Question

find the smallest number by which each of the following number must be divided to obtain a perfect cube 1536
​

Answer 1

Answer:

6

explanation

after getting lcm

clearly we can see.

2*2*2*2*2*3

after dividing by 6 it is a perfect square

Answer 2

Answer➝

We have,

We have, 1536 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3

We have, 1536 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 After grouping the prime factors in triplets, it’s seen that one factor 3 is left without grouping.

We have, 1536 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 After grouping the prime factors in triplets, it’s seen that one factor 3 is left without grouping. 1536 = (2 × 2 × 2) × (2 × 2 × 2) × (2 × 2 × 2) × 3

We have, 1536 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 After grouping the prime factors in triplets, it’s seen that one factor 3 is left without grouping. 1536 = (2 × 2 × 2) × (2 × 2 × 2) × (2 × 2 × 2) × 3 So, in order to make it a perfect cube, it must be divided by 3.

Thus, the smallest number by which 1536 must be divided to obtain a perfect cube is 3.

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