The number of natural number solutions to the | vedclass

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(A) Given the equation $xyz = 2^5 	imes 3^2 	imes 5^2$.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

Vedclass · Accepted Answer

$756$

Vedclass · Answer

(A) Given the equation $xyz = 2^5 	imes 3^2 	imes 5^2$.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 $x, y, z$ are natural numbers,$a_i, b_i, c_i$ must be non-negative integers.The exponents must satisfy:$a_1 + a_2 + a_3 = 5$$b_1 + b_2 + b_3 = 2$$c_1 + c_2 + c_3 = 2$Using the stars and bars formula,the number of non-negative integer solutions for $x_1 + x_2 + \dots + x_k = n$ is given by $\binom{n+k-1}{k-1}$.For $a_1 + a_2 + a_3 = 5$: $\binom{5+3-1}{3-1} = \binom{7}{2} = 21$.For $b_1 + b_2 + b_3 = 2$: $\binom{2+3-1}{3-1} = \binom{4}{2} = 6$.For $c_1 + c_2 + c_3 = 2$: $\binom{2+3-1}{3-1} = \binom{4}{2} = 6$.The total number of solutions is $21 	imes 6 	imes 6 = 756$.

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

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