4. Two chess players played 100 games against each other. 11 points were awarded for a win, $x$ points for a draw, and no points for a loss. Find all possible values of $\mathrm{x}$, if the chess players scored a total of 800 points, and $x$ is a natural number.
Answer: 3 and 4.

Sketch of the solution. Let $n$ games be decisive, then the number of draws is $100-n$, and the players scored $11 n+2 x(100-n)$. Since a total of 800 points were scored, we get the equation $11 n+2 x(100-n) = 800. 11 n-2 n x=800-200 x . n(11-2 x)=200(4-x) 3 n=200(4-x)-2 n(4-x)$. From which $3 n=2(4-x)(100-n)$. It remains to find $n$ for the five possible values of $x$. Since 800 is not divisible by 11, the case $x=0$ is impossible.

If $x=4$, then $n=0$ (there were only draws).

If $x=3$, then $n=40$.

If $x=2$, then $7 n=400, n$ is not an integer.

If $x=1$, then $3 n=200, n$ is not an integer.