An IMO $1971$ longlisted problem I have been stuck on this beautiful problem for a long time, but I have no idea as to where to start. The problem is as follows:  
One Martian, one Venusian and one Human reside on Pluto. One day they make the following conversation:
Martian: I have spent $1/12$ of my life in Pluto.
Human: I also have.
Venusian: Me too.
Martian: But Venusian and I have spent much more time here than you, Human.
Human: However Venusian and I are of the same age.
Venusian: I have lived $300$ Earth years.
Martian: Venusian and I have been on Pluto for the last $13$ years.  
It is known that Human and Martian together have lived for $104$ Earth years. Find their respective ages. 
While solving this problem, I have a hint that the numbers are not necessarily expressed in base $10$ but I was unable to find any use of that statement. Any help is appreciated. Thank you.
 A: We get a hint towards the different bases because otherwise the human would be three hundred years old, which is somewhat unrealistic.
So let $M$ and $V$ be the bases used by martians and venusians (whereas humans work in base ten) and let $t_M,t_H,t_V$ be the number of (presumably Earth) years each has spent on Pluto, and let $a_M,a_H,a_V$ be their ages in Earth years. We learn (all my numbers are in base ten, of course):
$$\begin{eqnarray}\tag1a_M &=& (M+2)t_M\\
\tag2a_H &=& 12t_H\\
\tag3a_V &=& (V+2)t_V\\
\tag5t_H&\ll&\begin{cases}t_M\\t_V\end{cases}\\
\tag6a_V &=&a_H\\
\tag7a_V &=& 3V^2\\
\tag8t_V&=&t_M=M+3
\end{eqnarray}$$
and of course $$\tag90<t_M\le a_M,\qquad0<t_H\le a_H,\qquad0<t_V\le a_V.$$
Also, the occurance of digits $2$ and $3$ tells us
$$ M>2,\qquad V>3.$$
From $(2)$, $(6)$, $(7)$, we have $12\mid 3V^2$, so $V$ is even, say $V=2W$. From $(3)$ and $(7)$, we have $V+2\mid 3V^2$, or $W+1\mid 6W^2$. As $\gcd(W+1,W)=1$, $W+1\mid 6$. Therefore $W\in\{0,1,2,5\}$, $V\in\{0,2,4,10\}$. 
From $(2)$, $(2)$, $(6)$ and $(5)$, we find $V<10$, hence together with $V>3$, we determined that 
$$V=4.$$
So $a_H=a_V=48$, $t_H=4$, and $t_V=t_M=8$. This makes $M=5$, so that $a_M=56$.

Incidentally, we indeed verify that $t_H+t_M=104$ - but we did not need that information to find the result.
