[Math Lair] Pillow-Problems: Problem #8

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For more information on this collection, see Pillow-Problems by Charles L. Dodgson (Lewis Carroll).


8. (20, 34)

Some men sat in a circle, so that each had 2 neighbours; and each had a certain number of shillings. The first had 1/ more than the second, who had 1/ more than the third, and so on. The first gave 1/ to the second, who gave 2/ to the third, and so on, each giving 1/ more than he received, as long as possible. There were then 2 neighbours, 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?



8. (2, 34)

7 men; 2 shillings.


8. (2, 20)

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’.