Number of natural solutions to the equation $x_1\cdot{x_2}\cdot{x_3}\cdot{x_4}=1000000$ I'm trying to tackle some question and I think I solved it, but I'm not sure and would like someone to check my work.
The question goes like this:

Find the number of non-negative solutions of the equation $x_1\cdot{x_2}\cdot{x_3}\cdot{x_4}=1000000$.

My try:
First I have written the equation as $x_1\cdot{x_2}\cdot{x_3}\cdot{x_4}=10^6=2^6\cdot{5^6}$, hence $x_i$ is of the form $\displaystyle 2^{a_i}\cdot{5^{b_i}}$ where $0\le a_i,b_i \le 6$.
Now we can write $$2^{a_1}5^{b_1}\cdot{2^{a_2}5^{b_2}}\cdot{2^{a_3}5^{b_3}}\cdot{2^{a_4}5^{b_4}}=2^{a_1+a_2+a_3+a_4}\cdot{5^{b_1+b_2+b_3+b_4}}=2^6\cdot{5^6}$$
Thus, we are looking to find the number of non-negative solutions to the equations $$\begin{cases}a_1+a_2+a_3+a_4=6\\b_1+b_2+b_3+b_4=6\end{cases}$$
We know that $0 \le a_i,b_i \le 6$, hence the generating function would be $$g(x)=(1+x+...+x^6)^4=\left(\frac{1-x^7}{1-x}\right)^4=(1-x^7)^4\cdot{\frac{1}{(1-x)^4}}\\g(x)=(1-x^7)^4\cdot{\sum_{n=0}^{\infty}{{n+3}\choose 3}x^n}$$
Now, $\displaystyle (1-x^7)^4=x^0-4x^7+6x^{14}-4x^{21}+x^{28}$, hence the only relavant term is $x^0$ (we want to find the coefficient of $x^6$). We find that the required coefficient is $\displaystyle {9\choose 3}=84$, so we have 84 optional solutions to any of the equations.
Edit: There is no dependence between the equations, hence the number of solutions is $84^2$.
Is my reasoning correct? Please help me fix my errors if there any. Thanks!

Let's try it for $100=x_1\cdot x_2$. Then the numbers of solutions to $a_1+a_2=2$ is $3$ and the number of solutions to $b_1+b_2=2$ is $3$. Your answer would give us $3+3=6$ answers.
Let's now enumerate:
$$100\cdot 1\\50\cdot 2\\25\cdot 4\\20\cdot 5\\10\cdot 10\\5\cdot 20\\4\cdot 25\\2\cdot 50\\1\cdot 100$$
That's nine solutions.
