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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’.

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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