(a) Given that $N ≥ 1$ is a natural number, prove that $⌊\log_{10} (N)⌋ + 1$ gives the number of digits of $N$.
(b) Determine the number of digits of Googol= $10^{100}$.
(c) The googolplex is the number given by $10^{(10^{100})}$. Find the number of digits of this googolplex.
(d) (i) Find the number of digits of $2^{74 207 211}$.
[Hint: Change of the base for logs is given by the formula $log_b (a) = \frac{log_c (a)}{\log_c (b)}$ .]
(ii) What is the number of digits of $2^{74 207 211}$ if we work with number base 2?
(a) As the natural number $N$ increases, the number of digits $d$ increments as $N$ grows from 9 to 10, then 99 to 100, then 999 to 1000.
So, for a number $N$ to have $d$ digits, the following must hold
$$ 10^{d-1} < N < 10^d $$
For example, 5 has 1 digits
$ 10^{1-1} = 1 < 5 < 10 = 10^1 $
Another example, 999 has 3 digits
$ 10^{3-1} = 100 < 999 < 1000 = 10^3$
The inequality gives us
$$ (d-1) < \log_{10} N < d $$
So $d-1$ is the largest integer less than $\log_{10} N$, which is the definition of the floor function,
$$ d -1 = \lfloor \log_{10} N \rfloor $$
That is, the number of digits $d$ of $N$ is
$$ d = \lfloor \log_{10} N \rfloor + 1 $$
(b) We apply the formula
$$ \lfloor \log_{10} (10^{100}) \rfloor + 1 = 100 + 1 = 101 $$
So googol has 101 digits. That is, 1 followed by 100 zeroes.
(c) We apply the formula
$$ \lfloor \log_{10} (10^{10^{100}}) \rfloor + 1 = 10^{100} + 1$$
So googolplex has $10^{100}+1$ digits. That's 1 then 99 zeroes then 1.
(d) (i) We apply the formula
$$ \begin{align} d & = \lfloor \log_{10} 2^{74 207 211} \rfloor + 1 \\ \\ & = \lfloor \frac{\log_2 2^{74 207 211}}{\log_2 10} \rfloor + 1 \\ \\ & = \lfloor \frac{74 207 211}{\log_2 10} \rfloor + 1 \\ \\ & = 22338597 \end{align}$$
The number of digits of $2^{74 207 211}$ is 22338597.
(ii) The equivalent formula for the number of digits, derived by the same rationale as above, is
$$ d = \lfloor \log_{2} N \rfloor + 1 $$
We apply the formula
$$ \begin{align} d & = \lfloor \log_{2} 2^{74 207 211} \rfloor + 1 \\ \\ & = \lfloor 74 207 211 \rfloor + 1 \\ \\ &= 74 207 212 \end{align}$$
The number of digits of $2^{74 207 211]$ in binary form (base 2) is 74 207 212.