TITLE: Prime divisors of $60!$
QUESTION [5 upvotes]: How to find all prime divisors of $60!$?, where $60!$ denotes the factorial of $60$, the product all integers from $1$ to $60$. 

Is there any solution to solve this? Thank you.

REPLY [3 votes]: Remember: if $n$ is a positive integer, then $n!$ is divisible by all the positive primes less than $n$. In this case, that means $2, 3, 5, 7, \ldots, 59$.
Now, do you need to find just the distinct prime divisors? If so, you're done. If you also need to count up how often each prime divisor occurs, you've got a bit of work to do. It's tedious but not difficult. Observe:

$2! = 2$
$3! = 2 \times 3$
$4! = 2 \times 3 \times 4 = 2^3 \times 3$. Observe that since $4 = 2^2$, we need to add $2$ to the (previously unwritten) exponent $1$ that $2$ had.
$5! = 2^3 \times 3 \times 5$. Since $5$ is prime, we can just carry over the previous factorization append the "$\times 5$" at the end.
$6! = 2^4 \times 3^2 \times 5$. Since $6$ is divisible by $2$ and $3$, we need to increment the exponents for both of those primes.
$7! = 2^4 \times 3^2 \times 5 \times 7$. Since $7$ is prime, we can just carry over the previous factorization append the "$\times 7$" at the end.
...
$60! = 2^{56} \times 3^{28} \times 5^{14} \times 7^{19} \times \ldots \times 47 \times \times 53 \times 59$.

Actually, come to think of it, I think there's a formula that will give you the exponent of each prime without having to go through this "reverse engineering" of the factorial. So yeah, there is a solution to this problem, more than one of them, in fact.