Question

the largest number 'n' such that (2016!)! is divisible by ((n!)!)! ​

Answer 1

ur answer is 2016 cz if u divide by any other large no. u will get point value .

Answer 2

It is known that, the largest possible integer, which divides a positive integer completely, is the integer itself. Such possible integer is known as the HCF of the integer, which is the integer itself.

Which means, as examples,

  ⇒ The largest possible integer which divides 10 completely, is 10 itself. The HCF of 10 is 10.

  ⇒ The largest possible integer which divides 12456 completely, is 12456 itself. The HCF of 12456 is 12456.

  ⇒ The largest possible integer which divides x completely, is x itself, for any positive integer x. The HCF of x is x.

Like this, the largest possible integer which divides (2016!)! completely is (2016!)! itself. The HCF of (2016!)! is (2016!)!.

Thus it can be stated that,

((n!)!)! = (2016!)!

We can cancel factorial sign each from both sides.

(n!)! = 2016!

Again it can be cancelled, thus,

n! = 2016

But 2016 is not the product of first any consecutive positive integers.

$\displaystyle \prod_{k=1}^{n}k \ \neq \ 2016$

But thus we can say that n! is the factor of 2016.

We have to take the prime factorization of 2016.

2016 = 2 × 2 × 2 × 2 × 2 × 3 × 3 × 7

Now we're going to make a factorial number from this prime factorization.

  ⇒ The value won't change even 1 is multiplied with it.

2016 = 1 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 7

  ⇒ Bring one 3 in the prime factorization at the third place.

2016 = 1 × 2 × 3 × 2 × 2 × 2 × 2 × 3 × 7

  ⇒ Two 2's are multiplied to get 4.

2016 = 1 × 2 × 3 × 4 × 2 × 2 × 3 × 7

  ⇒ But there's no 5. So conclude here. From this we get that,

2016 = 4! × 2 × 2 × 3 × 7

This displays that the largest factorial number as factor of 2016 is 4!.

Thus n! = 4!

⇒ n = 4.

So 4 is the answer.