Problem 9.6. On a line, two red points and several blue points are marked. It turned out that one of the red points is contained in exactly 56 segments with blue endpoints, and the other - in 50 segments with blue endpoints. How many blue points are marked?
Answer: 15.

Solution. Let there be $a$ blue points to the left of the first red point, and $b$ blue points to the right; $c$ blue points to the left of the second red point, and $d$ blue points to the right. Then $a b=56, c d=50$. Additionally, $a+b=c+d$ - the number of blue points.

Notice that among the numbers $c$ and $d$, exactly one is even, since the number 50 is divisible by 2 but not by 4. This means that among the numbers $a$ and $b$, exactly one is also even. There are exactly two ways to represent 56 as the product of an even number and an odd number: $56=8 \cdot 7$ and $56=56 \cdot 1$.

In the first case, $c=10, d=5$ work. As an example, you can place 5 blue points on the line, then a red one, then 3 more blue points, then another red one, and finally 7 more blue points.

In the second case, the total number of blue points is 57. Then it only remains to understand that the following system of equations has no integer solutions:

$$
\left\{\begin{array}{l}
c+d=57 \\
c d=50
\end{array}\right.
$$

One can check that the quadratic trinomial $t^{2}-57 t+50=0$, whose roots by Vieta's theorem should be the numbers $c$ and $d$, does not have a pair of natural solutions (for example, it is sufficient to verify that its values at $t=0$ and $t=1$ have different signs, meaning one of the roots lies in the interval $(0 ; 1))$.

Alternatively, one can simply enumerate the factorizations of the number 50 into $c$ and $d$.