Question

✌️✌️plzz take out the HCF of 60,84,108 plzz answer fast ❤️❤️✌️​

Answer 1

The prime factorisation of 60, 84 and 108 is :60 = 2 × 2 × 3 × 584 = 2 × 2 × 3 × 7108 = 2 × 2 × 3 × 3 × 3 HCF = 2 × 2 × 3 = 12

Answer 2

Step-by-step explanation:

$\mathfrak{H.C.F.}$

The full form of H.C.F. is Highest Common Factor. i.e. the number which is a factor of all the given numbers.

So, we need to find the factors of 60, 84 and 108.

So, let's get into it.

60=2×2×3×5

84=2×2×3×7

108=2×2×3×3×3

So, the factors are given.

We see that, 7 is a factor of 84 only, so this isn't the H.C.F.

Similarly, 5 is a factor of 60, not of 84 and 108.

But, 2 is a factor of all the given numbers which is present twice

And 3 is a factor of all the given numbers.

Therefore, the H.C.F. of 60, 84 and 108 is 2×2×3=12