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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
5 months, 3 weeks ago

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Here is an interesting result: \(\lceil \log([\log(2^{10000})]!) \rceil = 9166\) Svatejas Shivakumar · 5 months, 3 weeks ago

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@Svatejas Shivakumar That was indeed surprising LOL Nihar Mahajan · 5 months, 3 weeks ago

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@Svatejas Shivakumar @Nihar Mahajan Let's create some problem using factorials and 9166 tomorrow. Till then I won't follow anyone :P Svatejas Shivakumar · 5 months, 3 weeks ago

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@Svatejas Shivakumar Sure! Nihar Mahajan · 5 months, 3 weeks ago

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I'll try to list all such integer till n=6 by tmmrw.(through coding obviously) Harsh Shrivastava · 5 months, 3 weeks ago

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@Nihar Mahajan There is no one-formula for this question. Pi Han Goh · 5 months, 3 weeks ago

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@Pi Han Goh If we include some relation between \(a\) and \(n\) then? For example: if \(a\) has \(n\) digits. Nihar Mahajan · 5 months, 3 weeks ago

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@Nihar Mahajan Then you can use Stirling's formula. Pi Han Goh · 5 months, 3 weeks ago

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