Yoak: Lewis Carroll – Passing Shillings

Since my last post, I actually dug up one of my books with Carroll problems.  I’ll present this one in Carroll’s own words and add a few notes:

Carroll writes:

   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?

Notes:

A ‘/’ is clearly to be read as a shilling and the ‘I’ is to be read as 1.  With that, I think the operations is clear.  It is also clear that eventually someone will not be able to pass along 1 more shilling than he was passed, given the finite number of shillings in the game.  When that state occurs, instead of passing that person retains the shillings he was just passed.  We are then told that it is true that someone now holds 4 times as many shillings as one of his neighbors and are asked how many men there are and how many shillings the poorest of the group must have had to start.

 

1 Comment »

  1. jyoak said,

    August 26, 2009 at 1:22 am

    I’ll provide the answer in Carroll’s words:

    [spoiler]
    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 shillings.  Hence, after k circuits, each is k shillings poorer, the last man now having nothing, and the moving heap containing 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 the first and last man are the only two neighbours whose possessions can be in the ratio of ‘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 – (7/m) [JY: Formated slightly], 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’.

    [/spoiler]

    This problem can be found in The Mathematical Recreations of Lewis Carroll – Pillow Problems and A Tangled Tale

    http://www.amazon.com/Mathematical-Recreations-Lewis-Carroll-Problems/dp/0486204936/ref=sr_1_1?ie=UTF8&s=books&qid=1251267521&sr=8-1

     

     

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