Question

Number of prime factors in $(216{)}^{\frac{3}{5}}\times (2500{)}^{\frac{2}{5}}\times (300{)}^{\frac{1}{5}}$ is

• A. 6
• B. 7
• C. 8
• D. None of these

Answer 1

The correct option is B 7

$(216{)}^{\frac{3}{5}}\times (2500{)}^{\frac{2}{5}}\times (300{)}^{\frac{1}{5}}$
$=(2^3\times 3^3{)}^{\frac{3}{5}}\times (2^2\times 5^4{)}^{\frac{2}{5}}\times (2^2\times 3\times 5^2{)}^{\frac{1}{5}}$
$=2^{\frac{9}{5}}\times 3^{\frac{9}{5}}\times 2^{\frac{4}{5}}\times 5^{\frac{8}{5}}\times 2^{\frac{2}{5}}\times 3^{\frac{1}{5}}\times 5^{\frac{2}{5}}$
$=2^{(\frac{9}{5}+\frac{4}{5}+\frac{2}{5})}\times 3^{(\frac{9}{5}+\frac{1}{5})}\times 5^{(\frac{8}{5}+\frac{2}{5})}$
$=2^{\frac{15}{5}}\times 3^{\frac{10}{5}}\times 5^{\frac{10}{5}}=2^3\times 3^2\times 5^2$
$\therefore$ The number of prime factors $=(3+2+2)=7$