# Thread: Tricky Word Problem

1. Hey, I'm interested in Math problems, and I solved this problem in "536 puzzles & curious problems":

Seven men engaged in play. Whenever a player won a game he doubled the money of each of the other players. That is, he gave each player just as much money as each had in his pocket. They played seven games and, strange to say, each won a game in turn in the order of their names, which began with the letters A, B, C, D, E, F, and G. When they had finished it was found that each man had exactly \$1.28 in his pocket. How much had each man in his pocket before play?

Here's The solution:       I've worked this solution using a table, and noticed the strange behavior. As you can see, 1.28\$ = 128 cents = 2^7, and there are 7 players, playing 7 games. To make sure this isn't a coincidence, I worked a scenario where 8 players play 8 games, and each end with 2^8 = 256 cents. Sure enough, the last player started with 8*1 +1 = 9 cents, the next 8*2 +1 = 17 cents, the third 8*4 +1 = 33 cents.

I've been fiddling with this for the last two days, and I still can't see why it behaves this way?!
This isn't homework, mere enthusiasm. If you could explain this mystery for me, I would be most grateful.  2.

3. I haven't worked it out, but I think I have a couple of potential useful ideas for doing so. First, and most important, I would work it out backwards. Start with everyone having 256 cents at the end and analyze the steps in reverse order. This way you at least know where you are starting. Second, I would try to do the analysis algebraically, although perhaps with the 256 = 28 put in explicitly. I think that the pattern might become clear quite quickly.

Then you could have more fun generalizing the number of players and the corresponding number of games played. Is the simple answer tied to having 2N+1 cents for N players, or is the answer simple for other combinations?

By all means tell us what you find.

On second thought, I think you saw the first of these suggestions. Still, the second might help.  4. For my information, is this a logarithmic function? I seem to recognize an exponential effect (result).  5. Originally Posted by Write4U For my information, is this a logarithmic function? I seem to recognize an exponential effect (result).
Remembering that I haven't actually solved it, I see exponentials [2n] but not logarithms.  6. I recently saw a TV lecture By Prof. Bartlett regarding the exponential function.
He explains the effects of steady growth rates on the exponential growth of anything.
For instance a steady growth rate of 1% produces a "doubling time" of almost 70 years. A steady growth rate of 2% produces a doubling time of 35 years. 5% = doubling time of 14years. These are derived from the logarithmic tables.

70 = 100Lm2 = 69.3

I have no idea what I am trying to propose. But below is the link where Bartlett explains the exponential function from several different perspectives. It just looked familiar to me.  7.  8. Originally Posted by Sandwitch Here's The solution:       I've worked this solution using a table, and noticed the strange behavior. As you can see, 1.28\$ = 128 cents = 2^7, and there are 7 players, playing 7 games. To make sure this isn't a coincidence, I worked a scenario where 8 players play 8 games, and each end with 2^8 = 256 cents.

128=64+32+16+8+4+2+1 +1

256=128+64+32+16+8+4+2+1 +1

In general 2^n = 1 + 1 + 2 + 4 ... +2^(n-1)

Prove it with 2^n = 2^(n-1) + 2^(n-1) , and do this recursively with only one term. 2^(n-1)=2^(n-2)+2^(n-2).. etc etc  Bookmarks
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