Question

find the smallest square number which is divisible by each of the numbers 6,9,27 and 36​

Answer 1

Let's start by writing out the prime factors of the above numbers.

6 → 2, 3

9 → 3, 3

27 → 3, 3, 3

36 → 4*9 → 2, 2, 3, 3

So any number divisible by each of 6, 9, 27 and 36 need have only 2's and 3's as prime factors. It must have as many of each as the largest number of them that appear in any of 6,9,27 and 36.

That would be three 3's (takes care of 27) and two 2's (for 36).

So the smallest number divisible by 6, 9, 27 and 36 is 2*2*3*3*3 = 108.

But the question asks for a square number. This must have an even number of each of its prime factors. Throwing in another 3 will do it → 2*2*3*3*3*3 = (2*3*3)^2 = 18^2 = 324..

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Answer 2

=>6 =2,3

=>9 =3,3

=>27=3,3,3

=>36=2,2,3,3

=>2×2×3×3×3 = 108

=>(2×2)×(3×3)×(3×3) = (2²)×(3²)×(3²)

=>324

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