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?

Lets say that there an $n$ men and the poorest (the last man) has $m$ coins at the start of the game.
The richest (the first man) has $n+m-1$ coins.
The first man passes one coin to the second and the second passes on 2, and so down the line, until the last man currently has all of the passed along coins.  Each man (except the last who holds the heap) passes along one more than he received from his predecessor.
the last man passes on the heap and now has $m-1$ coins, and we go around again.  And the the $m^{th}$ time the heap comes to the last man, and he passes it along,  he is has no coins.  But the heap keeps going around until the last man now holds the heap.  He can't pass it on because he will now owe one more coin than is in his possession.  But, he is also the richest man.  The person before him is flat broke.  And the person after him is the first man. 
Every man but the last man has passed the heap $m+1$ times.
the last man only passed it $m$ times.
The last man holds $n(m+1) - 1$ coins.
The first man holds $n-2$ coins.
$n(m+1) - 1 = 4(n-2)\\
(3-m)n = 7$
Since 7 is prime.
$3-m = 1\\
n = 7$
Is the only integer solution.
