Question

The smallest natural no. by which 135 must be divided to obtain a perfect cube is​

Answer 1

Prime factorising 135, we get,

135=5×3×3×3 =$3^{3}$×5

We know, a perfect cube has multiples of 3 as powers of prime factors.

Here, number of 3's is 3 and number of 5's is 1.

So we need to divide the factorization by 5 to make 135 a perfect cube.

Hence, the smallest number by which 135 must be divided to obtain a perfect cube is 5.