Question

How to find the number of common odd positive divisors between two numbers?

the two numbers are 27900 and 20700. I've put them into their prime factorisation but I can't remember what to do next?

Answer 1

\(27900 = 2^2\times 3^2 \times 5^2 \times 31\)

\(20700 = 2^2 \times 3^2 \times 5^2 \times 23\)

Answer 2

Since you're interested in odd common divisors, discard the even prime "2"

Answer 3

do u know number of divisors function ?

Answer 4

as in Euler's Totient function?

Answer 5

I think it is represented using \(\large \sigma \)

Answer 6

let \(N = p_1^{a_1} p_2^{a_2}p_3^{a_3}.... \) be the prime factorization of some number \(N\)

then, number of divisors of \(N\) is given by :
\(\large \sigma = (a_1+1)(a_2+2)(a_3+3)... \)

Answer 7

take the GCF of both numbers and find the number of divisors,
you're done

Answer 8

so assumingly if the number of divisors is written in such a way, then there has to be some sort of ordering
for example:
\[27900=2^2 \times 3^2 \times 5^2 \times 31\] but as you said we discard the 2, how do we know which power, so to speak, to add the 1 to, which to add 2 to, etc...as surely you will get a different value depending which power you take to be \[\alpha_1\] etc

I may have just completely overcomplicated that, I haven't done this in a long time

Answer 9

lets finish solving the problem first,
and get back to ur question maybe :)

Answer 10

\(27900 = 2^2\times 3^2 \times 5^2 \times 31\)

\(20700 = 2^2 \times 3^2 \times 5^2 \times 23\)

GCF excluding even primes = \( 3^2 \times 5^2\)

So, number of odd common divisors = \((2+1)(2+1) = 9\)

Answer 11

oh I get what you mean now!! Sorry lol

Answer 12

good, easy to see it for oneself... than explaining :)

Answer 13

why is it plus one for both ganeshie?(personal curiosity)

Answer 14

just combinatorics @FibonacciChick666

Answer 15

Thank you so much!!

Answer 16

\(N = p_1^{a_1} p_2^{a_2}p_3^{a_3}.... \)

How many ways you can choose \(p_1\) ?

Answer 17

ahh, ok I thought based on the phi function that the second would be plus 2. Thanks Ganeshie

Answer 18

one way, alpha sub 1 times

Answer 19

powers of \(p_1\) can be : \(0, 1, 2, ..... a_1 \)

that gives \(a_1 + 1\) ways to choose a power of \(p_1\)

Answer 20

Ok, but why is \[\large \sigma = (a_1+1)(a_2+2)(a_3+3)...\] Saying to use the progression 1,2,3... ?

Answer 21

sorry, i made a mistake earlier
it should be :

\(\large \sigma = (a_1+1)(a_2+1)(a_3+1)... \)

Answer 22

I see what you mean now, I've just answered my own question trying to give an example haha

Answer 23

:) thanks for catching the mistake @FibonacciChick666

Answer 24

ok so I do not need to help you

Answer 25

sorry just quick question, given that it specifies for POSITIVE common odd divisors, when would you not get positive divisors? and if you didn't would you then just have to half it?

Answer 26

ahh ok, I was confused thanks. That clears it up. @ganeshie8 (I was following this because I just couldn't remember how to solve it. Can't believe I lost that musch number theory in a semester!)

Answer 27

@FibonacciChick666 I'm exactly the same...I read the question and I was like whaaaaT? haha

Answer 28

Bye:) and good luck

Answer 29

same wid me lol
@sarahusher negative divisor = -positive divisor

Answer 30

@sarahusher haha yea, it's never fun to forget a class you need which was difficult to learn in the first place, but Ganeshie has ya covered.

Answer 31

so if u want all the divisors,
just take twice of positive divisors

Answer 32

and for just the total number of common positive divisors, would I do the same but leave the 2 in there too!

Answer 33

yup

Answer 34

Do you want me to help or NOT

Answer 35

yup take it on the complete factorization

Answer 36

so in this case the total number of common divisors will be 27?

Answer 37

ahhhhhhhhhhh just forget it!!

Answer 38

\(27900 = 2^2\times 3^2 \times 5^2 \times 31\)

\(20700 = 2^2 \times 3^2 \times 5^2 \times 23\)

GCF = \( 2^2\times3^2 \times 5^2\)


# of common divisors = \((2+1)(2+1)(2+1)\)

Answer 39

great, thank you for clearing this up for me!! Helped a lot!! :-)

Answer 40

np :)