In a condominium, 29 families live, each of them has either 1 cat or 3 cats or 5 cats. The number of families that have only 1 cat is the same as the number of families that have 5 cats. How many cats are there in this condominium?
Let:

$$
\begin{aligned}
& x=\text { number of families that have only } 1 \text { cat; } \\
& y=\text { number of families that have exactly } 3 \text { cats; } \\
& z=\text { number of families that have } 5 \text { cats. }
\end{aligned}
$$

It follows that $x+y+z=29$ and $x=z$. Therefore, $2 x+y=29$. On the other hand, the number of cats is $x+3 y+5 z$. Thus, we have:

$$
\text { number of cats }=x+3 y+5 z=6 x+3 y=3(2 x+y)=3 \times 29=87 \text {. }
$$