<s>[INST] Number of integral solutions of $\text{xyz}=3000$ I want to find the number of integral solutions of the equation $$xyz=3000$$
I have been able to solve similar sums where the number on the right hand side was small enough to calculate all the factors of. Such as $xyz=24$ or $xyz=30$.
What is the proper method to solve such a problem when the number is too big to consider all the factors of it?
 [/INST] Note that $3000 = 2^3 \cdot 3^1 \cdot 5^3$. Let $x=2^{x_1}3^{x_2}5^{x_3}$, $y=2^{y_1}3^{y_2}5^{y_3}$ and $z=2^{z_1}3^{z_2}5^{z_3}$. Then obtain the number of solutions to
$$x_1+x_2+x_3 = 3$$
$$y_1+y_2+y_3 = 1$$
$$z_1+z_2+z_3 = 3$$
The product of the number of non-negative solutions will give us the total number of solutions. Hence, in our case, the number of non-negative solutions to the first one is $\dbinom{3+3-1}{3-1} =10$, the second one is $\dbinom{1+3-1}{3-1} =3$ and the third one is $\dbinom{3+3-1}{3-1} =10$. Hence, total number of solutions is $10\cdot 3 \cdot 10 = 300$.
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