Task 1. There are 3 types of installations totaling no less than 100. 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 22 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

$$
\begin{gathered}
x_{1}+x_{2}+x_{3} \geq 100 \\
x_{2}=4 x_{1} \\
x_{3}=k x_{1} \\
5 x_{3}=x_{2}+22 \\
k, x_{1}, x_{2}, x_{3} \in \mathbb{N}
\end{gathered}
$$

From (2)-(4) we get $(5 k-4) x_{1}=22$, from which it follows that $5 k-4$ and $x_{1}$ are natural divisors of the number 22, i.e., these values belong to the set $\{1,2,11,22\}$.

If $x_{1}=1$, then $5 k=22+4$, which is impossible since $k$ is an integer.

If $x_{1}=2$, then $5 k=11+4, k=3, x_{2}=8, x_{3}=6$, (1) is not satisfied.

If $x_{1}=11$, then $5 k=2+4$, which is impossible.

If $x_{1}=22$, then $5 k=1+4, k=1, x_{2}=88, x_{3}=22$, all conditions are satisfied.

Answer: $22,88,22$.