The South African Mathematical Olympiad Third Round

Consider, for a positive integer $k$, the multiple $M_k=k\times 20$. If $k=2^{a_1}\cdot 3^{a_2}\cdot 5^{a_3}\dots$ is the prime factorization of $k$, where $a_i\ge 0$ for all $i\ge 1$, then $M_k=2^{a_1+2}\cdot 3^{a_2}\cdot 5^{a_3+1}\cdot 7^{a_4}\dots$ is the prime factorization of $M_k$. For $M_k$ to have $20$ positive divisors, we need to have $(a_1+3)(a_2+1)(a_3+2)\times {\prod }_{i\ge 4}(a_i+1)=20=2^2\cdot 5$. This forces $a_1+3\in \{4,5,10\}$ and $a_3+2\in \{2,4,5\}$. Also, since we want the smallest such $k$, we may assume that $k$ has no prime divisors larger than $5$. (Any $a_i+1=2$, with $i\ge 4$, can be replaced by $a_2+1=2$, resulting in a smaller $k$, but without changing the number of divisors of $M_k$.) Henceforth, we assume that $a_i=0$ for all $i\ge 4$. All the possibilities are summarised in the following table:

$a_1+3$	$a_2+1$	$a_3+2$	$a_1$	$a_2$	$a_3$	$k=2^{a_1}\cdot 3^{a_2}\cdot 5^{a_3}$
4	1	5	1	0	3	250
5	1	4	2	0	2	100
5	2	2	2	1	0	12
10	1	2	7	0	0	128

We conclude that the smallest positive multiple of $20$ with exactly $20$ positive divisors is $12\times 20=240$.

---

Alternative solution.

Of course, one can try to be lucky, and simply start to determine the numbers of divisors of $1\times 20,2\times 20,3\times 20,\dots$, and soon end up with $12\times 20$ having $20$ positive divisors (for the first time in this sequence). Hence, $12\times 20$ is the smallest such multiple.