III OBM

Call the containers $L_8$, $L_5$, $L_3$. Fill $L_5$ from $L_8$, then fill $L_3$ from $L_5$, leaving $2$ in $L_5$. Empty $L_3$ into $L_8$. Empty $L_5$ into $L_3$ (so now $L_8$ has $6$, $L_5$ has $0$, $L_3$ has $2$). Fill $L_5$ from $L_8$. Fill $L_3$ from $L_5$. Empty $L_3$ into $L_8$. Now $L_5$ and $L_8$ each contain $4$.

Now consider the general case. It is an easy induction that the amount in each container is always a multiple of $\gcd (m,n)$. Use induction on the number of steps, and note that the only possible move is to replace $a,b$ by $D,a+b-D$, where $D$ is one of $0,m,n,m+n$. So it is certainly a necessary condition that $\frac{m+n}{2}$ is divisible by $\gcd (m,n)$. In particular, it must be an integer and so $m+n$ must be even. So it remains to show that if $m+n$ is even and $\frac{m+n}{2}$ is a multiple of $\gcd (m,n)$ then we can get $\frac{m+n}{2}$ into $L_m$. If $m=n$, then that is trivial. So assume $m>n$. Put $d=m-n$. Now suppose that after some moves we have got $k$ in the $L_n$ and the rest $m+n-k$ in the $L_{m+n}$. Fill $L_m$ from $L_{m+n}$, then fill $L_n$ from $L_m$. That gives $m-(n-k)=k+d$ in $L_m$. Now $k+d=qn+r$ for some $0\le r<n$. Repeatedly (for more precisely $q$ times) fill $L_n$ from $L_m$ and empty it into $L_{m+n}$, finally pour the remainder of $r$ from $L_m$ into $L_n$. So starting with all the wine in $L_{m+n}$ (i.e. $k=0$), and iterating this process we get $[hd]$ in $L_n$ where $[hd]$ denotes the residue of $hd$ mod $n$. Now we may put $\frac{m+n}{2}=Qn+R$, where $0\le R<n$. Since $n$ and $\frac{m+n}{2}$ are multiples of $\gcd (m,n)$, so is $R$. But $\gcd (m,n)=\gcd (d,n)$. So $R$ is a multiple of $\gcd (d,n)$. But that means we can write $R=hd-h^{\prime }n$ for some

(1981)