |
| 10 | 17 | 24 | 31 | $\cdots$ |
| 13 | 22 | 31 | 40 | $\cdots$ |
| $\vdots$ | $\vdots$ | $\vdots$ | $\vdots$ |  |

The numbers in each row in the table form an arithmetic sequence. The numbers in each column in the table form an arithmetic sequence. The first four entries in each of the first four rows and columns are shown.
Determine the number in the 50th row and 40th column.

Solution: First, we determine the first entry in the 50th row.

Since the first column is an arithmetic sequence with common difference 3, then the 50th entry in the first column (the first entry in the 50th row) is $4+49(3)=4+147=151$.

Second, we determine the common difference in the 50th row by determining the second entry in the 50th row.

Since the second column is an arithmetic sequence with common difference 5 , then the 50 th entry in the second column (that is, the second entry in the 50th row) is $7+49(5)$ or $7+245=252$.

Therefore, the common difference in the 50th row must be $252-151=101$.

Thus, the 40th entry in the 50th row (that is, the number in the 50th row and the 40th column) is $151+39(101)=151+3939=4090$.

Final answer: 4090