Divisors of 55,125: Counting Principle

[weiji]

How many divisors does 55,125 have? For example, 55,125 = (3)^2 . (5)^3 . (7)^2

 

[tiny-tim]

Welcome to PF!

Hi weiji! Welcome to PF!

(try using the X² tag just above the Reply box )

How many divisors does 55,125 have? For example, 55,125 = (3)^2 . (5)^3 . (7)^2

Well, each divisor has to be of the form 3^a5^b7^c, wiht a b and c integers > 0 …

so how many is that? ​

 

[weiji]

I did a very long calculation by assume a=1, b=1 ; a=1,c=1 ; b=1,c=1, from here, I know 1575x35 = 2625x21 = 3675x15. Then I calculate each possible answer, I got 36 divisors. But is there any faster way? I really have no idea. :(

 

[JSuarez]

If the prime factorization of $$n$$ is $$n=p_1^{e_1}p_2^{e_2}...p_n^{e_n}$$. Now, in any divisor, each prime factor's exponent $$a$$ range from $$0\leq a \leq e_i$$.

 

[weiji]

Thanks for sharing. By the way, I'm new here and nice to meet you all.