TITLE: Total digits number of $2^{1971}$ and $5^{1971}$
QUESTION [2 upvotes]: I think this is very interesting question. I'm trying to solve this problem for 3 hours. But now I accept the defeat...

Question:
$2^{1971}$ and $5^{1971}$  numbers are consistently (continuously) written. Find the total digits.

REPLY [11 votes]: Let there is $k$ digit in $2^{1971}$ and there is $m$ digit in $5^{1971}$.
We must find $k+m$.
Given:
$$10^{k-1}<2^{1971}<10^{k}$$ 
$$10^{m-1}<5^{1971}<10^{m}$$ 
The above two relations imply:
$$10^{k+m-2}<10^{1971}<10^{k+m}$$
Finally, $k+m-1=1971 \Rightarrow k+m=1972$
Answer: $1972$ digits are written.

REPLY [0 votes]: Let's take the base-10 logarithm of $2^{1971}$.
This is equal to $1971\log2$.
We have that $\log 2 = 0.30103$, so $1971\log 2 = 593.33$
So our number is $10^{593.33}$. We know that $10^{6}<x<10^{7}$ has $7$ digits.
Using that logic, this number has exactly $\boxed{594}$ digits.
Similarly, we rewrite $5^{1971}$ as $1971 \log 5 = 1377.67$, which has $
\boxed{1378}$ digits.
So $594+1378=\boxed{1972}$ digits.
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