Answer 3: Men in a circle

7 men, 2 shillings

Reasoning

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 contains mk shillings. Hence the thing ends when the last man is again called upon 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 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=3-7/m, which evidently gives no integral values other than m=7, k=2.

The second gives 4mk=2-3m, which gives no positive integral values.