# Intriguing problem

Today, I , Svatejas and Vighnesh were chatting on slack and I accidentally came across this problem:

Find the largest integer $$n$$ such that $$n!<a$$ for some positive integer $$a$$ having $$n$$ digits.

Find a method to do this. Hoping for numerous approaches :)

Note by Nihar Mahajan
2 years, 7 months ago

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Here is an interesting result: $$\lceil \log([\log(2^{10000})]!) \rceil = 9166$$

- 2 years, 7 months ago

That was indeed surprising LOL

- 2 years, 7 months ago

@Nihar Mahajan Let's create some problem using factorials and 9166 tomorrow. Till then I won't follow anyone :P

- 2 years, 7 months ago

Sure!

- 2 years, 7 months ago

I'll try to list all such integer till n=6 by tmmrw.(through coding obviously)

- 2 years, 7 months ago

- 2 years, 7 months ago

There is no one-formula for this question.

- 2 years, 7 months ago

If we include some relation between $$a$$ and $$n$$ then? For example: if $$a$$ has $$n$$ digits.

- 2 years, 7 months ago

Then you can use Stirling's formula.

- 2 years, 7 months ago