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Counting Principle

  1. Jan 23, 2010 #1
    How many divisors does 55,125 have? For example, 55,125 = (3)^2 . (5)^3 . (7)^2
  2. jcsd
  3. Jan 23, 2010 #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:)
    Well, each divisor has to be of the form 3a5b7c, wiht a b and c integers > 0 …

    so how many is that? :smile:
  4. Jan 23, 2010 #3
    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. :(
  5. Jan 23, 2010 #4
    If the prime factorization of [tex]n[/tex] is [tex]n=p_1^{e_1}p_2^{e_2}...p_n^{e_n}[/tex]. Now, in any divisor, each prime factor's exponent [tex]a[/tex] range from [tex]0\leq a \leq e_i[/tex].
  6. Jan 24, 2010 #5
    Thanks for sharing. By the way, I'm new here and nice to meet you all.
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