Divisors of 55,125: Counting Principle

AI Thread Summary
The discussion focuses on determining the number of divisors of the number 55,125, which is expressed in its prime factorization as (3)^2 . (5)^3 . (7)^2. Each divisor can be represented in the form 3^a * 5^b * 7^c, where a, b, and c are integers greater than or equal to zero. A participant calculated the total number of divisors to be 36 through various combinations of the exponents. The conversation also touches on the formula for calculating divisors based on prime factorization, highlighting the range of exponents for each prime factor. The discussion concludes with a welcome to new members and an invitation for further insights.
weiji
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How many divisors does 55,125 have? For example, 55,125 = (3)^2 . (5)^3 . (7)^2
 
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Welcome to PF!

Hi weiji! Welcome to PF! :smile:

(try using the X2 tag just above the Reply box :wink:)
weiji said:
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 3a5b7c, wiht a b and c integers > 0 …

so how many is that? :smile:​
 
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. :(
 
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.
 
Thanks for sharing. By the way, I'm new here and nice to meet you all.
 

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