Obtaining the number of factors from prime factorization

  • Thread starter danne89
  • Start date
  • #1
180
0

Main Question or Discussion Point

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:
 

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
41,833
955
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.
 
  • #3
180
0
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?
 
  • #4
lurflurf
Homework Helper
2,432
132
danne89 said:
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:
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
p1^n1*p2^n2*p3^n3*...*pk^nk*...
a general factor (including improper ones) is
p1^m1*p2^m2*p3^m3*...*pk^mk*...
where pk is a prime and 0<=mk<=nk
Thus the number of such factors is
(1+n1)(1+n2)(1+n3)...(1+nk)...
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
http://mathworld.wolfram.com/DivisorFunction.html
 

Related Threads on Obtaining the number of factors from prime factorization

Replies
19
Views
1K
Replies
3
Views
1K
Replies
12
Views
3K
  • Last Post
Replies
6
Views
10K
  • Last Post
Replies
2
Views
2K
Replies
2
Views
733
  • Last Post
Replies
3
Views
675
Replies
7
Views
534
  • Last Post
Replies
7
Views
3K
  • Last Post
Replies
20
Views
9K
Top