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Do you know this one?
June 3, 2012 at 11:17 am
(This post was last modified: June 3, 2012 at 11:17 am by Whateverist.)
I enjoyed working on this little chestnut. Maybe you will too.
Use unit blocks (1 by 1 by 1 cubes) to build a larger cube with an odd edge length at least three unit blocks long. Now remove one stack of blocks (1 by 1 by n) from the larger cube. Explain why the number of blocks remaining must always be divisible by 24.
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RE: Do you know this one?
June 3, 2012 at 11:26 am
I can explain mathematically, in the form of a Proof by Induction.
If more of us valued food and cheer and song above hoarded gold, it would be a merrier world. - J.R.R Tolkien
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RE: Do you know this one?
June 3, 2012 at 12:29 pm
There's nothing wrong with proof by induction.
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RE: Do you know this one?
June 3, 2012 at 7:14 pm
Drop the modular arithmetic on this sucker! It's super-straightforward.
So these philosophers were all like, "That Kant apply universally!" And then these mathematicians were all like, "Oh yes it Kan!"
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RE: Do you know this one?
June 3, 2012 at 7:33 pm
I've been wracking my seemingly limited brains for hours...still nothing.
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RE: Do you know this one?
June 4, 2012 at 5:48 pm
Here it is:
NB: I haven't written every step of working, to save time and effort.
no. of blocks forming one stack = 2n + 1.
total no. of blocks = (2n + 1)^3
Therefore blocks remaining = (2n + 1)(4n^2 + 4n) = f(n)
Basis: for n=1. No. blocks remaining = 3x8 = 24. Therefore divisible by 24.
Assume true for n=k. i.e. (2k + 1)(4k^2 +4k) is divisible by 24.
For n= k + 1.
f( k + 1) = 4(2k +3)(k^2 + 4k + 4)
f(k + 1) - f(k) = 24k^2 + 48k + 48
= 24(k^2 + 2k + 2)
Therefore, f(k + 1) = f(k) + 24(k^2 + 2k + 2)
Both terms of this expression are divisible by 24, therefore f(k + 1) is divisible by 24. It therefore stands by induction that f(n) is divisible by 24 for all integers n>=1.
If more of us valued food and cheer and song above hoarded gold, it would be a merrier world. - J.R.R Tolkien
I feel stupid now...
