Number of natural solutions of the equation x | vedclass

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Question

(A) The given equation is $xyz = 2^5 	imes 3^2 	imes 5^2$.We need to find the number of natural number solutions $(x, y, z)$.Let $x = 2^{a_1} 3^{b_1

Vedclass · Accepted Answer

$756$

Vedclass · Answer

(A) The given equation is $xyz = 2^5 	imes 3^2 	imes 5^2$.We need to find the number of natural number solutions $(x, y, z)$.Let $x = 2^{a_1} 3^{b_1} 5^{c_1}$,$y = 2^{a_2} 3^{b_2} 5^{c_2}$,and $z = 2^{a_3} 3^{b_3} 5^{c_3}$,where $a_i, b_i, c_i \ge 0$.Since $x, y, z$ must be natural numbers,we require $x, y, z \ge 1$.For the exponents of prime factors,we have the following equations:$a_1 + a_2 + a_3 = 5$ (where $a_i \ge 0$)$b_1 + b_2 + b_3 = 2$ (where $b_i \ge 0$)$c_1 + c_2 + c_3 = 2$ (where $c_i \ge 0$)Using the stars and bars formula,the number of non-negative integer solutions for $x_1 + x_2 + \dots + x_k = n$ is $\binom{n+k-1}{k-1}$.For $a_i$: $\binom{5+3-1}{3-1} = \binom{7}{2} = 21$.For $b_i$: $\binom{2+3-1}{3-1} = \binom{4}{2} = 6$.For $c_i$: $\binom{2+3-1}{3-1} = \binom{4}{2} = 6$.The total number of solutions is $21 	imes 6 	imes 6 = 756$.

Number of natural solutions of the equation $xyz = 2^5 \times 3^2 \times 5^2$ is equal to

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