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Enigma 1716 - Pyramid selling
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Chappy
Nov 2, 2012, 6:12:37 AM11/2/12
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Enigma 1716 - Pyramid selling
New Scientist magazine, 22 September 2012.
By Susan Denham.
I commissioned a goldsmith to make some identical
solid gold tetrahedrons, in which the original
specification gave the length of each of the
sides. He calculated what the cost of the gold
would be and, as I could afford a certain whole
number percentage more, I asked him to increase
their size by that percentage. He started making
the tetrahedrons, but made the error of increasing
each side by that percentage. After he had made
one he realised that if he continued in this way
he would use much more gold than expected. So he
made the second tetrahedron with each side reduced
by that same percentage from its original
specification. The rest he made to the original
specification.
I still got the required number of tetrahedrons
and in fact the overall effect was to increase
the amount of gold used by the percentage I wanted.
How many tetrahedrons did I order, and what
percentage increase did I ask for?
Ciao,
Chappy.
New Scientist magazine, 22 September 2012.
By Susan Denham.
I commissioned a goldsmith to make some identical
solid gold tetrahedrons, in which the original
specification gave the length of each of the
sides. He calculated what the cost of the gold
would be and, as I could afford a certain whole
number percentage more, I asked him to increase
their size by that percentage. He started making
the tetrahedrons, but made the error of increasing
each side by that percentage. After he had made
one he realised that if he continued in this way
he would use much more gold than expected. So he
made the second tetrahedron with each side reduced
by that same percentage from its original
specification. The rest he made to the original
specification.
I still got the required number of tetrahedrons
and in fact the overall effect was to increase
the amount of gold used by the percentage I wanted.
How many tetrahedrons did I order, and what
percentage increase did I ask for?
Ciao,
Chappy.
Dave Baker
Nov 2, 2012, 7:33:32 AM11/2/12
to
"Chappy" <petergreg...@hotmail.com> wrote in message
news:0efaad35-f68d-4956...@googlegroups.com...
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)
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
Dave Baker
Nov 2, 2012, 8:23:47 PM11/2/12
to
"Dave Baker" <Nu...@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.
> 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)
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
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