Find the number of digits, $D$, in the decimal expansion of the large number $N=4^{4^{4^{4}}}$

Question

The full question is:

Find the number of digits, $D$, in the decimal expansion of the large number

$$N=4^{4^{4^{4}}}$$

Try and find the most efficient ways of finding $D$.

I know that $4^{4^{4^{4}}}$ computes a very large number i.e. $4^{4^{256}}$. I'm not sure how to go about working out both the decimal expansion and finding the number of digits $D$ in the question. Please can someone kindly help! thank you

Answer 1

The number of digits is

$$\eqalign{&80723047260282253793826303970853990300713679217387430318\cr&67082828418414 48156830914919 891181470122948345198\cr&15575747711564964572385352990874812 44990261351117\ .\cr}$$

I got this by asking for ${\tt IntegerPart[2^{513}\ Log[10,2]]+1}$ in Mathematica.

Answer 2

Hint: since the number of digits of any number $m$ is $\lfloor \log_{10} m \rfloor +1$, you can compute the base $10$ log of $N$, which is a much smaller number.

Answer 3

$\log_{10}N=4^{4^4}\log_{10}4$, so $\log_{10}\log_{10}N=4^4\log_{10}4+\log_{10}\log_{10}4\approx153.906997548$, and $\log_{10}N\approx 10^{153.906997548}\approx8.072304726028\times 10^{153}$; this is the approximate number of decimal digits in $N$. I did this on the Excalibur RPN calculator app, and I see that it agrees with Christian Blatter’s exact result within the limits of its precision; you’re probably not going to do much better than this without using a very high precision package.