# Largest Integer Expressible by 3 Digits/Number of Digits

## Theorem

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

## Proof

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:

 $\displaystyle \map {\log_{10} } {9^{9^9} }$ $\approx$ $\displaystyle 369 \, 693 \, 099 \cdotp 63157 \, 03685 \, 87876 \, 1$ Largest Integer Expressible by 3 Digits: Logarithm Base 10 $\displaystyle \leadsto \ \$ $\displaystyle n$ $=$ $\displaystyle 1 + \floor {369 \, 693 \, 099 \cdotp 63157 \, 03685 \, 87876 \, 1}$ $\displaystyle$ $=$ $\displaystyle 1 + 369 \, 693 \, 099$ Definition of Floor Function $\displaystyle$ $=$ $\displaystyle 369 \, 693 \, 100$

$\blacksquare$

## Historical Note

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