The question is:

> 
> Find the last two digits of the expansion of $2^{12n}-6^{4n}$ where $n$
>  is any positive integer.
> 
> 
> 

If we put the value of $n=1$ we would get $2800$. For $n = 2$ the result will be $15097600$ so the last two digits are $00$. Now this approach (using induction) is fairly simple but a bit tedious and time consuming. In my module there is another approach given which proceeds by representing $2^{12n}-6^{4n}$ as $64^{2n}-36^{2n}$ and then as this is divisible by $(64+36)=100$, it says hence the last two digits will be $00$. Now I suppose this is just because anything that is divisible by 10 or 100 would have last digits ending with zeroes. If so then it is not possible to get the last digits of expansions which don't have $10$ or $100$ as a factor. That is, last digits would be anything but zero. Am I correct?