[Guest]

I liked the algebra behind the small puzzle Cockfit, so I thought I'd extend on it (and post new as that topic seems dead).

The original puzzle summarized:

There are gambling arenas which have all have both and entry and exit fee of $10. Each time you enter one of these arena's you are able to double your money in hand. How much money would you need to start with if you were to enter and exit three of these arenas and end up with nothing.

Simple algebra yields a solution of $26.25.

I'd like to extend the puzzle by taking away its constants? Say the entry and exit fees are $b, each time you enter an arena you are able to multiply your money by a, and you go through n arenas (and still have $0 in the end).

Can you derive a formula for the starting amount in terms of a, b, and n?

If we stick with the original problem and only modify n, we have a=2 and b=10.

Writing out the equations for 1 to 4 arenas, then simplifying them yields:

n: (equation)

1: 2x-30 = 0

2: 4x-90 = 0

3: 8x-210 = 0

4: 16x-450 = 0

Analyzing this allowed me to come up with the generic equation (2^n)*x - 10*( 2^(n+1) + 2^n - 3) = 0.

Using basic algebra, one can derive the formula for x to be x = 30*(1-1/(2^n))

It seems obvious that the 2 is directly determined from a.

Thus, the end formula should be x = g(a,b)*(1-1/(a^n)) where g(2,10) = 30.

Trying out a=3 and b=8 yields:

n: (equation)

1. 3x - 32 = 0 -> x = 16*(1-(1/3))

2. 9x - 128 = 0 -> x = 16*(1-(1/9))

Mentally trying a couple other a's allowed me to confirm that g(a,b) = b*((a+1)/(a-1))

Thus, the final formula is x = b*((a+1)/(a-1))*(1-1/(a^n))

Note that this solution isn't a true derivation as I only found patterns and derived a formula from them. See if you can top it with a true derivation!

Edited January 20, 2009 by vinays84

[akaslickster]

I liked the algebra behind the small puzzle Cockfit, so I thought I'd extend on it (and post new as that topic seems dead).

The original puzzle summarized:

There are gambling arenas which have all have both and entry and exit fee of $10. Each time you enter one of these arena's you are able to double your money in hand. How much money would you need to start with if you were to enter and exit three of these arenas and end up with nothing.

Simple algebra yields a solution of $26.25.

I'd like to extend the puzzle by taking away its constants? Say the entry and exit fees are $b, each time you enter an arena you are able to multiply your money by a, and you go through n arenas (and still have $0 in the end).

Can you derive a formula for the starting amount in terms of a, b, and n?

If we stick with the original problem and only modify n, we have a=2 and b=10.

Writing out the equations for 1 to 4 arenas, then simplifying them yields:

n: (equation)

1: 2x-30 = 0

2: 4x-90 = 0

3: 8x-210 = 0

4: 16x-450 = 0

Analyzing this allowed me to come up with the generic equation (2^n)*x - 10*( 2^(n+1) + 2^n - 3) = 0.

Using basic algebra, one can derive the formula for x to be x = 30*(1-1/(2^n))

It seems obvious that the 2 is directly determined from a.

Thus, the end formula should be x = g(a,b)*(1-1/(a^n)) where g(2,10) = 30.

Trying out a=3 and b=8 yields:

n: (equation)

1. 3x - 32 = 0 -> x = 16*(1-(1/3))

2. 9x - 128 = 0 -> x = 16*(1-(1/9))

Mentally trying a couple other a's allowed me to confirm that g(a,b) = b*((a+1)/(a-1))

Thus, the final formula is x = b*((a+1)/(a-1))*(1-1/(a^n))

Note that this solution isn't a true derivation as I only found patterns and derived a formula from them. See if you can top it with a true derivation!

Well, giving the answer this early is not recommended but, I looked because I forgot all my algebra in 1982 and never made sense of it. Edited January 21, 2009 by akaslickster

[Guest]

Can you derive a formula for the starting amount in terms of a, b, and n? ?????

Well this is the best i could do.....

solution.bmp