There are positive integers $x$ and $y$ that satisfy the system of equations 
\begin{align*} \log_{10} x + 2 \log_{10} (\text{gcd}(x,y)) &= 60\\ \log_{10} y + 2 \log_{10} (\text{lcm}(x,y)) &= 570. \end{align*}
Let $m$ be the number of (not necessarily distinct) prime factors in the prime factorization of $x$, and let $n$ be the number of (not necessarily distinct) prime factors in the prime factorization of $y$. Find $3m+2n$.
Add the two equations to get that $\log x+\log y+2(\log(\gcd(x,y))+2(\log(\text{lcm}(x,y)))=630$.
Then, we use the theorem $\log a+\log b=\log ab$ to get the equation, $\log (xy)+2(\log(\gcd(x,y))+\log(\text{lcm}(x,y)))=630$.
Using the theorem that $\gcd(x,y) \cdot \text{lcm}(x,y)=x\cdot y$, along with the previously mentioned theorem, we can get the equation $3\log(xy)=630$.
This can easily be simplified to $\log(xy)=210$, or $xy = 10^{210}$.
$10^{210}$ can be factored into $2^{210} \cdot 5^{210}$, and $m+n$ equals to the sum of the exponents of $2$ and $5$, which is $210+210 = 420$.
Multiply by two to get $2m +2n$, which is $840$.
Then, use the first equation ($\log x + 2\log(\gcd(x,y)) = 60$) to show that $x$ has to have lower degrees of $2$ and $5$ than $y$  (you can also test when $x>y$, which is a contradiction to the restrains we set before). Therefore, $\gcd(x,y)=x$. Then, turn the equation into $3\log x = 60$, which yields $\log x = 20$, or $x = 10^{20}$.
Factor this into $2^{20} \cdot 5^{20}$, and add the two 20's, resulting in $m$, which is $40$.
Add $m$ to $2m + 2n$ (which is $840$) to get $40+840 = \boxed{880}$.
~minor mistake fix by virjoy2001
~minor mistake fix by oralayhan
Remark: You can obtain the contradiction by using LTE. If $\nu_2{(x)}\geq{\nu_2{(y)}}, \nu_2{(y^2x)}=60$. However, $\nu_2{(xy)}=210$ a contradiction. Same goes with taking $\nu_5{(x,y)}$

Easier Approach to Finish
After noting that $xy=10^{210},$ notice that we can let $x=10^a$ and $y=10^b.$ Thus, we have from the given equations (1) and (2) respectively, that:
\[a+2a=60\]
\[b+2b=570\]
Solving, we get $(a,b)=(20,190).$ This matches with our constraint that $xy=10^{210},$ so we finish from here.
$x=2^{20}\cdot 5^{20}, y=2^{190}\cdot 5^{190}.$ Thus, the answer desired is $3(20+20)+2(190+190)=880.$ ~mathboy282