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 $7$

Lets decompose the numbers to their prime factorisation form,

$216=2^3{.3}^3$

$2500=2^2{.5}^4$

$300=2^2.3{.5}^2$

Combining powers of each, we get

$2^{3.\left(3/5\right)+2.\left(2/5\right)+2.\left(1/5\right)}=2^{\left(15/5\right)}=2^3$

$3^{3.\left(3/5\right)+1.\left(1/5\right)}=3^{\left(10/5\right)}=3^2$

$5^{4.\left(2/5\right)+2.\left(1/5\right)}=5^{\left(10/5\right)}=5^2$

Total possible prime factors $=$ sum of prime powers $=3+2+2=$$7$.