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(B) First,find the prime factorization of $3000$: $3000 = 3^1 	imes 2^3 	imes 5^3$. To find the number of positive integral solutions for $xyz = n$,

Vedclass · Accepted Answer

$300$

Vedclass · Answer

(B) First,find the prime factorization of $3000$: $3000 = 3^1 	imes 2^3 	imes 5^3$. To find the number of positive integral solutions for $xyz = n$,where $n = p_1^{a_1} p_2^{a_2} \dots p_k^{a_k}$,we use the stars and bars formula for each prime factor. The number of ways to distribute $a_i$ identical items into $3$ distinct variables is given by $\binom{a_i + 3 - 1}{3 - 1} = \binom{a_i + 2}{2}$. For $3^1$: $a_1 = 1$,ways = $\binom{1+2}{2} = \binom{3}{2} = 3$. For $2^3$: $a_2 = 3$,ways = $\binom{3+2}{2} = \binom{5}{2} = 10$. For $5^3$: $a_3 = 3$,ways = $\binom{3+2}{2} = \binom{5}{2} = 10$. Total number of solutions = $3 	imes 10 	imes 10 = 300$.

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

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