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.