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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?

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3  
How did you get 360? Showing your method is helpful here. –  Ian Coley Mar 21 at 19:12
    
copper.hat has said your answer is incorrect. After you try working the problem again, you can check your answer by seeing if the answer you get has exactly $12$ positive integer divisors. –  Dave L. Renfro Mar 21 at 19:24
    
Do you need to count the divisors or list them? –  Tim S. Mar 21 at 19:56
    
@Dave Only 12? I got 60 (including 1 and 7!). –  Adam Brown Mar 21 at 22:11
    
It sounds like you are really just looking for wolframalpha.com/input/?i=positive+divisors+of+7!&dataset= ?! –  Marco13 Mar 22 at 0:49

3 Answers 3

up vote 12 down vote accepted

360 is incorrect.

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

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

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There is an easier way than counting by hand. The divisor function is multiplicative. –  Ross Millikan Mar 21 at 19:27
    
@RossMillikan, I think it may be copper meant to use that property of the divisor function... –  DonAntonio Mar 21 at 19:30
4  
@RossMillikan: I was using 'counting' in a liberal sense... –  copper.hat Mar 21 at 19:31

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

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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.

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