6. Boys are dividing the catch. The first one took $r$ fish and a seventh part of the remainder; the second - $2 r$ fish and a seventh part of the new remainder; the third - $3 r$ fish and a seventh part of the new remainder, and so on. It turned out that in this way all the caught fish were divided equally. How many boys were there?
# Solution:

Let $x$ be the number of boys; $y$ be the number of fish each received. Then the last boy took $x r$ fish (there could be no remainder, otherwise there would not have been an even distribution), so $y=x r$. The second-to-last boy took $(x-1) r+\frac{x r}{6}=y$; i.e., $x r$ is $\frac{6}{7}$ of the second-to-last remainder, so the last remainder is $\frac{7}{6} x r$, and its seventh part is $\frac{x r}{6}$.

Then $(x-1) r+\frac{x r}{6}=x r ; 6 r=x r ; x=6$.

Answer: 6.