Common Factor trick?

[AlphaBlaziken]

The question is how many factors of 144,000 that are also factors of 450,000. I think there's a trick to it. I tried to use the prime factors {2,2,2,2,3,3,5,5,5} because all the factor are made of them(?) I tried to use 9!/4!*3!*2! but it's wrong..Please help me. Thanks

 

[pka]

The question is how many factors of 144,000 that are also factors of 450,000. I think there's a trick to it. I tried to use the prime factors {2,2,2,2,3,3,5,5,5} because all the factor are made of them(?) I tried to use 9!/4!*3!*2! but it's wrong..Please help me. Thanks

First FACTOR as you did. Here is a web resource that will help you.
\(\displaystyle 144,000=2^7\cdot3^2\cdot5^3~\&~450,000=2^4\cdot3^2\cdot5^5\)

The GCF is \(\displaystyle 2^4\cdot3^2\cdot5^3\). Thus the number of common divisors is \(\displaystyle (4+1)(2+1)(3+1)=~?\)
Why add 1 to each common exponent?

 

[AlphaBlaziken]

I'm sorry but could you explain why the common divisor is (4+1)(2+1)(3+1)= ? Is it some kind of formula?

 

[pka]

I'm sorry but could you explain why the common divisor is (4+1)(2+1)(3+1)=? Is it some kind of formula?

I would not call it a formula. It is a basic counting rule.

If \(\displaystyle N=3^{6}\cdot5^3\cdot7^4\cdot11\) then the number \(\displaystyle N\) has \(\displaystyle (6+1)(4)(5)(2)\) divisors.

If \(\displaystyle p^n\) is a in the prime factorization of \(\displaystyle N\) then then +\(\displaystyle p^j\) for \(\displaystyle 0\le j\le n\) is a divisor of \(\displaystyle N\).
That is \(\displaystyle j+1\) values for the exponent: \(\displaystyle j=0,1,2,\cdots,n-1,n\) or \(\displaystyle n+1\) values.
The is true for each exponent in the prime factorization of \(\displaystyle N\).

 

[Steven G]

I'm sorry but could you explain why the common divisor is (4+1)(2+1)(3+1)=? Is it some kind of formula?

Come on pka clearly said Thus the number of common divisors is (4+1)(2+1)(3+1) not the common divisor is (4+1)(2+1)(3+1)

 

[AlphaBlaziken]

Come on pka clearly said Thus the number of common divisors is (4+1)(2+1)(3+1) not the common divisor is (4+1)(2+1)(3+1)

Yeah, I meant how do you know the number is that. I didn't get it why would you add 1 to the exponent.
@pka Thanks, it's all clear now.