I'm trying to do the problem Mersenne asked in 1644. Find the smallest positive integer that has 60 divisors.
So far I have found the prime factors,
$60=2 \times 2 \times 3 \times 5 $
I know that if you subtract unity to obtain $1 \times 1 \times 2 \times 4$ to use as the exponents $a^1 \times b^1\times c^2\times d^4$ however I do not know how to find $a,b,c,d$.
I even know the answer is $5040$ with $a=7, b=5, c=3, d=2$ but I do not know what theorems are used, why unity is subtracted and how the values are found. Any help would be greatly appreciated! Thanks!
Best Answer
This follows from properties of the divisor function. Say we have some number $n$ and we want to calculate how many divisors it has. This is the same as calculating $\sigma_0(n)$, where
$$\sigma_0(n)=\sum_{d|n}1.$$
If we write $n=p_1^{a_1}\cdots p_k^{a_k}$ as a product of primes, then one may show that we have
$$\sigma_0(n)=\prod_{i=1}^k(a_i+1).$$
You want to find the least $n$ such that $\sigma_0(n)=60$. So let's factorize 60 into primes. As you noted, this is $2\times 2\times 3\times 5$. Now it just remains for you to argue that we must have $k=4$, and $\{a_1+1,\dots,a_4+1\}=\{2,2,3,5\}$ under some ordering, then say why the choice you give in the question is minimal.