Answer: 211.

Solution. Let there be $x$ people in front of Peter, $y$ people in front of Nicholas, and $z$ people in front of Denis. There are three possible cases.

![](https://cdn.mathpix.com/cropped/2024_05_06_1d29d5cb430346a58eb5g-13.jpg?height=245&width=1280&top_left_y=2493&top_left_x=388)

- $x=4 y$. Then $4 y+50=y+100$, from which $y$ is not an integer, which is impossible.
- $y=4 z$. Then $4 z+100=z+170$, from which $z$ is not an integer, which is impossible.
- $x=4 z$. Then $4 z+50=z+170$, from which $z=40$. In total, there are 211 recruits.
7-7. The recruits stood in a row, facing the same direction. Among them were three brothers: Peter, Nikolai, and Denis. In front of Peter were 50 people, in front of Nikolai 100, and in front of Denis 170. On the command "About face!" everyone turned to face the opposite direction. As a result, it turned out that in front of one of the brothers there are now four times as many people as in front of another. How many recruits could there be, including the brothers? List all possible options.