Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Obtaining the number of factors from prime factorization

  1. Jul 2, 2005 #1
    Hi! How do I determine the number of distinct factors of a number, say, 2520?
    2520 = 2*2*2*3*3*5*7
    So we've 8 different primes. The number of combinations of those is, according to me:
    C(8,1)+C(8,2)+...+C(8,8)=155 (I think, calculated it by hand; but it isn't important)
    Obviously those aren't distinct. (Pick the fist 2 and the second 2 = 4, but pick the second 2 and the third 2 also = 4.) :yuck:
  2. jcsd
  3. Jul 2, 2005 #2


    User Avatar
    Science Advisor

    I don't understand what you are doing. 2520= 2*2*2*3*3*5*7 has 4 distinct prime factors, not "8 different primes". And I don't see what being "prime factors" has to do with number of combinations. Are you asking "of those 8 numbers (not all distinct) how many combinations can I make"? Wouldn't that be the same as asking "of the 8 letters "aaabbcd", how many different combinations can I make?" There are only 3 different one letter combinations: "a", "b", and "c", not C(8,1)= 8.
  4. Jul 2, 2005 #3
    Isn't the combinations of the prime factors = all factors?
    I mean, i pick 2*2 (a product of the primes in the positions specified by the combination) or 12
    i pick 2*2 or 23
    i pick 2*3 or 24

    and so on.
    Do you see?
  5. Jul 2, 2005 #4


    User Avatar
    Homework Helper

    This is a special case of the divisor functions studied in number theory where you sum the kth power of the divisors.
    In this case the 0th power.
    say your number factors as
    a general factor (including improper ones) is
    where pk is a prime and 0<=mk<=nk
    Thus the number of such factors is
    In particular 1 has one factor and and p^n (p prime) has n+1
    2520 has 4*3*2*2=48
    see this link for more info
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook