Largest Integer Expressible by 3 Digits/Number of Digits

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Theorem

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


Proof

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

From Number of Digits in Number:

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

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.


Sources