Divisors of 55,125: Counting Principle

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SUMMARY

The number of divisors of 55,125 is determined through its prime factorization: 55,125 = (3)^2 . (5)^3 . (7)^2. Using the formula for counting divisors, where each exponent in the prime factorization is increased by one and then multiplied together, the total number of divisors is calculated as (2+1)(3+1)(2+1) = 3 * 4 * 3 = 36. This method provides a systematic approach to finding divisors efficiently without exhaustive enumeration.

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  • Familiarity with the divisor counting formula
  • Basic knowledge of integer properties
  • Ability to manipulate exponents in mathematical expressions
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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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