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Dad's factorial problem

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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.


[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.

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