Friday Puzzler -- Five Beggars

Time for a hint?

Alex Y

The puzzle does not state what happens if one of the beggars is unable to put the required amount into the bag, even considering the possibility of borrowing from other players. And for good reason--that will not occur.

Hint 2

Alex Y

How many pennies are in the bag at each passing of the bag?

Re: Hint 2

Larry Barrett

No answer so far. This is the way I think the game goes as the bag is passed from beggar to beggar.

All five beggars start with the same number of pennies, say N. Lets call the five A, B, C, D, and E, knowing that A=1, B=2 and so on, because it will get confusing using numbers for the beggars and for the pennies they put into and take out of the bag.

So A puts N pennies into the bag and passes it to C.

C takes out 2 pennies and gives to B who now has N+2 and takes out 4 pennies and gives to D who now has N+4. C then puts 1/2N pennies into the bag and passes it to E; C now has N-1/2N = 1/2N pennies and the bag has N-6 +1/2N = 3/2N -6 pennies.

E takes out 1 penny and gives to A, who now has 1 penny. E takes out 4 pennies and gives to D who now has N+8 pennies. E puts into the bag 1/2(3/2N -6) and passes the bag to B. E now has N- 1/2(3/2N -6) = 1/4N +3 pennies and the bag has 3/2N -11 +(3/4N -3) = 9/4N -14 pennies.

B takes out 1 penny and gives it to A, who now has 1+1=2 pennies and takes out 3 pennies and gives to C, who now has 1/2N +3 pennies. B now puts 1/2(9/4N -14) pennies into the bag and passes it to D. B now has N+2 - (9/8N -7) = -1/8N +9 pennies and the bag has 9/4N -18 + (9/8N -7) =27/8N -25 pennies.

If N =72 then at this point B will have 0 pennies and the bag will have 243. If N is less than 72 then B will have had to borrow extra pennies from C, D, or E. We also know that after D completes his turn he will pass the bag to A, and A will have only 2 pennies, which could be the end of the game. What seems confusing to me (and I may already be confused by keeping track of what is in the bag after each turn) is that when D receives the bag he takes out pennies and gives them to C and takes out 5 pennies and gives to E. So even if B had to borrow pennies from C, D, and E, C and D will still have some pennies, which contradicts the story that A leaves with the bag plus his 2 pennies and all the rest have none.

Nearly there

Alex Y

You wrote:

No answer so far. This is the way I think the game goes as the bag is passed from beggar to beggar.

All five beggars start with the same number of pennies, say N. Lets call the five A, B, C, D, and E, knowing that A=1, B=2 and so on, because it will get confusing using numbers for the beggars and for the pennies they put into and take out of the bag.

So A puts N pennies into the bag and passes it to C.

C takes out 2 pennies and gives to B who now has N+2 and takes out 4 pennies and gives to D who now has N+4.

Right so far. But a corollary to hint 2 is don't worry about what is in each player's pile, since they can draw from other's piles (except A's) as needed). The piles of A, B, C, and D are common property. There's no harm in tracking individual piles, and it might be interesting once you get the answer as a confirmation, but it's not necessary.

C then puts 1/2N pennies into the bag and passes it to E;

Just to clarify the notation, I think you saying that C puts (1/2)N, or N/2 pennies in the bag. That is correct.

C now has N-1/2N = 1/2N pennies and the bag has N-6 +1/2N = 3/2N -6 pennies.

Correct, and your calculations on through D getting the bag, with 27N/8-25 pennies is all correct.

Then you say:

If N =72 then at this point B will have 0 pennies and the bag will have 243. If N is less than 72 then B will have had to borrow extra pennies from C, D, or E. We also know that after D completes his turn he will pass the bag to A, and A will have only 2 pennies, which could be the end of the game. What seems confusing to me (and I may already be confused by keeping track of what is in the bag after each turn) is that when D receives the bag he takes out pennies and gives them to C and takes out 5 pennies and gives to E. So even if B had to borrow pennies from C, D, and E, C and D will still have some pennies, which contradicts the story that A leaves with the bag plus his 2 pennies and all the rest have none.

But D will have to come up with (27N/8-25)/2 pennies to put in the bag before passing it to A. Doing so will wipe out all other player's holdings, other than A's

Whoops!

Alex Y

I said that the piles of A,B,C, and D could be considered to be communal property. WRONG! A is the one whose pile cannot be borrowed from. I should have said B, C, D, and E.

Re: Nearly there

Larry Barrett

So when D passes the bag to A, it must contain

(27N/8 - 25 - + 1/2(27N/8 - 25) pennies. The story says that when A receives the bag he puts the 2 pennies that he had received into the bag and then walks off, and all the other beggars have nothing. Since they all started with N pennies, the total number of pennies is 5N when A received the bag it also must have contained 5N-2 pennies.

Therefore, (27N/8 - 25 - + 1/2(27N/8 - 25)=5N-2

Reducing terms, 54N/8 + 27N/8 -66 -25 = 10N - 4,

multiplying by 8, 81N - 728 = 80N - 32

combining terms, 81N - 80N = 728 - 32,

and therefore N = 696.

Correct!

Another way to look at it

Alex Y

At each stage, the previous balance is multiplied by 1.5 and an amount is pulled out of the bag. Chronologically, it happens in the opposite order, which is needed so that D doesn't pull out and distribute pennies to E and C before passing the bag to A. But mathematically, it can be looked at in this order.

A's original N into the bag gets increased 4 times, by C, E, B, and D, so its contribution to the final amount in the bag is

N*1.5^4 = (5 1/16)*N

C's withdrawal of 6 gets increased 3 times, so its contribution to the final amount is

-6*1.5^3 = -(3 3/8)*6 = -20 1/4

Other withdrawals have analogous effects on the total:

E: -5*1.5^2 = -(2 1/4)*5 = -11 1/4

B: -4*1.5 = -6

D: -8

A: +2

So the final amount in the bag is

(5 1/16)N - 20 1/4 - 11/1/4 - 6 - 8 + 2

=(5 1/16)N - 43 1/2

Set this equal to 5N and solve to get

(1/16) N = 43 1/2, or

N = 16 * (43 1/2) = 696

Re: Another way to look at it

Larry Barrett

Not an easy answer to intuit no matter how you look at it. I started trial and error with 10 pennies each to see if there was a pattern, then 12, then quit. That was not a useful approach. I wonder how Charles Dodgson came up with this problem.