BxMO Team Selection Test

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 $n-a_{1}k\equiv n-a_{2}k(\mathrm{mod}\ell )$. Then we also have $(a_1-a_2)k\equiv 0(\mathrm{mod}\ell )$. 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 $n-a_{1}k\not\equiv n-a_{2}k(\mathrm{mod}\ell )$. 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 $\gcd (k,l)=1$.

$\square$