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(B) The prime factorization of $3000$ is $3000 = 3^1 	imes 2^3 	imes 5^3$.We need to distribute the prime factors among $x, y,$ and $z$.For the prim

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$300$

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(B) The prime factorization of $3000$ is $3000 = 3^1 	imes 2^3 	imes 5^3$.We need to distribute the prime factors among $x, y,$ and $z$.For the prime factor $3^1$,the number of ways to distribute it among $3$ variables is given by the formula $\binom{n+r-1}{r-1}$,where $n=1$ and $r=3$. This is $\binom{1+3-1}{3-1} = \binom{3}{2} = 3$.For the prime factor $2^3$,the number of ways to distribute it among $3$ variables is $\binom{3+3-1}{3-1} = \binom{5}{2} = 10$.For the prime factor $5^3$,the number of ways to distribute it among $3$ variables is $\binom{3+3-1}{3-1} = \binom{5}{2} = 10$.Therefore,the total number of positive integral solutions is $3 	imes 10 	imes 10 = 300$.

The number of positive integral solutions of the equation $xyz = 3000$ is

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