Simple but interesting mathematics

On cube roots of the cubes of two-digit integers.

This problem appeared on an Australian television program "The Panel" in 2001.  A guest was able to determine, in his head, the cube root of the cube of any two-digit integer.  For example, a member of the panel would choose a two digit integer (59 say), determine its cube using a calculator (59x59x59 = 205379) and would then give this number to the guest, who would instantly respond with its cube root, 59.  How did the guest do this?

Actually, its very simple (of course it had to be, or else the guest wouldn't have been able to do it!).  It relies on the fact that there is a one-to-one mapping between integers 0 to 9 and the last digit of their cubes. i.e.
   0^3 =    0     0->0
   1^3 =    1     1->1
   2^3 =    8     2->8
   3^3 =   27     3->7
   4^3 =   64     4->4
   5^3 =  125     5->5
   6^3 =  216     6->6
   7^3 =  343     7->3
   8^3 =  512     8->2
   9^3 =  729     9->9
The mapping turns out to be very easy to remember also, when one notices that the last digit of the cube is the same as the original for 0,1,4,5,6, and 9, and is the 10's complement for 2,3,7 and 8.  That is, given the last digit of the cube, one can instantly determine the last digit of the original two-digit integer.  The first digit is of course determined by the range in which the cube falls.

e.g. 79507 lies between 64000 and 125000, and therefore the first digit of the two-digit integer is 4. The last digit of the cube is 7, so the last digit of the two-digit integer is 3.  That is 79507^{1/3} = 43.

I wonder what this fellow would have thought the cube root of 79508 was.  I bet he would have said 42, which is of course nonsense.

As an aside, similar one-to-one mappings exist for all odd-numbered powers.  The above mapping holds for powers of 3, 7, 11, ...  For powers of 5, 9, 13, ... the mapping is even simpler, with the last digit of the result being the same as the original integer.  i.e.
So, one could just as easily do any odd-numbered power (except that typical calculators don't display enough digits!)

Round-Robin Tournament Schedules

This one came up when one of my students was trying to make a schedule for a twelve-team pool competition.  Essentially, she wanted every team to be able to play every other team exactly once, each playing one game per round, over an eleven-round competition.

For fewer teams, this is a simple exercise.  For example, with only two teams its trivial: Team 1 v Team 2.  For four teams a suitable schedule is:

Round 1
1 v 2
3 v 4

Round 2
1 v 3
2 v 4

Round 3
1 v 4
2 v 3

Six teams is a little trickier, but can be done by trial and error with pen and paper in just a few minutes.  As the number of teams in the competition is increased, this becomes very difficult very quickly --- try doing it for just 10 teams!

Although the solution isn't unique (different schedules can be produced by reordering rounds or relabelling teams), the total number of possible orderings of these games is huge for even a small number of teams.   If you had to do it by trial and error, this would take an excessively long time.

Fortunately, there exist some simple algorithms for generating these schedules quickly. 

Schedules for n=2,4,...,24, can be found here.

A small program for computing such schedules (and those for up to 10000 teams!) can be found here.


David Scullen