16. The product of all natural divisors of a natural number $n$ (including $n$) ends with 120 zeros. How many zeros can the number $n$ end with? (List all possible options and prove that there are no others.)
$\square$ Answer: $1,2,3,4,5$.

The square of the product of all divisors of $n$ (it ends with 240 zeros) is equal to $n^{k}$, where $k$ is the number of divisors of $n$. This can be understood by pairing the divisors as $d \cdot(n / d)$. Thus, if $n$ ends with $s$ zeros, then $k s = 240$. Note that $k$ is divisible by $s+1$ (for example, if $n$ is divisible by $2^{s}$ but not by $2^{s+1}$, then all divisors can be grouped into $x, 2 x, \ldots, 2^{s} x$, where $x$ is odd, similarly, if $n$ is divisible by $5^{s}$ but not by $5^{s+1}$.) Moreover, $k \geqslant(s+1)^{2}$, since each of the factors 2,5 in the factorization of $n$ is at least in the $s$-th power. Thus, 240 is divisible by $s(s+1)$, but $240 \geqslant s(s+1)^{2}$. This is possible for $s=1,2,3,4,5$.

Examples: $2 \cdot 5^{119}, 2^{2} \cdot 5^{39}, 2^{3} \cdot 5^{19}, 2^{4} \cdot 5^{11}, 2^{5} \cdot 5^{7}$.

(based on Kangaroo 2013)

![](https://cdn.mathpix.com/cropped/2024_06_04_c99b173e74a82b0c88e3g-07.jpg?height=587&width=676&top_left_y=1159&top_left_x=386)

Fig. 5: to problem 17