$5 \cdot 61$ A group of young people went to dance disco. Each dance costs 1 yuan, and each boy danced with each girl exactly once, then they went to another place to dance. Here, they paid with subsidiary coins, and they spent the same amount of money as before. The entry fee for each person is one subsidiary coin, and each dance costs one subsidiary coin. Each person danced with every other person exactly twice (regardless of gender), and there was 1 subsidiary coin left at the end. How many subsidiary coins does 1 yuan equal?
[Solution] Let there be $b$ boys and $g$ girls. Suppose 1 yuan can be exchanged for $x$ coins, then initially they dance $b g$ times, and later they dance $2 C_{b+g}^{2}=(b+g)(b+g-1)$ times.
According to the problem, $b g x=(b+g)(b+g-1)+(b+g)+1$, which means
$$
(b+g)^{2}+1=b g x .
$$

Let $u=x-2$, then $g^{2}-u b g+b^{2}+1=0$.
Assume $b, g$ are positive integers that satisfy (1), and $b+g$ is the smallest, without loss of generality, let $b \leqslant g$.
It can be verified that $u b-g, b$ are solutions to equation (1), and since $(u b-g) g=b^{2}+1>$ 0, $u b-g, b$ are positive integer solutions to (1). Moreover, $u b-g \geqslant g$.
Thus, we have
$$
(b+1)^{2}>b^{2}+1=g(u b-g) \geqslant g^{2} \text {. }
$$

Considering
$b \leqslant g$,
it must be that
$g=b$.
Substituting back into (1) gives
$$
2 g^{2}-u g+1=0
$$

Thus, $g$ divides 1, so $g=1, x=5, u=3$. That is, 1 yuan can be exchanged for 5 coins.