We call a natural number 3-partite if the set of its divisors can be partitioned into 3 subsets each with the same sum. Show that there exist infinitely many 3-partite numbers.
Multiply such a number by a prime it doesn't have as a divisor to get another such number as you get the old divisors plus their corresponding multiples and can hence form the subgroups accordingly.
OccamsAftershave wrote:
Multiply such a number by a prime it doesn't have as a divisor to get another such number as you get the old divisors plus their corresponding multiples and can hence form the subgroups accordingly.
This has to be combined with the fact that there is at least one 3-partite number, and there is one: 120.
You can form infinitely many such 3-partite numbers by taking 120 and multiplying it by some prime p > 5.