I am proud with myself
Mar. 18th, 2007 02:38 pm![[personal profile]](https://www.dreamwidth.org/img/silk/identity/user.png){kind=small}
fr_defenestratoAfter several years of intermittently pondering and being wholly stumped by a certain math problem, I figured out the solution the other day. This problem is part of Ronald Hoeflin's Power Test (Hoeflin is a philosopher working principally in "truth theory" and human intelligence, and he's created a number of really really difficult IQ tests and founded a number of snooty smart-people societies.) Anyway, this is No. 18, The Crystal Problem, from the Power Test:
"Suppose a tetrahedral-shaped crystal is formed, like a giant pile of apples or oranges at a greengrocer's store, consisting of one atom on the top layer, three on the next-to-top layer, six on the third layer, ten on the fourth layer, and so forth... If there are exactly 1,000,000 layers, specify the total number of atoms in the entire crystal. Give an exact answer, not an approximate one or a formula for making the calculation." (emphasis my'un)
(Fuck atoms and crystals; it's been billiard balls in my head since I first started working on the problem.)
For the longest time I couldn't for the life of me figure out how to get at the total number of billiard balls without making a million calculations or at least half a million. The rules for the Power Test forbid the use of any and all calculating tools besides pen and paper, so an Excel spreadsheet is right out.
So amongst any of the really smart people I cherish on LJ, has anybody ever tackled such a problem, or does a methodology for answering it seem perfectly obvious to anyone? Take your time... I've got it figured out and am about to put my methodology down on (e-)paper for posterity (advising, of course, that reading the answer disqualifies you from taking Dr. Hoeflin's test in future... for the record, I have not taken the whole test; the seeming impenetrability of this one problem convinced me I'm nowhere near smart enough... maybe I'll reconsider now). UPDATE: Here's the solution.
I should also note that I solved this problem without any recourse to higher math, which I don't remember at all... It's all algebra and logic.
Full disclosure of cheating activities: I looked up "tetrahedron" on Wikipedia, but learned nothing useful in the solution of the problem. I already knew from my own work, for example, that tetrahedra cannot tessellate space, and from an old plastic puzzle from childhood that a regular tetrahedron can be sliced in two in such a way as to create two halves of identical shape and size, with six vertices and five faces.
"Suppose a tetrahedral-shaped crystal is formed, like a giant pile of apples or oranges at a greengrocer's store, consisting of one atom on the top layer, three on the next-to-top layer, six on the third layer, ten on the fourth layer, and so forth... If there are exactly 1,000,000 layers, specify the total number of atoms in the entire crystal. Give an exact answer, not an approximate one or a formula for making the calculation." (emphasis my'un)
(Fuck atoms and crystals; it's been billiard balls in my head since I first started working on the problem.)
For the longest time I couldn't for the life of me figure out how to get at the total number of billiard balls without making a million calculations or at least half a million. The rules for the Power Test forbid the use of any and all calculating tools besides pen and paper, so an Excel spreadsheet is right out.
So amongst any of the really smart people I cherish on LJ, has anybody ever tackled such a problem, or does a methodology for answering it seem perfectly obvious to anyone? Take your time... I've got it figured out and am about to put my methodology down on (e-)paper for posterity (advising, of course, that reading the answer disqualifies you from taking Dr. Hoeflin's test in future... for the record, I have not taken the whole test; the seeming impenetrability of this one problem convinced me I'm nowhere near smart enough... maybe I'll reconsider now). UPDATE: Here's the solution.
I should also note that I solved this problem without any recourse to higher math, which I don't remember at all... It's all algebra and logic.
Full disclosure of cheating activities: I looked up "tetrahedron" on Wikipedia, but learned nothing useful in the solution of the problem. I already knew from my own work, for example, that tetrahedra cannot tessellate space, and from an old plastic puzzle from childhood that a regular tetrahedron can be sliced in two in such a way as to create two halves of identical shape and size, with six vertices and five faces.
