In a certain condominium, 29 families live, each of which has either one, three, or five cats. The number of families that have only one cat is the same as the number of families that have five cats. How many cats are there in this condominium?
Let $x$ be the number of families that have only one or exactly five cats, and $y$ be the number of families that have exactly three cats. It follows that $x+y+x=29$ and, therefore, $2 x+y=29$. Since the number of cats is $x+3 y+5 x=6 x+3 y$, we obtain

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