Problem
NT-B1-M01-P011 Smallest Number with 12 Divisors
#11
★★☆☆☆ Level 2 of 5
Find the smallest positive integer with exactly \(12\) positive divisors.
Write \(12\) as a product of numbers \(\alpha_i+1\), then put larger exponents on smaller primes.
Possible products \((\alpha_i+1)\) are \(12\), \(6\cdot2\), \(4\cdot3\), and \(3\cdot2\cdot2\). They give candidates \(2^{11}\), \(2^5\cdot3=96\), \(2^3\cdot3^2=72\), and \(2^2\cdot3\cdot5=60\). The smallest is \(60\). Indeed, \(60=2^2\cdot3\cdot5\) has \((2+1)(1+1)(1+1)=12\) divisors.
First optimisation use of the divisor-counting formula.