Question

number of prime factors in (216)^3/5×(2500)^2/5×(300)^1/5 is​

Answer 1

Answer:

7

Step-by-step explanation:

Lets decompose the numbers to their prime factorisation form,

216 =2³.3³

2500 =2³.5².5²

300 =2.2.3.5²

 Combining powers of each, we get

2  ^3.(3/5)+2.(2/5)+2.(1/5) =2 ^ (15/5) =2 ^ 3

33.(3/5)+1.(1/5) 3(10/5) 3²

54.(2/5)+2.(1/5) 5(10/5) 5^2

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