This is a follow up to First digit of 3^2020
Can you find the first digit of 2020! (factorial) without a computer?
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Sign up to join this communityThis is a follow up to First digit of 3^2020
Can you find the first digit of 2020! (factorial) without a computer?
I believe it's
3
Because
We know, that for sufficiently large $n$ we have $n!\approx \sqrt{2\pi n}\ n^n/\exp n$. Applying logarithm, we get $\log n! \approx n\log n - n + \frac12 \log 2\pi n$, or $\log_{10} n! \approx n(\log n - 1)/\log 10+\frac12\log_{10} 2\pi n$. For $n=2020$, we get $\log_{10} 2020!\approx 5801.58667 $. Raising 10 to the power of mantissa (0.58667), we get about 3.861, so the first digit is 3.
Notice
Checking in Python (for example) does show that $2020!$ does indeed start with 3, but nevertheless, the $n!\approx \sqrt{2\pi n}\ n^n/\exp n$ approximate equality must be used with great care, since it says that it will hold for sufficiently large $n$, but in no way does it specify how large must $n$ be to achieve reasonable accuracy. Additionally, numbers may be very close but nevertheless start with different digits, when both are "close" to a digit multiplied by a power of 10 (e.g. 29999 and 30001 are within 0.007% of each other, but start with different digits, both being close to 30000).
$3$
Because:
$\ln 2020!=\sum_\limits{k=2}^{2020} \ln k \approx13358.65$.
$e^{13358.65}\approx 3.8724041499\times10^{5801}$.