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 :)

## Comments

Sort by:

TopNewestHere is an interesting result: \(\lceil \log([\log(2^{10000})]!) \rceil = 9166\) – Svatejas Shivakumar · 1 year, 2 months ago

Log in to reply

– Nihar Mahajan · 1 year, 2 months ago

That was indeed surprising LOLLog in to reply

@Nihar Mahajan Let's create some problem using factorials and 9166 tomorrow. Till then I won't follow anyone :P – Svatejas Shivakumar · 1 year, 2 months ago

Log in to reply

– Nihar Mahajan · 1 year, 2 months ago

Sure!Log in to reply

I'll try to list all such integer till n=6 by tmmrw.(through coding obviously) – Harsh Shrivastava · 1 year, 2 months ago

Log in to reply

@Pi Han Goh @Ivan Koswara @Sharky Kesa @Brian Charlesworth – Nihar Mahajan · 1 year, 2 months ago

Log in to reply

– Pi Han Goh · 1 year, 2 months ago

There is no one-formula for this question.Log in to reply

– Nihar Mahajan · 1 year, 2 months ago

If we include some relation between \(a\) and \(n\) then? For example: if \(a\) has \(n\) digits.Log in to reply

Stirling's formula. – Pi Han Goh · 1 year, 2 months ago

Then you can useLog in to reply