Make a list of all 120 five-digit numbers created by permuting the digits 1,2,3,4,5.
Consider it as a set of numbers in base-10 system, ranging from 12345 to 54321.

Show that it is possible to partition this set into 2 subsets possessing equal sums of squares of their respective members.

bonus question: Is the result dependent of the base we choose?

I just spent 20 minutes typing this in and then hit a wrong buttonand deleted the whole thing.  Here is the abbreviated version with little proof.

It is possible, independent of base or choice of digits.

There are two subsets of 60.  The first subset starts with these 12

abcde
abecd
abdec

acbde
aebcd
adbec

acdbe
aecbd
adebc

acdeb
aecdb
adecb

Note a is fixed, in each trio b is shifted, and cde rotate.
The other four sets of 12 have a in each of the other positions, each trio has b in each of the four other spaces, and within each trio cde rotate again.

The other subset of 60 mirrors these but d and e are swapped.  (the cde cycle becomes ced)

Posted by Jer on 2011-04-25 10:59:05