Sharing Circles

[BMAD]

Some card players sat in a circle, so that each had two neighbors, and each had a certain number of dollars. The 1st player had $1 more than the 2nd player, who had $1 more than the third, and so on. The first player gave $1 to the second, who gave $2 to the third, and so on, each giving $1 more than they received, around and around the table as long as possible. There were then 2 neighbors, one having 4 times as much money as the other.

(a) How many players were there?

(b) How much money did the first player start with?

[Prime]

This way it's more interesting. I understand, one person has 4 times the money of his neighbor at the end of the game.

Fractional amount solutions involve fractions of a penny, therefore we should not consider those.

Then the game is over when the money is passed to the man who has $0 left, hence he cannot make a transfer and keeps the money. That would be the man who started with the lowest amount of money.
The only whole number solution where one man has 4 times more than his neighbor may exist between the first man and the last at the end of the game.

If there are N gamblers and the last has X dollars, then the first one has X+N-1.
The money transfers shall make X full rounds plus one more round stopping at the last man. With each transfer, the pot gains $1. Thus, in the end, the last man shall receive N*(X+1) - 1 = NX + N - 1 dollars. The first man will have his bank reduced to (X+N-1) - (X+1) = N - 2 dollars.
There are two possibilities:
either NX + N – 1 = 4*(N - 2),
or 4*( NX + N - 1) = N -2.
The second equation does not have positive whole number solutions. The first one yields a single whole number positive solution: N = 7; X = 2.
Therefore,
7 players, the first one started with $8, the last – with $2. They ended: $5, $4, $3, $2, $1, 0, and $20. ($20 and $5 being 4:1 ratio between two neighbors.)

They don't need to go through the motions. Each could just pay to the last man in accordance with the above formula and head to a nearby bar.

[Sairakan]

2 players, player 1 has $3, player 2 has $2.

Edit: When you say, "as long as possible," what exactly does that entail? That may change the solution.

Edited March 31, 2013 by Sairakan

[BMAD]

2 players, player 1 has $3, player 2 has $2.

Edit: When you say, "as long as possible," what exactly does that entail? That may change the solution.

There are more than 2 people. The phrase as long as possible means that each person passed until the last person received money.

[Sairakan]

4 people, with player 1 having $4.

[Prime]

2 players, player 1 has $3, player 2 has $2.

Edit: When you say, "as long as possible," what exactly does that entail? That may change the solution.

There are more than 2 people. The phrase as long as possible means that each person passed until the last person received money.

If the "last" person does not pass money to the "first", why do they sit in a circle and not in row?

Making an assumption, the game does not continue in circles, as the OP suggests and the clarification denies.

On each turn, the gambler receiving money has as much as the first one had initially. As long as there is a player in the lineup with the amount of money 4 times less than the "first person," the 1:4 ratio between two neighbors will occur.

If the first player has a whole number of $$, then it must be divisible by 4. E.g. First has $4, fourth has $1. When the 3rd player gets his $2, his total is $4 and his neighbor, the the 4th, has $1. So 4 people will do. Or if the first player has $8 and there is a gambler in the lineup who has $2, again the condition of 4:1 will be met. Then there must be at least 7 players (8 will also work.) For the first player with $12, there must be one with $3 -- at least 10 men total. And so on.

Therefore, the first player must have N $$ divisible by 4 and the number of players must be at least N + 1 - N/4.

In case, those gamblers do not use cash, rather computer record of their account, fractional solutions are also possible. E.g., the first person has $5 and 1/3.

Edited March 31, 2013 by Prime

[BMAD]

2 players, player 1 has $3, player 2 has $2.

Edit: When you say, "as long as possible," what exactly does that entail? That may change the solution.

There are more than 2 people. The phrase as long as possible means that each person passed until the last person received money.

If the "last" person does not pass money to the "first", why do they sit in a circle and not in row?

Making an assumption, the game does not continue in circles, as the OP suggests and the clarification denies.

On each turn, the gambler receiving money has as much as the first one had initially. As long as there is a player in the lineup with the amount of money 4 times less than the "first person," the 1:4 ratio between two neighbors will occur.

If the first player has a whole number of $$, then it must be divisible by 4. E.g. First has $4, fourth has $1. When the 3rd player gets his $2, his total is $4 and his neighbor, the the 4th, has $1. So 4 people will do. Or if the first player has $8 and there is a gambler in the lineup who has $2, again the condition of 4:1 will be met. Then there must be at least 7 players (8 will also work.) For the first player with $12, there must be one with $3 -- at least 10 men total. And so on.

Therefore, the first player must have N $$ divisible by 4 and the number of players must be at least N + 1 - N/4.

In case, those gamblers do not use cash, rather computer record of their account, fractional solutions are also possible. E.g., the first person has $5 and 1/3.

my clarification was poorly worded. I apologize. What i meant by the last person that receives money was not meant to imply a last person in a sequence. It was meant until the last person receives money and the 'rule' for passing can't be continued.

[dark_magician_92]

7 players, with 2$,3$.....8$.

In the end the one with 2$ ends up with 20$, and the neighbour (with 8$) ends with 5$.

[dark_magician_92]

Though Prime has already solved it.