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I know the answer is more than $100$, because $100!$ is clearly less than $100^{100}$. And I know the answer is less than $10000$, because ($10000 \cdot 9999 \cdot 9998 \cdot \dots\cdot 9901) \cdot (100 \cdot 99 \cdot 98 \cdot \dots \cdot 1)$ is greater than $100^{200}$. But I don't know how to find the answer. Can this be done without testing every value?

user64742
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Dan
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    It's no coincidence that the answer is approximately $100e$: that's what Stirling's formula suggests as an approximation. But finding the exact answer seems trickier. – Eric Wofsey Dec 13 '16 at 00:28
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    I think we all have an ordinary language sense of what "without technology" means here… I.e., can this be done without extensive, tedious computation, of the sort one would normally resort to an electronic computer for. – Sridhar Ramesh Dec 13 '16 at 00:33
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    "without technology" may be a poor choice of words, but dismissing and/or criticising the question solely on that seems pointless and a little mean. Obviously we can do this "without technology" by calculating n! for each and every value of n until we find the limit. But it's equally obvious that this heavy calculation is not an acceptable answer either. – fleablood Dec 13 '16 at 00:37
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    No, I get that and I don't disagree. I just don't think it's useful, fair, or relevant to have that discussion now. It's pretty clear the OP meant how to solve this without actually calculating it (which the OP assumed could be done "with technology"). That the OP's wording was poor doesn't merit at worst getting pedantic and condescending, and at best going off on a tangent. – fleablood Dec 13 '16 at 00:58
  • This question seems somewhat similar, although it is a bit unclear. – Martin Sleziak Dec 13 '16 at 01:23

1 Answers1

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Stirling says $n! \sim \sqrt{2\pi} n^{n+1/2} e^{-n}$. $(n/e)^n = 100^n$ for $n = 100 e \approx 271$. Not far off from the actual value, but getting something much better without technology might be difficult.

EDIT: I take that last sentence back slightly. Generalizing to $n! = t^n$, a better solution asymptotically is $n \sim t e - \ln(2\pi t)/2 - 1/2$, as this makes $ \ln(n!) - \ln(t^n) = O(\ln(t)^2/t)$.

For $t = 100$ this would be $n = 100\; e - \ln(200 \pi)/2 - 1/2 \approx 268.107$. Indeed this is quite close to the actual solution of $\Gamma(x+1) = 100^x$, which is approximately $268.087$, close enough so the next higher integer, $269$, is the correct value.

Of course some low-tech "technology" would be convenient for numerically evaluating $100 e - \ln(200 \pi)/2 - 1/2$.

I hope Rob Arthan likes this version better.

Robert Israel
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    @RobArthan : Ask the OP. – Eric Towers Dec 13 '16 at 00:31
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    I think we all have an ordinary language sense of what "without technology" means here: done without extensive, tedious computation, of the sort one would normally resort to an electronic computer for. – Sridhar Ramesh Dec 13 '16 at 00:33
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    So? Are you the OP? What relevance does that have to the question or this excellent answer? – fleablood Dec 13 '16 at 00:59
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    @fleablood: the question should be fixed to avoid a gross abuse of the word "technology". However, the inability of those commenting on this thread to understand this very simple terminological point leaves me in a state of despair. So I have deleted my other comments and leave only this one. Why you think Robert's answer is excellent is beyond me. – Rob Arthan Dec 13 '16 at 02:20
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    We understand it. We just don't care. The op asked a good question and to have coniptions and berate the op because he used a word, which has nothing to do with mathematics, in a way not to your liking is mean and tangential and irrelevant. And this answer is excellent in that it clearly gives a distinct and easy to understand result and cites the nescessary theorem for further study. – fleablood Dec 13 '16 at 07:55
  • @TheGreatDuck: thank you for sharing your opinions (which bring nothing useful to bear on this interesting question). I have at no time written anything in this thread that could be constituted as rude. Your claim that "nobody wants [me] here" is incompatible with the upvote for my comment above or the fact that Robert was good enough to edit his answer. – Rob Arthan Dec 14 '16 at 00:54