Number of divisors of a number ...

[danielvnl]

How can I know the number of the natural divisor of a number ??

Ex.: 18 = 3^2 . 2¹ --> number of naturals divisors --> (2+1).(1+1) = 6

and if there is 2 or more numbers at the same time, how can I do ?

Ex.: The number of the natural divisors of 18 and 48 at the same time ...

Sorry for my English

THX

 

[pka]

How can I know the number of the natural divisor of a number ??
Ex.: 18 = 3^2 . 2¹ --> number of naturals divisors --> (2+1).(1+1) = 6
and if there is 2 or more numbers at the same time, how can I do ?
Ex.: The number of the natural divisors of 18 and 48 at the same time ...

I think that you are asking something like this: How many common divisors do 240 & 400 have?

If that is correct then find the number of divisors the least common divisor has,
\(\displaystyle LCD(400,240)=40\), so how many divisors of 40?

 

[danielvnl]

I was trying to solve this problem :

n (E) is the number of elements of a set E. If A is the set of natural dividers 18 and B is the set of natural dividers 48, then n (A ∪ B) is a number:
Answer: greater than 10

Because n(E) is 12.

So I asked that to know how can I do to know the number of a divisors of a number
Ex.: 18 = 3^2 . 2¹ --> number of naturals divisors of 18 --> (2+1).(1+1) = 6
48 = 2^4 . 3¹ --> number of naturals divisors of 48 --> (4+1).(1+1) = 10

I don't know why this equation "(2+1).(1+1) = 6". Actually 18 has 1,2,3,6,9 and 18, exatelly 6 divisors.
The same happens to 48, "(4+1).(1+1) = 10". Why to do (4+1).(1+1)=10 ??? Where we find this ??? And what we can do to know how many divisors 18 and 48 have at the same time ???


THX again ...

 

[HallsofIvy]

If \(\displaystyle p= a^mb^n\), a and b distinct primes, then each of \(\displaystyle a^0= 1\), \(\displaystyle a^1= a\), \(\displaystyle a^2\), ..., \(\displaystyle a^m\) is a factor: m+1 factors. Similarly, each of \(\displaystyle b^0= 1\), \(\displaystyle b^1= b\), \(\displaystyle b^2\), ..., \(\displaystyle b^n\) is a factor: n+1 factors. And any product of a power of a and a power of b is a factor for a total of (m+ 1)(n+ 1).

 

[danielvnl]

Very good explanation HallsofIvy, now how can I put what you said to do 18 and 48 at the same time ?

I'm trying MMC 18,48 = (2^4).(3^2) so (4+1).(2+1) = 15. I'm wrong, why ???

 

[pka]

to do 18 and 48 at the same time ? I'm trying MMC 18,48 = (2^4).(3^2) so (4+1).(2+1) = 15. I'm wrong, why ???

As I said before \(\displaystyle LCD(48,18)=6\) now any divisor of \(\displaystyle 6\) is a divisor of \(\displaystyle 18~\&~48\) there are no others. Thus there are four divisors of both \(\displaystyle 18~\&~48\)

Now \(\displaystyle 6+10-4=12\) so there are 12 divisors of \(\displaystyle 18\text{ or }48\).

 

[danielvnl]

OK, now I understand

thanks :mrgreen:

 

[lookagain]

I think that you are asking something like this: How many common divisors do 240 & 400 have?

If that is correct then find the number of divisors the >>> least common divisor <<< has,
LCD(400,240) = 40, so hoamy divisors of 40?

1) "LCD" is not what is meant. You are looking for the GCF (greatest common factor)
or GCD (greatest common divisor). Another name I have seen for it is the
HCF (highest common factor). If divisors are meant to belong to the set of
positive integers, then the LCD (least common divisor) of two or more positive integers
is always 1.


2) Even if you had typed "GCF(400, 240)," that would be 80, not 40.

- - - - - - - - - - - - - -

As I said before>>> \(\displaystyle LCD(48,18)=6\) <<<now any divisor of \(\displaystyle 6\) is a divisor of \(\displaystyle 18~\&~48\) there are no others. Thus there are four divisors of both \(\displaystyle 18~\&~48\)

Now \(\displaystyle 6+10-4=12\) so there are 12 divisors of \(\displaystyle 18\text{ or }48\).

The same is true for the above from a later post. It is not the LCD(48,18). It is supposed to be the
GCF(48,18) or GCD(48,18), for instance.