Help with combinations problem? 
Initially there are $m$ balls in one bag, and $n$ in the other, where $m,n>0$. Two different operations are allowed:  
a) Remove an equal number of balls from each bag;
  b) Double the number of balls in one bag.  
Is it always possible to empty both bags after a finite sequence of operations?
Operation b) is now replaced with
b') Triple the number of balls in one bag.
Is it now always possible to empty both bags after a finite sequence of operations?

This is question 4 on Round $1$ of the $2011/2012$ British Mathematical Olympiad. 
I suck at combinatorics and the like but need to practise to try and improve my competition mathematics. If anyone could give me a hint on where to start I'd be most grateful :D 
EDIT: Never mind guys, I just completely mis-read the question. I thought it said you had to double the numbers of balls in both bags. Thanks for the help! 
 A: For the first problem, if $m=n$, just take everything. Otherwise, apply operation (a) to reduce the smaller bag to a single ball, and let $a$ be the number of balls in the other bag at that point. Double the $1$ and take $1$ ball from each bag, so that you now have $a-1$ balls in one bag and $1$ in the other. Repeat until you have one ball in each bag and then empty the bags with operation (a).
For the second question I conjecture that $m=1$, $n=2$ is a position from which the bags cannot be emptied, though I don’t immediately see an argument.
Added: The conjecture is correct, and the other answer was on the right track. Suppose that $m$ and $n$ have opposite parity.Subtracting the same amount from each leaves two numbers of opposite parity, and multiplying a number by $3$ does not change its parity, so you always have two numbers of opposite parity. In particular, you cannot empty both bags.
A: Regarding the first part...
Let $m>n$
Remove $n-1$ balls from each bag so that you have $m-n+1$ balls in one bag and $1$ ball in the other bag.
Now repeat the algorithm of doubling the balls in the bag which has $1$ ball and then taking away $1$ from each bag till you have $1$ ball in each bag. Finally remove $1$ ball from each bag and you have emptied them in finite number of steps.
For the second part
Again suppose $m>n$
Remove $n-1$ balls from each bag so that you have $m-n+1$ balls in one bag and $1$ ball in the other bag.
Now repeat the algorithm of tripling the balls in the bag which has $1$ ball and then taking away $2$ from each bag. If $2|(m-n)$, then you'll end up with $1$ ball in each bag in some steps. Otherwise you'll end up with $2$ balls in one bag and $1$ ball in the other bag and it seems like no matter what you do from here, you'll always end up with this arrangement.
Hence, in the second part, it may be conjectured that the bags cannot be emptied in finite steps if $m-n\equiv 1\pmod2$.
