Passing Shillings

Question

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?

Answer 1

Instead of thinking of it as each man gets a bunch of shillings, puts them in his pocket, then takes out a bunch of shillings from his pocket and gives them to the next man; think of it as a man with $k$ shillings gets a basket containing $w$ shillings. The man adds a shilling to it and passes the basket on. The man now has $k -1$ shillings and the basket now has $w + 1$ shillings in it.

Let it start with there being $m$ men, the last man is the poorest with $k$ shillings (and the first man is the richest with $m+k -1$ shillings). The basket starts with $0$ in it. But each man puts in $1$ shilling so at the end of a "circuit" (when it gets to the last man) every man has put in a shilling so the basket has $m$ shillings and each man is a shilling poorer. The last/poorest man has $k -1$ shilling and the first richest has $m+k -2$.

Each round the basket grows by $m$ and each man (in particular the last man) gets poorer by $1$. So after $j$ rounds the basket with have $m*j$ shillings and the last man will have $k - j$ shillings.

At the end of $k$ rounds the last man puts his last shilling in, and the basket has $mk$ shillings. He passes it on for the $k+1$ round. At the end of that round, the last man can't do anything and the whole thing ends. The basket has $km + (m-1)$ shillings as everybody but the last man put a last shilling in.

The last man breaks the basket open and keeps the $km + m -1$ shillings. The man before him has no shillings as that second man had only one more than the last man. The first man started with $m+k -1$ shillings and he put in $k+1$ shillings in so the first man has $m-2$ shillings.

We are told a neighbor has $4$ times as much as another, but each neighbor has $1$ more that the neighbor ahead of him. Except the last man and the first man.

So either the first man has $4$ times as much as the last man and $m-2 = 4(km + m - 1$ or the last man has $4$ times as mch as the first and $km + m -1 = 4(m-2)$.

So $k$ is presumably positive it must be the latter. And

$km + m - 1 = 4(m-2)$

$(k+1)m -1 = 4m - 8$

$(k-3)m = -7$

$(3-k)m = 7$

Either $3-k = 7$ and $m =1$ Which is ... silly... as we know there were more than one man. Or $3-k = 1; k = 2$ and $m=7$.

So: Beginning: First man has $8$ shillings and last man has $2$. Basket has $0$ shillings.

End of first round: Basket has $7$ shillings. First man has $7$ and last an has $2$.

End of second round: Basket has $14$ shilings. First man has $6$ and last has $0$.

Final round: First man has $5$ shillings. Men $1-6$ have added shillings so the basket has $20$ shillings in it. It gets passed to the last man who is supposed to add a shilling to it but he can't because he doesn't have any.

So the last man keeps the basket. The last man has $20$ shillings. The first man has $5$. (And men $2-6$ have $4,3,2,1,0$ respectively.

Answer 2

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

Answer 3

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.