The book Pillow-Problems: Thought Out During Wakeful Hours, by Charles Dodgson, better known by the pseudonym Lewis Carroll, was first published in 1893. It contains 72 problems that Carroll thought of while lying awake at night over the course of a few decades. Carroll's intent was that the puzzler would, like he did, work out the answers to the questions mentally.
This is problem #8 in the book.
Previous problem: Problem #5.
Next problem: Problem #16
Note about pre-decimal British currency: Before decimalisation, the pound (symbol £) was divided into 20 shillings, (symbol s) and each shilling into 12 pence (singular penny; symbol d), making 240 pence to the pound.
Usually, prices were written with a / (called a solidus) between the amounts. For example, a sum of four shillings and eight pence was written as 4/8, and pronounced "four and eight". An even sum of shillings, say six shillings, was written as 6/-. A sum of one pound, nineteen shillings, and eleven pence would be written as £1/19/11 and pronounced "one pound, nineteen and eleven".
8.
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?
[3/89
Answer: men, shillings.
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 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 −
(7)/
(m)
, 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.
