Counting candies in boxes There are $5$ boxes containing $80$ candies. After taking $\frac{1}{5}$ of the candies in the first box and putting them in the seconf one, we take  $\frac{1}{5}$ of the candies in the second box and putting them in the third one (after we added the $\frac{1}{5}$ candies from the first box) and so on untill we take $\frac{1}{5}$ of the candies in the fifth box and put it in the first one.
After this process, each box contained the smae number of candies.  How many candies were initially in each box?  

So I can say $a+b+c+d+e=80$ and then slowly and brutally start calculating how many candies were in each box after the process- which will yield 6 equations (including the first one). But is there a better way?
We have: $\dfrac{4a}{5} + \dfrac{e}{5} = \dfrac{a}{5} + \dfrac{4b}{5} = \dfrac{b}{5} + \dfrac{4c}{5} = \dfrac{c}{5} + \dfrac{4d}{5} = \dfrac{d}{5} + \dfrac{4e}{5}= \dfrac{80}{5} = 16$, and $a+b+c+d+e=80$. Can you solve this system?
Suggestion:

Can you work backwards?  At the end the distribution is $(16,16,16,16,16)$.  The step before, the distribution would be $(12,16,16,16,20)$.