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 :)

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
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Comments

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

A Former Brilliant Member - 5 years, 2 months ago

That was indeed surprising LOL

Nihar Mahajan - 5 years, 2 months ago

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

A Former Brilliant Member - 5 years, 2 months ago

Sure!

Nihar Mahajan - 5 years, 2 months ago

@Pi Han Goh @Ivan Koswara @Sharky Kesa @Brian Charlesworth

Nihar Mahajan - 5 years, 2 months ago

There is no one-formula for this question.

Pi Han Goh - 5 years, 2 months ago

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

Nihar Mahajan - 5 years, 2 months ago

@Nihar Mahajan – Then you can use Stirling's formula.

Pi Han Goh - 5 years, 2 months ago

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

Harsh Shrivastava - 5 years, 2 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$