Problem
NT-B1-M09-P012 Exactly Three Divisors
#12
★★☆☆☆ Level 2 of 5
Find all positive integers less than \(100\) with exactly \(3\) positive divisors.
A number with three divisors has the form \(p^2\).
If \(\tau(n)=3\), then \(n=p^2\) for a prime \(p\). We need \(p^2<100\), so \(p<10\). The primes are \(2,3,5,7\). The numbers are \(4,9,25,49\).
Classical inverse problem.