Friday Puzzle -- Jabberwock

Re: Friday Puzzle -- Jabberwock

Larry Barrett

The initial poorest person started with the same amount that the initial richest person ended up with. And the number of people in the circle is one more than that number.

  I don't think so

Alex Y

Ad least that doesn't match Rev. Dodgson's answer, and I can't see an answer in which that would be true.

Question

Larry Barrett

Is this the correct interpretation of the problem?

Suppose there are 4 players, A, B, C, D, and A starts with 6 dollars, B with 5, C with 4, D with 3.

Round 1. A gives 1 to B, B then gives 2 to C, C gives 3 to D, D gives 4 back to A.

So after Round 1, A has 9 (6 - 1 + 4), B has 4, C has 3, D has 2.

To start round 2, A then gives 5 to B, B gives 6 to C, C gives 7 to D, D gives 8 to A.

So after Round 2, A has 12, B has 3, C has 2, D has 1.

This continues until one player, after receiving $$ from the player on his left, and adding that to what he has, does not have enough to pass on to the player on his right.

Correct

Re:   I don't think so

Larry Barrett

After further scratching of my head and several more back of the envelopes, I think there are 7 in the circle. If you multiply the amount the poorest person in the first round started with times the amount the richest person in the first round ended up with in the last round, it is half what the richest person in the last round ended up with.

That's better

Alex Y

Lewis Carroll presents a solution that is pretty straight-forward high-school algebra. The setup is not that easy to come up with, though.

Actually, he solved directly for the two numbers the puzzle asks for.

I think one paradigm shift makes it immensely easier: imagine a basket that is passed around, with each person adding a dollar before passing it to the next. Adding that little bit of wicker clearly doesn't change the substance of the problem, but it makes it a lot easier for me to see the solution.

Solution

Alex Y

M people in the circle

$k held by the poorest person at the start.

I start with M+k-1 dollars.

After k complete rounds, everyone is $k poorer, except for the person holding the pot. The pot being passed from the poorest person to me has Mk dollars in it. The poorest person has $0 after making this pass, and I have M-1 dollars before accepting the pot.

It is obvious at this point that the person to my right, who now has $0 will not be able to add to the pot on the next round, so the process will stop after M-1 more passes.

At that point, I will have M-2 dollars, and the originally poorest person will have Mk+M-1. It's pretty easy to convince yourself that the only 4:1 ratio possible is between the originally poorest person and me. So

Mk +M -1 = 4*(M-2)

Mk =3M -7

k=3-7/M

And the only non-negative integral solutions to that are k=2 and M=7