Problem
NT-B1-M01-P022 Fifteen Divisors and Divisibility by 12
Find the smallest positive integer that is divisible by \(12\) and has exactly \(15\) positive divisors.
Write \(15\) as \(15\) or \(3\cdot5\), and \(12\mid n\) means \(v_2(n)\ge2\), \(v_3(n)\ge1\).
If \(\tau(n)=15\), then the possible structures are \(p^{14}\) or \(p^4q^2\). The first is not minimal and, under divisibility by \(12\), would be very large. In the second case, the larger exponent should be placed on the smaller prime. Since the number must be divisible by \(12=2^2\cdot3\), take \(2^4\cdot3^2=144\). It has \((4+1)(2+1)=15\) divisors and is divisible by \(12\). The alternative \(2^2\cdot3^4=324\) is larger, and adding other primes only increases the number. The answer is \(144\).
This already includes optimisation over prime exponents.