Solution c1a57230-445e-4985-9824-293e8f28d62d

Points: 3 pts

Lemma 1: $\operatorname{div}(n)=\operatorname{div}_S(n)\cup\operatorname{div}_L(n)$

We define
\[
\operatorname{div}_S(n)\subseteq\{1,2,\ldots,\lfloor\sqrt{n}\rfloor\}
\]
to be the set of divisors $d_1$ of $n$ satisfying
\[
d_1\leq\sqrt{n}.
\]
Then, we define
\[
\operatorname{div}_L(n)\subseteq\{\lceil\sqrt{n}\rceil,\ldots,n\}
\]
to be the set of divisors $d_2$ of $n$ satisfying
\[
d_2\geq\sqrt{n}.
\]
Together, these two sets contain every divisor of $n$, giving the desired union.

Lemma 2: $\#\operatorname{div}_S(n)=\#\operatorname{div}_L(n)$

For every $d_1\in\operatorname{div}_S(n)$, the divisor $d_2=n/d_1$ belongs to
$\operatorname{div}_L(n)$, and
\[
d_1d_2=n.
\]
Conversely, every $d_2\in\operatorname{div}_L(n)$ gives a divisor
$d_1=n/d_2\in\operatorname{div}_S(n)$. Thus, every divisor in
$\operatorname{div}_S(n)$ is paired with a unique divisor in
$\operatorname{div}_L(n)$, and vice versa. When $n$ is a perfect square,
$\sqrt{n}$ belongs to both sets and is paired with itself. Hence, this
correspondence gives
\[
\#\operatorname{div}_S(n)=\#\operatorname{div}_L(n).
\]

Lemma 3: $\#\operatorname{div}(n)\leq 2\sqrt{n}$

By the definition of $\operatorname{div}_S(n)$,
\[
\#\operatorname{div}_S(n)\leq\lfloor\sqrt{n}\rfloor\leq\sqrt{n}.
\]
Using Lemmas 1 and 2, we obtain
\[
\begin{aligned}
\#\operatorname{div}(n)
&\leq\#\operatorname{div}_S(n)+\#\operatorname{div}_L(n)\\
&=2\,\#\operatorname{div}_S(n)\\
&\leq 2\sqrt{n}.
\end{aligned}
\]

Final answer:
\#\operatorname{div}(n)\leq 2\sqrt{n}