Problem 2353

If $m = n$, then attach to chair $i$ the leaf with number $i+1$. Everyone then moves up a chair every clap and for everyone, the first time they return to the chair they started on is after $n$ claps. So after $n$ claps for the first time, everyone has returned to the chair they started on. Now suppose $m < n$.

Since $m$ is not a prime power, we can write $m$ as $m = k\ell$ with $\gcd(k, \ell) = 1$. We claim that there are $a, b > 0$ such that
$$ak + b\ell = n.$$
Note that the numbers $n-ak$ with $a \in \{1, 2, \dots, \ell\}$ are all different modulo $\ell$. We show this by contraposition. Suppose that there are $a_1$ and $a_2$ in $\{1, 2, \dots, \ell\}$ such that $n - a_1k \equiv n - a_2k \pmod \ell$. Then we also have $(a_1 - a_2)k \equiv 0 \pmod \ell$. As $\gcd(k, \ell) = 1$, it follows that $\ell \mid (a_1 - a_2)$. Since $a_1$ and $a_2$ both are in $\{1, 2, \dots, \ell\}$, it follows that $a_1 = a_2$. Therefore if $a_1$ and $a_2$ are in $\{1, 2, \dots, \ell\}$ and $a_1 \ne a_2$, then $n - a_1k \not\equiv n - a_2k \pmod \ell$. So the numbers $n-ak$ with $a \in \{1, 2, \dots, \ell\}$ are indeed all different modulo $\ell$.

It follows that the $n-ak$ for $a \in \{1, 2, \dots, \ell\}$ are all the $\ell$ residue classes modulo $\ell$. Since $n-ak \ge n-\ell k = n-m > 0$, these $\ell$ numbers are also all greater than $0$. Now choose the $a$ for which $n - ak$ is congruent to $0$ modulo $l$. Since $n - ak$ is greater than $0$, there is now a $b > 0$ such that $n - ak = bl$. This gives the $a, b > 0$ such that $ak + bl = n$.

Now we divide the chairs into $a$ groups of $k$ chairs and $b$ groups of $l$ chairs. In each group, we arrange the chairs in a circle and attach the leaves to chairs in such a way that each leaf has the number of the next chair in the circle. The players on a chair in a group with $k$ chairs, return to the chair they started on every $k$ claps (and not on any other clap). The players on a seat in a group with $l$ seats, return to the chair they started on every $l$ claps (and not on any other clap). So the first time everyone returns to the chair they started on is after $\text{lcm}(k, l)$ claps. But
$$\text{lcm}(k, l) = k\ell = m,$$
because $\gcd(k, l) = 1$.

$\square$