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Poor Man's Society
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The Poor Man's Society had a meeting yesterday. Each member of the Society brought with them all the money they had in terms of one-dollar coins and nothing else.
The richest member of the Society had exactly one dollar more that the second-richest man, who had exactly one dollar more than the third-richest man, and so on.
The men sat at a circular table so that each man had two neighbours. To the richest man's right sat the second-richest man, who had the third-richest man sitting to his right, and so on, until the richest man sat to the poorest man's right.
The richest man gave exactly one one-dollar coin to the second-richest man, who gave exactly two coins to the third-richest man, and so on, each man giving the man to his right exactly one coin more than they had received for as long as possible.
As soon as it was all over, one man had exactly four times as much money as his neighbour.
How much money did the richest man have at first?
The richest member of the Society had exactly one dollar more that the second-richest man, who had exactly one dollar more than the third-richest man, and so on.
The men sat at a circular table so that each man had two neighbours. To the richest man's right sat the second-richest man, who had the third-richest man sitting to his right, and so on, until the richest man sat to the poorest man's right.
The richest man gave exactly one one-dollar coin to the second-richest man, who gave exactly two coins to the third-richest man, and so on, each man giving the man to his right exactly one coin more than they had received for as long as possible.
As soon as it was all over, one man had exactly four times as much money as his neighbour.
How much money did the richest man have at first?
Hint
To solve this problem, you must know how many members there were in total, and how much the poorest man had at first.Answer
The richest member had exactly 8 dollars at first.Let m be the number of members of the Poor Man's Society. Let c be the number of one-dollar coins owned by the poorest man at first. This means that, at first, the richest man had m+c-1 dollars.
After exactly one cycle, each man is one dollar poorer, and the poorest man is giving the richest man exactly m coins. Therefore, after c cycles, each man is c dollars poorer, and the now-penniless poorest man is giving the richest man m*c coins.
The circuit ends when the second-poorest man gives the poorest man m*c+(m-1) dollars, but the poorest man is unable to give the richest man m*c+m dollars. At this point, the richest man has lost c-1 dollars and now has (m+c-1)-(c-1), or m-2, dollars.
The richest man and the poorest man are now the only two neighbours who can have a 4:1 ratio of dollars. Therefore, either 4*(m*c+m-1)=m-2 (meaning 4*m*c=2-3*m) or m*c+m-1=4*(m-2) (meaning m*c=3*m-7)
There are no solutions to the first of these two equations that give positive integral values to both m and c, but the only positive integral solution to the second equation is m=7 and c=2.
Therefore, there are 7 members of the Poor Man's Society, the poorest man had 2 dollars at first, and the richest man had 8 dollars at first.
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