TITLE: What is the sum of number of digits of the numbers $2^{2001}$ and $5^{2001}$
QUESTION [7 upvotes]: What is the sum of number of digits of the numbers $2^{2001}$ and $5^{2001}$? (Singapore 1970)
I attempted to solve this question by working out what each digit must be, and maybe find some pattern, but I couldn't find any, apart from the fact that $2^{2001}\mod{10}\equiv 4$ and $2^{2001}\pmod{10}\equiv 5$. Could you please explain to me how to solve this question? This question is multiple choice with options $1999, 2003, 4002, 6003, 2002$

REPLY [2 votes]: Generalization of the problem:

What is the sum of number of digits of the numbers $2^N$ and $5^N$?

$$10^m<\underbrace {2^{N}}_{m+1 ~ \text{digits}}<10^{m+1}$$
$$10^n<\underbrace{5^{N}}_{n+1 ~ \text{digits}}<10^{n+1}$$
$$10^{m+n}<10^{N}<10^{m+n+2}$$
$$N= m+n+1$$
$$ m+n=N-1$$
The sum of digits of the numbers $2^{N}$ and $5^{N}$ will be equal :
$$\begin{align}\color {gold}{\boxed {\color{black}{m+1+n+1=m+n+2\\ \qquad \qquad \qquad\thinspace=N-1+2 \\\qquad \qquad \qquad \thinspace=N+1.}}}\end{align}$$

Short answer: $N+1$ digits.