Solution 8fb58e14-b926-403a-87aa-1d70727a53a4

Points: 5 pts

Lemma 1: Any divisor $d$ has the form $d=p_1^{b_1}p_2^{b_2}\cdots p_r^{b_r}$

Any divisor $d$ of $n$ can be written as
\[
d=p_1^{b_1}p_2^{b_2}\cdots p_r^{b_r}.
\]

We see this because
\[
n=dm,
\]
where
\[
m=p_1^{a_1-b_1}p_2^{a_2-b_2}\cdots p_r^{a_r-b_r}
\]
is an integer.

Lemma 2: We must have $0\leq b_i\leq a_i$

Since $d$ must be an integer,
\[
0\leq b_i.
\]

But $m$ must also be an integer, so
\[
0\leq a_i-b_i.
\]

Combining these inequalities gives
\[
0\leq b_i\leq a_i
\]
for all $i$.

Lemma 3: Each divisor $d_i$ corresponds to a tuple $t_i$

Let $T$ be the set of tuples $t_i$ such that
\[
t_i=(b_1,b_2,\ldots,b_r).
\]

The order of a tuple records which exponent belongs to each prime $p_i$. By the fundamental theorem of arithmetic, each tuple therefore corresponds to a single divisor.

Lemma 4: $\#B_i=a_i+1$

Let $B_i$ be the set of possible values for $b_i$ in the exponent. Counting, we find
\[
b_i\in\{0,1,2,\ldots,a_i\}.
\]

Hence there are
\[
\#B_i=a_i+1
\]
possible values.

Lemma 5: $\#T=\#B_1\#B_2\cdots\#B_r$

This follows from the multiplication principle for independent choices. Therefore,
\[
\#T
=\#B_1\#B_2\cdots\#B_r
=(a_1+1)(a_2+1)\cdots(a_r+1).
\]

Since each tuple in $T$ corresponds to exactly one divisor of $n$, we can read off the answer.

Final answer:
\#\operatorname{div}(n)=(a_1+1)(a_2+1)\cdots(a_r+1)