Problem of the Month (July 2014)

This month we extend the problem of the December 1998 Math Magic to 3 dimensions. Consider the ways in which an n×n×n cube can be cut into smaller integer-sided cubes. Define:

f(n) = the largest possible size of the smallest cube

g(n) = the smallest number of cubes

h(n) = the smallest possible value of the largest multiplicity of cubes needed

When n is composite with smallest prime factor p, it appears that f(n) = n/p f(p) and g(n) = g(p). Can you find counter-examples? Thus we mostly concern ourselves with finding the values of f(n) and g(n) for n prime.

Can you beat the best known tilings below?

In particular, what is the smallest n for which f(n) = k? How fast does g(n) grow? What is the smallest n for which different packings are needed to illustrate f(n) and g(n)? What is limn→∞ h(n) ? It is known that no n has h(n)=1. Are there n for which h(n)=2?

And a meta-question: what is the best way to visualize complicated cube packings?


ANSWERS

Answers were received from Joe DeVincentis.

Here are the best known values for f(n) and g(n), the cubes needed for that tiling, and top and bottom views (with the unit squares omitted):

nf(n)g(n)
218
1(8)
 
3120
2(1) 1(19)
 
5150
3(1) 2(7) 1(42)
 
7171
4(1) 3(4) 2(15) 1(51)
 
(Joe DeVincentis)
11196
6(1) 5(3) 4(1) 3(12) 2(39) 1(40)
 

Bryce Herdt showed f(29)≥2, f(41)≥3, f(71)≥4, and f(97)≥5. Are these the smallest prime cubes with those values of f?

Here are the best known values for h(n), and the cubes needed for that tiling:

nh(n)Cubes
281(8)
3192(1) 1(19)
482(8)
5423(1) 2(7) 1(42)
683(8)
7514(1) 3(4) 2(15) 1(51)
(Joe DeVincentis)
884(8)
9196(1) 3(19)
(Bryce Herdt)
1085(8)
11406(1) 5(3) 4(1) 3(12) 2(39) 1(40)
1286(8)


If you can extend any of these results, please e-mail me. Click here to go back to Math Magic. Last updated 7/1/14.