Passing Shillings I am trying to solve one of Carroll's pillow problems, but even with the solution I can't really graps it.
The problem is as follow

Some men sat in a circle, so that each had 2 neighbours; and each had
  a certain number of shillings.  The first had I/ more than the second,
  who had I/ more than the third, and so on.  The first gave I/ to the
  second, who gave 2/ to the third, and so on, each giving I/ more than
  he received, as long as possible.  There were then 2 neighbors, one of
  whom had 4 times as much as the other.  How many men were there?  And
  how much had the poorest man at first?

And this is the given solution

Let m = No. of men, k = No. of shillings possessed by the last (i.e.
  the poorest) man. After one circuit, each is a shilling poorer, and
  the moving heap contains m shgs. Hence, after k circuits, each is k
  shillings poorer, the last man now having nothing, and the moving heap
  contains mk shillings. Hence the thing ends when the last man is again
  called on to hand on the heap, which then contains (mk + m − 1)
  shillings, the penultimate man now having nothing, and the first man
  having (m − 2) shillings.
It is evident that that the first and last man are the only 2
  neighbours whose possessions can be in the ratio ‘4 to 1’. Hence
  either
                             mk + m − 1 = 4(m − 2), 

                or else      4(mk + m − 1) = m − 2.

The first equation gives mk = 3m − 7, i.e. k = 3 − 7m, which evidently
  gives no integral values other than m = 7, k = 2.
The second gives 4mk = 2 − 3m, which evidently gives no positive
  integral values.
Hence the answer is ‘7 men; 2 shillings’.

I am having trouble understanding starting from when he says that the thing ends when the last man is again called on to hand on the heap, and then I can't follow the other equations.
Can someone give me another explanation for the problem?

After $k$ circuits, i.e. $mk$ steps, the amount most recently passed was $mk$ shillings from the last man to the first man, and the last man has nothing left.  
After a further $m-1$ steps the amount most recently passed was $mk+m-1$ shillings from the penultimate man to the last man, and so is the amount held by the last man.  
He should then pass on $mk+m$ shillings to the first man (i.e. $m(k+1)$ shillings to complete the $k+1^\text{th}$ circuit) but cannot. So the process stops.
The equations then look at the ratio between what the last man holds of $mk+m-1$ shillings and the amount held by the first man of $m-2$ shillings (i.e. starting with $k+m-1$ and having lost $k+1$) , and sets this ratio equal to $4$ for the two possibilities 
