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The total amount of positive integral (x, y, and z) solutions. for x y z=384 is

Option: 1

111


Option: 2

110


Option: 3

109


Option: 4

108


Answers (1)

best_answer

It is given that x y z=384

Then it can also be written in the form x . y . z=2^{7} .3^{1}
Then \mathrm{x}, \mathrm{y}, and \mathrm{z} can be written in the form

x=2^{a_{1}} \cdot 3^{b_{1}}
y=2^{a_{2}} \cdot 3^{b_{2}}
z=2^{a_{3}} \cdot 3^{b_{3}}

\left(a_{1}, a_{2}, a_{3}\right) \varepsilon(0,1) and \left(b_{1}, b_{2}, b_{3}\right) \varepsilon(0,1)

also a_{1}+a_{2}+a_{3}=7

Then the number of solutions will be ={ }^{7+3-1} C_{3-1}
={ }^{9} C_{2}
=36
and b_{1}+b_{2}+b_{3}=1

Then the number of solutions will be given as ={ }^{(1+3-1)} C_{(3-1)}
={ }^{3} C_{2}
=3 The total number of solutions will be =36 \times 3
=108

The correct option is (d).

Posted by

Nehul

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