Can you solve this rather difficult problem?

Question

John is sitting around a round table with some other men. He has one more dollar than the person to his right and that person in turn has one more dollar than the person to his right and so on around the table. Then, John gives one dollar to the person to his right and he in turn gives 2 dollars to the person to his right and that person gives 3 dollars to the person to his right and so on. This process continues around the table as many times as is necessary until someone has no money left. At that time, John has 9 times the money as the person to his right. How many men are there along with John, how much money did he start with, and how much money does he have at the end?
Can you prove the uniqueness of the solution?

O well, I dont think anyone is going to get this right.........................

Answer 1

There were 16 along with John (17 total). John started with $23. At the end he had $135.

Here's the proof of uniqueness. There are k men around the table. Suppose Martin is to John's right, and Abraham is to John's left. (Abraham, Martin, & John, but not in order.) The men are numbered 1 to k counterclockwise beginning with John (#1) and ending with Abraham (#17). A "round" of passing money begins with John and ends with Abraham giving money to John. All money John receives is set aside in a separate pile.

Abraham starts with m dollars. That means John starts with k+m-1 dollars, and Martin starts with k+m-2 dollars. In each round, all k men give away one dollar, and those k dollars end up in John's pile.

After m rounds, Abraham goes broke. He started with m dollars, and gave away m dollars, one on each round.

Excluding the pile he's building up separately, John started with k+m-1 dollars, and gave away m dollars in the m rounds, so when Abraham goes broke, John has (k+m-1) - m = k-1 dollars in his "working" pile.

After each round, Martin always has one dollar less in his working pile than John has in his. So when Abraham goes broke, Martin has k-2 dollars.

In his "side" pile, after m rounds, John has mk dollars. Pooling all his money (both piles), John has (k-1) + mk dollars, and that is nine times what Martin has.

Now let's do algebra.

(k-1) + mk = 9(k-2)
k - 1 + mk = 9k - 18
17 = 8k - mk = k(8 - m)
8 - m = 17/k

But 17 is a prime number divisible only by 1 and 17.
So k = 17.
8 - m = 1 ==> m = 7

Abraham started with $7; John started with 17+7-1 = $23; Martin started with $22.

John ended with $16 in the working pile and 17*7 = $119 in the "side" pile, for a total of $135.

Martin ended with $15, one-ninth of what John had.

I have a unique solution for both m and k, so the entire solution is unique.

Answer 2

Every full time around the table, everyone looses one dollar except for John which gains all of those dollars.

Because the person on John's left is the one with the least amount of money, they will always run out of money first.

If John starts with x dollars, and there are n people around the table, then the first full time around the table John will have x+n-1 dollars. The person to his right will have x-2 dollars. The second time around John will have x+2(n-1) dollars and the person to his right will have x-3 dollars, etc.

Let c equal the number of times around the table. The game will end when the person to John's left has 0 dollars x - (n - 1) - c = 0 and when John has 9 times the person on his right x + c(n - 1) = (9)(x-1-c)

Question: Find all solutions for n, x, and how much he has at the end, which is = x + c(n - 1), using these formulas.
x - (n - 1) - c = 0
x + c(n - 1) = (9)(x - 1 - c)

(n - 1) = x - c
x + c(x - c) = (9)(x - 1 - c)
x + cx - c^2 = 9x - 9 - 9c
-c^2 + 9c + 9 = 8x - cx
x = (-c^2 + 9c + 9) / (-c + 8)
x = (c^2 - 9c - 9) / (c - 8)
Synthetic division yields
x = c - 1 - 17/(c - 8)
x and c must be whole numbers. Any whole c where +/- 17/(c - 8) is an integer are possible solutions. The only way a fraction can be an integer is when the denominator is evenly divisable by the numerator. Since the numerator is prime, then the solutions must be the following c-8 = 17, c-8 = -17, c-8 = 1, c-8 = -1.
These formulas yield c = 25, -9, 7, 9. -9 can't be an equation since c must be a positive integer.

Lets check the results
Using x = c - 1 - 17/(c - 8)
If c = 9 then x = -9 (Invalid)
If c = 25 then x = 7
If c = 7 then x = 23

Using n = x - c + 1
If c = 25, and x = 7 then n = -17 (Invalid)
If c = 7, and x = 23 then n = 17

Therefore the only answer is
17 men including John,
John started with 23 dollars
John has (9)(x-1-c) = 135 dollars in the end.

Answer 3

I must not be reading the question right.

From what information is presented. John has some money and each person to his right has a dollar more. This holds true until we get to the person seated on John's left. The person seated on John's left has just as much money as John has, plus a dollar more for every person seated at the table. If we subtract what John has from the person on his left, then we know how many people is seated at the table. John has the least amount of money. John will reach zero before anyone else does. The person on John's right will take John's dollar and add his to it and give it to the person to his right. He is only out a dollar from his original amount. The difference between John's money and everyone else (with the exception of who was given the money last) remains the same.

If John will reach zero first, the game is over. Nothing times 0 will give 9 and the last part of the question is false.

Some more clarification is needed.

Answer 4

Edit: I see bpiguy already gave a tidy solution.

There are a total of 17 people at the table. John started with $23 and finished with $135

If there are N men, then John gains N-1 dollars per round and everyone else loses 1 dollar per round. If John starts with X bucks then the guy on the left will have X-N+1 bucks to start , and the guy to his right has X-1 bucks. It will take X-N+1 rounds to break the guy on his left, and at that time John will have X+(N-1)(X-N+1) = -N^2 +2N+NX-1 bucks and the guy to his right will have X-1-(X-N+1) = N-2 bucks.
We are given -N^2+2N+NX-1 = 9(N-2) and so N(N+7-X) =17. Which means N=17 (since N not 1), and N+7-X=1. So X = $23. And John finished with $135. The man on John’s right finished with 17-2 = $15 = $135/9; check.

Answer 5

I stopped at the fact that "He has one more dollar than the person to his right and that person in turn has one more dollar than the person to his right and so on around the table"

That statement kills all solutions, because ultimately, the person to John's left will be looking to have one more dollar than John, and if John started the trend, or is somewhere near it, that cannot be true.

Answer 6

False alarm, it's not impossible... my math was wrong, but I have since fixed it. Here's the solution:

(Note: the following proof is based on the assumption that john has more money than the person to his left [question's unclear])

Observations:

John cannot have more money than there are people at the table. If John has the same amount of money than there are people at the table, then the last person, the one to his right, will have 1 dollar. So, if John has more money than there are people at the table, the person to his left will have 0 dollars and the experiment will not even begin.

The person to John's left will have the least amount of money.

The experiment will not end until everyone at the table has passed money an equal number of times. Let's say John has 10 dollars, and there are five people at the table (including him). John will give one dollar to the person on his right, whom having 9 dollars will now have 10. John will now have 9 dollars. The person on his right will give 2 dollars to the person on his right, whom having 8, now has 10. The person to John's right will now have 8 dollars. A pattern emerges. Every time a player is given money, then gives it away, he or she will have 1 less dollar than before and the maximum amount at any time in the round will be the amount John started with at the beginning of the round (a round being from the time John gives money to the time he receives it again, thus everyone has played [taken and given money] once). Therefore, because everyone’s money decreases by 1 every round, the person to John’s left must run out first. Thus, the game will only end at the end of a given round.

The number of rounds will be equal to the amount of money the person to the left of John starts with. If he loses one dollar every round, then starting with r dollars will cause r rounds of play until he is out of cash.

The amount of money the person to John’s left has (as well as the number of rounds, we’ll call it r) will be dictated by the number of people at the table, as well as how much money John starts with. The amount of money John starts with (J) subtract the number of players in the game other than John (E) will be equal to the number of rounds of play (r).
r = J – E

The amount of money that the person to John’s right has at the end (R) will be equal to the number of people in the game (n) minus 2, because he will have one dollar for every person between John and the person to his left.
R = n – 2

The number of people other than John is equal to the number of people in the game minus one (John).
n – 1 = E

Since every person other than John loses 1 dollar every round, John must gain one dollar from each of them every round. If we multiply the number of people other than John by the number of rounds, we will find how much money John will make at the end of the game. Add this to John’s original bankroll and we’ll have how much John will have at the end.
J(end) = J + rE

We want John to have 9 times the money that the person to his right has at the end.
J(end) = 9R

Equations:
r = J – E
R = n – 2
n – 1 = E
J(end) = J + rE
J(end) = 9R

J(end) = 9R
J + rE = 9(n – 2)
J + (J – E)E = 9n – 18
J + JE – E^2 = 9n – 18
J + J(n – 1) – (n – 1)^2 = 9n – 18
J + Jn – J – n^2 + 2n – 1 = 9n – 18
Jn – n^2 – 7n + 17 = 0

Since J cannot be less than n, Jn will always be equal to or more than n^2, thus Jn – n^2 will be non-negative. This means that either -7n + 17 = 0, which is impossible, because 17 is prime, or 7n will be a multiple of 17 and J – n will be the number multiples of 17 that 7n is over 17. As little sense as this makes, it works.

n must be a multiple of 17, so we’ll say n is 17. -7(17) + 17 = -6(17), therefore J must be 6 more than n. Thus J has to be 23.

23(17) – 17^2 – 7(17) + 17 = 0
391 – 289 – 119 + 17 = 0
0 = 0 It works!

But is it unique? Let’s see… n must be a multiple of 17, so let’s try 34. -7(34) + 17 = -6.5(34). Hmm… no good. 6.5 must equal J – n, but J and n are integers, so this won’t work. In order to get the 6.5 though, we merely divide -7n + 17 by n:

(-7n + 17)/n = J – n
The only way this can be an integer is if there is a factor of n in the numerator on the left hand side. In order for that to occur, there must be a factor of n in 17. Since 17 is prime, the only possible value for n is 17. And that’s it.

So, if John starts the game with 23 dollars, and there are 17 players (including John), the game will end with John having 9 times the cash that the person to his right has. And yes, this solution is unique.

That took way too long...

Answer 7

9 Persons and 27 Dollars.

Answer 8

9 people were there, assuming he started with 1 dolar, and each person that give a dolar plus what they had will have nothing, so he ended with 9 dolar (9 times more than the person at his right)

Answer 9

I don't know a solution off the top of my head, but I suspect that there cannot be a unique solution using only that information.

Answer 10

No one has time for this!