Dad's factorial problem
Photo credit to wal_172619
Goals of the article
This is an article for my father about a size of large numbers question.
Dear Dad,
How many digits does the following number have?
\begin{equation}\label{eq:start}
x = \left((9^9)!\right)!
\end{equation}
To start \( 9^9 = 387,420,489\). Next we need Sterling's approximation, a ubiquitous formula of statistical mechanics [1,2,3].
\begin{equation}
\log(z!) = z\log(z) - z
\end{equation}
or in log base 10
\begin{equation}
\log_{10}(z!) \approx z\log_{10}(z) - z\log_{10}(e)
\end{equation}
Next for \(z = 387,420,489\).
\begin{equation}
\log_{10}(z!) \approx 3,158,983,316
\end{equation}
So \((9^9)!\) is a number with approximately 3.16 billion digits. What happens when we add another factorial?
\begin{align}
z &= 10^{3,158,983,316}\\
\log_{10}(z!) &\approx z\log_{10}(z) - z\log_{10}(e)\\
\end{align}
Therefore, if \(b = 3.158983316\), then
\begin{align}
\log_{10}\left[\left((9^9)!\right)!\right] &\approx 10^{b\times 10^9} \log_{10} \left(10^{b\times 10^9} \right) - 10^{b\times 10^9}\log_{10}(e)\\
\log_{10}\left[\left((9^9)!\right)!\right] &\approx 10^{b\times 10^9} \left( \log_{10} \left(10^{b\times 10^9} \right) - \log_{10}(e)\right)\\
\log_{10}\left[\left((9^9)!\right)!\right] &\approx 10^{b\times 10^9} b\times 10^9\\
\log_{10}\left[\left((9^9)!\right)!\right] &\approx b \left(10^{b\times 10^9+9}\right)\\
\log_{10}\left[\left((9^9)!\right)!\right] &\approx 3.158983316 \times 10^{3,158,983,325}\\
\end{align}
So very roughly, \( \left((9^9)!\right)! \) is a number with \(3\times 10^\text{3 to 3.2 billion}\) digits.
References
[1] D. A. McQuarrie, "Introduction and Review," in Statistical Thermodynamics, pp. 1—34, 2000.
[2] K. A. Dill, and S. Bromberg, "Entropy and the Boltzmann Law," in Molecular Driving Forces, pp. 81—92, 2011.
[3] M. S. Shell, "Equilibrium and Entropy," in Thermodynamics and Statistical Mechanics: An Integrated Approach, pp. 6—20, 2015.
