Question

What is the smallest natural number n such that H! is
divisible by 990?​

Answer 1

Answer:

11 will be divisible by 110

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Answer 2

We're asked to find smallest natural number n such that n! is exactly divisible by 990.

Let us prime factorise 990.

990 = 2 × 3² × 5 × 11

And take the greatest prime factor among them. Here it's 11.

We see 11! contains all 2, 3² = 9, 5 and 11.

So we can say 11! can exactly be divisible by 990.

Therefore, n = 11.

If we look for possible value of 'n' less than 11, that's not possible because there's no natural number n < 11 such that n! includes 11.