Question

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

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  • [a)\text{\textcolor{#4257b2}{a)}}]Let nn be a positive integer satisfying τ(n)=10\tau(n)=10. If n=p1a1p2a2prarn=p_1^{a_1}p_2^{a_2}\ldots p_r^{a_r}, then

τ(n)=(1+a1)(1+a2)(1+ar)\begin{align*} \tau(n)=(1+a_1)(1+a_2)\ldots(1+a_r) \end{align*}

Since 10=2510=2\cdot 5 has exactly two prime factors, we see that nn can have at most two prime factors (else it would have too many factors). From this we have the two following possibilities for nn:

n=p9n=pq4\begin{align*} n&=p^9\\ n&=pq^4 \end{align*}

for prime p,qp,q. The smallest number for which τ(n)=10\tau(n)=10 of the first form is 292^9 and of the second form is 3243\cdot 2^4. It is readily seen that the other number is smaller, thus n=48n=48 is the smallest number such that τ(n)=10\tau(n)=10.

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