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find no of positive integral solution of x1x2x3=10800

find no of positive integral solution of x1x2x3=10800

Grade:12

2 Answers

moumi roy
91 Points
7 years ago
can you make the question a bit clear i cant understand its meanig.please clear it 
can you make the question a bit clear i cant understand its meanig.please clear it 
jagdish singh singh
173 Points
7 years ago
\hspace{-0.70 cm}$Given $x_{1}\cdot x_{2}\cdot x_{3} = 10800 = 2^4 \cdot 3^3 \cdot 5^2.$ Let $x_{1} = 2^{a_{1}}\cdot 3^{b_{1}}\cdot 5^{c_{1}}$\\\\ and $x_{2} = 2^{a_{1}}\cdot 3^{b_{2}}\cdot 5^{c_{2}}$ and $x_{3} = 2^{a_{3}}\cdot 3^{b_{3}}\cdot 5^{c_{3}}$ $\\\\ $So $2^{a_{1}+a_{2}+a_{3}}\cdot 3^{b_{1}+b_{2}+b_{3}}\cdot 5^{c_{1}+c_{2}+c_{3}} = 2^4 \cdot 3^3 \cdot 5^2$\\\\ So $a_{1}+a_{2}+a_{3}=4$ and $b_{1}+b_{2}+b_{3}=3$ and $c_{1}+c_{2}+c_{3} = 2$\\\\ where $a_{i}\in \{0,1,2,3,4\}$ and $b_{i}\in \{0,1,2,3\}$ and $c_{i}\in \{0,1,2\}$\\\\
 
\hspace{-0.70 cm}$So for calculation of all non negative integer ordered triplets of \\\\ $a_{1}+a_{2}+a_{3}=4$ . put $a_{1}=0$ get $5$ ordered pair of $(a_{2},a_{3})$ \\\\ Similarly put $a_{2}=0$ get $3$ ordered pair of $(a_{2},a_{3})$.\\\\ So Total $5+4+3+2+1 = 15$\\\\ Similarly for $b_{1}+b_{2}+b_{3}=3$ we get $4+3+2+1=10$\\\\ Similarly for $c_{1}+c_{2}+c_{3} = 2$ we get $3+2+1=6$\\\\
 
\hspace{-0.70 cm}$ So we get total triplets $ = 15 \times 10 \times 6 = 900$

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