Question

Let $m$ and $n$ be positive integers such that $m$ has exactly 5 positive divisors, $n$ has exactly 6 positive divisors, and $mn$ has exactly 14 positive divisors. How many distinct prime factors does $mn$ have

Answer 1

If
`A` is the set of positive divisors of
`m` and
`B` the set of positive divisors of
`n`, then
`A\cup B` is the set of positive divisors of
`mn`.

Use the inclusion/exclusion principle:

`|A \cup B| = |A| + |B| - |A \cap B|`

where
`|\cdot|` denotes set cardinality (the number of elements the set contains). The set
`A\cap B` is the set of common divisors of
`m` and
`n`. Then

`14 = 5 + 6 - |A\cap B| \implies |A\cap B| = 3`

so that
`m` and
`n` share 3 divisors
`d_1,d_2,d_3`; let
`k=d_1d_2d_3` be their product. They must be prime

This means we can write

`m = p_1 p_2 k`

`n = p_3 p_4 p_5 k`

`\implies mn = p_1 p_2 p_3 p_4 p_5 k^2 = p_1 p_2 p_3 p_4 p_5 {d_1}^2 {d_2}^2 {d_3}^2`

so that
`mn` has up to 8 distinct prime factors.