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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
1 year, 7 months ago

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

Brilliant Member - 1 year, 7 months ago

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That was indeed surprising LOL

Nihar Mahajan - 1 year, 7 months ago

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

Brilliant Member - 1 year, 7 months ago

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Sure!

Nihar Mahajan - 1 year, 7 months ago

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

Harsh Shrivastava - 1 year, 7 months ago

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

Nihar Mahajan - 1 year, 7 months ago

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There is no one-formula for this question.

Pi Han Goh - 1 year, 7 months ago

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

Nihar Mahajan - 1 year, 7 months ago

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

Pi Han Goh - 1 year, 7 months ago

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