#### Question

(a) Find the form of all positive integers $n$ satisfying $\tau(n)=10 .$ What is the smallest positive integer for which this is true? (b) Show that there are no positive integers $n$ satisfying $\sigma(n)=10$ .

#### Solution

Verified#### Step 1

1 of 3- [$\text{\textcolor{#4257b2}{a)}}$]Let $n$ be a positive integer satisfying $\tau(n)=10$. If $n=p_1^{a_1}p_2^{a_2}\ldots p_r^{a_r}$, then

$\begin{align*} \tau(n)=(1+a_1)(1+a_2)\ldots(1+a_r) \end{align*}$

Since $10=2\cdot 5$ has exactly two prime factors, we see that $n$ can have at most two prime factors (else it would have too many factors). From this we have the two following possibilities for $n$:

$\begin{align*} n&=p^9\\ n&=pq^4 \end{align*}$

for prime $p,q$. The smallest number for which $\tau(n)=10$ of the first form is $2^9$ and of the second form is $3\cdot 2^4$. It is readily seen that the other number is smaller, thus $n=48$ is the smallest number such that $\tau(n)=10$.