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If \mathrm{x, y,z} are positive integers and \mathrm{x \times y \times z=72},  how many possible integral solutions are there?

Option: 1

2456


Option: 2

2374


Option: 3

2496


Option: 4

3464


Answers (1)

best_answer

To find the number of possible integral solutions for \mathrm{x, y, z} such \mathrm{x \times y \times z=72}, we need to consider the prime factorization of 72.

The prime factorization of 72 is \mathrm{2^3 \times 3^2}.
Now, we need to distribute these prime factors among \mathrm{x, y} and \mathrm{z}.

Since \mathrm{x, y,and\, z} are positive integers, the prime factors can be distributed in the following ways:

Case 1: \mathrm{\quad x=2^{\mathrm{a}} \times 3^{\mathrm{b}}, \mathrm{y}=2^{\mathrm{c}} \times 3^{\mathrm{d}}, \mathrm{z}=2^{\mathrm{c}} \times 3^{\mathrm{f}}}

where a,b,c,d,e,f are non-negative integers.

The possible values for a,b,c,d,e,f are:

0,1,2,3 for a,c,e

0,1,2 for b,d,f

Therefore, for Case 1, there are \mathrm{4 \times 3 \times 4 \times 3 \times 4 \times 3=1,728}  possible solutions.

Case 2:  \mathrm{x=1, y=2^{\mathrm{a}} \times 3^{\mathrm{b}} \times 2^{\mathrm{c}} \times 3^{\mathrm{d}}, \mathrm{z}=2^{\mathrm{e}} \times 3^{\mathrm{f}}}
where a,b,c,d,e,f  are non-negative integers.

The possible values for a,b,c,d,e,f are:

0,1,2,3 for a,b,c,d

0, 1 for e,f

Therefore, for Case 2, there are \mathrm{4 \times 4 \times 4 \times 4\times 2\times 2 =512}

possible solutions.

Case 3:  \mathrm{\mathrm{x}=1, \mathrm{y}=1, \mathrm{z}=2^{\mathrm{a}} \times 3^{\mathrm{b}} \times 2^{\mathrm{c}} \times 3^{\mathrm{d}}}

where a,b,c,d  are non-negative integers.

The possible values for a,b,c,d  are:

0,1,2,3 for a,b,c,d

Therefore, for Case 3, there are \mathrm{4 \times 4 \times 4 \times 4 =256} possible solutions.
Hence, the total number of possible integral solutions for \mathrm{x,y,z} such that
\mathrm{x \times y \times z= 72 \, is \, 1,728+512+256= 2,496}.
 

Posted by

vishal kumar

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