Problem 802

Solution:
Answer: 042

Note that $\frac{1}{998} + \frac{1}{2} = \frac{250}{499}$ repeats every 498 digits because 499 is prime, so $\frac{1}{998}$ does as well (after the first 498 block). Now we need to find $38^{\text{th}}$ to $40^{\text{th}}$ digits. We expand this as a geometric series
$$\frac{1}{998} = \frac{\frac{1}{1000}}{1 - \frac{2}{1000}} = .001 + .001 \times .002 + .001 \times .002^{2} + \cdots$$
The contribution to the $36^{\text{th}}$ through $39^{\text{th}}$ digits is 4096, the $39^{\text{th}}$ through $42^{\text{nd}}$ digits is 8192, and $41^{\text{st}}$ through $45^{\text{th}}$ digits is 16384. We add these together:

The remaining terms decrease too fast to have effect on the digits we are looking at, so the $38^{\text{th}}$ to $40^{\text{th}}$ digits are 042.