What are all positive divisors of 7 factorial?

Question

I need to determine all the positive divisors of 7!. I got 360 as the total number of positive divisors for 7!. Can someone confirm, or give the real answer?

Answer 1

360 is incorrect.

$7! = 2^4 3^2 5^1 7^1$. Now start counting...

Note: Count $\{0,1,2,3,4\} \times \{0,1,2\} \times \{0,1\} \times \{0,1\}$.

Answer 2

Once you factorize a number as $N=p_1^{a_1}p_2^{a_2}p_3^{a_3}...p_n^{a_n}$, $p_i$ prime for every $i$, $a_i>0$ for every $i$ the number of divisors is given by $(a_1+1)(a_2+1)(a_3+1)...(a_n+1)$.

It is easy to see why this formula works from a combinatorial point of view, the divisors of $N$ are also of the form $p_1^{b_1}p_2^{b_2}p_3^{b_3}...p_n^{b_n}$, with $b_i\leq a_i$ for every $i$, but this time some (or all) of the $b_i$ can be $0$, this mean we can pick $a_i+1$ values for $b_i$, from $0$ to $a_i$.

In your case $7!=2^43^25^17^1$ so it has $(4+1)(2+1)(1+1)(1+1)=60$ divisors

Answer 3

Just to generalize what others have said, it's a neat little fact that the number of distinct factors of $n!$ is given by:

$$ \prod_{p \in primes}\left( 1 + \sum_{k=1}^{\infty}\left \lfloor \frac{n}{p^k} \right \rfloor \right) $$

Note that the sum is simply a shortcut to calculating the exponent for an individual prime factor that only works with factorials. The product and "1+" part are adequately explained by Alessandro's answer.