Task 1. There are 3 types of installations, no more than 200 in total. The number of installations of type 2 is 4 times the number of type 1, and the number of installations of type 3 is a multiple of the number of installations of type 1. If the number of installations of type 3 were 5 times more, then they would be 99 more than the number of installations of type 2. Find the number of installations of each type.
Solution. If $x_{1}, x_{2}, x_{3}$ are the quantities of installations of types $1,2,3$, then the conditions are represented by the relations

$$
\begin{gathered}
x_{1}+x_{2}+x_{3} \leq 200 \\
x_{2}=4 x_{1} \\
x_{3}=k x_{1} \\
5 x_{3}=x_{2}+99 \\
k, x_{1}, x_{2}, x_{3} \in \mathbb{N}
\end{gathered}
$$

From (2)-(4) we get $(5 k-4) x_{1}=99$, from which it follows that $5 k-4$ and $x_{1}$ are natural divisors of the number 99, i.e., these values belong to the set $\{1,3,9,11,33,99\}$.

If $x_{1}=1$, then $5 k=99+4$, which is impossible since $k$ is an integer.

If $x_{1}=3$, then $5 k=33+4$, which is impossible.

If $x_{1}=9$, then $5 k=11+4, k=3, x_{2}=36, x_{3}=27$, all conditions are satisfied.

If $x_{1}=11$, then $5 k=9+4$, which is impossible.

If $x_{1}=33$, then $5 k=3+4$, which is impossible.

If $x_{1}=99$, then $5 k=1+4, k=1, x_{3}=99, x_{2}=4 \cdot 99$, inequality (1) is not satisfied.

Answer: $9,36,27$.