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Factors of product of n distinct primes

  1. Mar 9, 2009 #1
    What will be the numbers of positive factors of product of n distinct prime numbers?

    i was able to get 2^n. pls how do i prove this?
    Last edited: Mar 9, 2009
  2. jcsd
  3. Mar 9, 2009 #2
    In any factor you come up with, it will either be divisible by one of the n unique primes, or not. Whether or not your particular factor is divisible by a prime is independent of whether your factor is divisible by another prime; so how many ways can you make a factor if you have to make n yes/no decisions?

    Think about it as a prime factor set. You're finding the cardinality of the power set...
  4. Mar 9, 2009 #3
    you mean there will exist a 1-1 correspondence between power set of a set A with cardinality n and the set of positive factors of the product of n primes
  5. Mar 9, 2009 #4
    Yes. Hence the 2^n.
  6. Mar 10, 2009 #5
    what about if the primes are not distinct? i don't the number of positive factor will be 2^n, it will be less than that.
  7. Mar 10, 2009 #6
    You said "distinct primes" in your first post. So your answer - and my reason - were correct.

    If you are now changing the question, that's fine.

    If the primes are not distinct, all you need to do is to calculate the number of equivalence classes of primes. For instance,

    P = {2, 2, 3, 3, 3, 5, 7, 7, 11, 11, 11, 11}


    P' = { {2, 2}, {3, 3, 3}, {5}, {7, 7}, {11, 11, 11, 11} }

    Then the number of factors is 2^|P'| = 32.

    If the primes are all unique, then clearly |P'| = |P|.
  8. Mar 10, 2009 #7

    matt grime

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    That is clearly incorrect: just consider a prime power. By your logic it always has exactly 2 divisors.

    Google for Euler's phi (or totient) function.
  9. Mar 10, 2009 #8
    Oh, yeah. Sorry, I was still thinking in terms of the first question.

    Matt Grime is correct for the second question. I was for the first one. Sorry for the confusion...
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