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From: "Dave Baker" <N...@null.com>
Newsgroups: rec.puzzles
Subject: Re: Enigma 1716 - Pyramid selling
Date: Sat, 3 Nov 2012 00:23:47 -0000
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"Dave Baker" <N...@null.com> wrote in message 
news:k70b2f$orr$1@news.datemas.de...
> Let the number of tetrahedrons ordered be N, the original volume of each 
> one be V and the percentage increase in volume requested be P.
>
> The desired amount of gold finally ordered is therefore NV(1 + P/100)
>
> What was delivered is V(1 + P/100)^3 + V(1 - P/100)^3 + V(N-2) which is 
> equal to NV(1 + P/100)

Just to run through the maths in more detail if anyone is interested. 
Firstly let P/100 = X to simplify the equations. Then:

V(1 + X)^3 + V(1 - X)^3 + V(N-2) = NV(1 + X)

V(1 + 3X + 3X^2 + X^3) + V(1 - 3X + 3X^2 - X^3) + NV - 2V = NV + NVX

Cancel out various factors

V(1 + 3X^2) + V(1 + 3X^2) - 2V = NVX

2V + 6VX^2 - 2V = NVX

6X = N i.e. 6P/100 = N


> Solving that out we get 6P/100 = N where P<100
>
> The only integer solution is therefore P = 50 and N = 3
>
> The large tetrahedron has a volume of  3.375V, the small one 0.125V and 
> the remaining one V for a total volume of 4.5V which equals the 3V plus 
> 50% ordered.
> -- 
> Dave Baker
>
>