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Number Theory Level 3

How many digits are there in $2018!$ ?

Note:- - ! denotes factorial

1 solution

Mrigank Shekhar Pathak
Dec 8, 2017

The number of digits in a number $m$ can be calculated using the formula $\lfloor{\log(m)}\rfloor +1$

Stirling's approximation for $n!$ says that $n! \approx \sqrt{2 \pi n}\left(\frac{n}{e}\right)^n.$

Thus number of digits in $n!$ can be calculated using $\lfloor M \rfloor +1$ where, $M$ is $\frac {\log(2 \pi n)}{2} + n\cdot \log(n/e)$

Putting $n=2018$ we get the number of digits as $\boxed{5795}$

Clarification: $e$ is the Euler's number $\approx2.71828$ , $\lfloor \cdot \rfloor$ is the Greatest Integer or Floor Function

Correction:

It seems to me that there is a small error in your solution:

The number of digits in the decimal expansion of a number $m \ne 0$ can be calculated using the formula $\lfloor{\log(m)}\rfloor + 1$

This formula is in reality equivalent to the one you have given, except for powers of 10:

For example, for $m= 10$ , your formula gives $\lceil{\log(10)}\rceil = \lceil{1}\rceil = \color{#D61F06}{1}$ .

Filip Rázek - 3 years, 6 months ago

Thank you . Please see the corrections.

Mrigank Shekhar Pathak - 3 years, 6 months ago

Great! Thanks for your problem!

Filip Rázek - 3 years, 6 months ago