Largest Integer Expressible by 3 Digits/Number of Digits

Theorem

$9^{9^9}$ has $369 \, 693 \, 100$ digits when expressed in decimal notation.

Let $n$ be the number of digits in $9^{9^9}$

$n = 1 + \floor {\map {\log_{10} } {9^{9^9} } }$

where $\floor {\ldots}$ denotes the floor function.

Then:

$\ds \map {\log_{10} } {9^{9^9} }$

$\approx$

$\ds 369 \, 693 \, 099 \cdotp 63157 \, 03685 \, 87876 \, 1$

$\ds \leadsto \ \ $

$\ds n$

$=$

$\ds 1 + \floor {369 \, 693 \, 099 \cdotp 63157 \, 03685 \, 87876 \, 1}$

$\ds $

$=$

$\ds 1 + 369 \, 693 \, 099$

Definition of Floor Function

$\ds $

$=$

$\ds 369 \, 693 \, 100$

$\blacksquare$

The number of digits in $9^{9^9}$ was demonstrated to be $369 \, 693 \, 100$ in $1906$ by Charles-Ange Laisant.