What is the sum of number of digits of the numbers $2^{2001}$ and $5^{2001}$ 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 $1999, 2003, 4002, 6003, 2002$
 A: It's $2002$.  It's asking for the sum of the number of digits of $2^{2001}$ and $5^{2001}$ in base $10$, so just take the log base $10$ of each, take the ceiling function and hey presto:
$603+1399=2002$
A: The answer is
$$
\overbrace{\lfloor2001\log_{10}(2)\rfloor+1}^\text{digits in $2^{2001}$}+\overbrace{\lfloor2001\log_{10}(5)\rfloor+1}^\text{digits in $5^{2001}$}
$$
However, we also have, using Iverson Brackets,
$$
\lfloor x\rfloor+\lfloor y\rfloor=\lfloor x+y\rfloor-[\{x\}+\{y\}\ge1]
$$
So we need to know $\{2001\log_{10}(2)\}+\{2001\log_{10}(5)\}$, but since $2001\log_{10}(2)+2001\log_{10}(5)=2001$, we know that the sum of their fractional parts is exactly $0$ or exactly $1$. Since the fractional parts are both positive, we must have exactly $1$.
Therefore,
$$
\begin{align}
\lfloor2001\log_{10}(2)\rfloor+1+\lfloor2001\log_{10}(5)\rfloor+1
&=\lfloor2001\log_{10}(2)+2001\log_{10}(5)\rfloor+1\\
&=2002
\end{align}
$$
A: 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.

A: It looks like you don't need logarithms or any calculator to solve this problem. Let's start.
First, observe that the following inequalities hold:
$$10^m<\underbrace {2^{2001}}_{m+1 ~ \text{digits}}<10^{m+1}$$
$$10^n<\underbrace{5^{2001}}_{n+1 ~ \text{digits}}<10^{n+1}$$
You get,
$$10^{m+n}<10^{2001}<10^{m+n+2}$$
$$2001=m+n+1$$
$$m+n=2000$$
Finally,  the sum of digits of $2^{2001}$ and $5^{2001}$ is equal :
$$\begin{align}\color {gold}{\boxed {\color{black}{m+1+n+1=m+n+2\\ \qquad \qquad \qquad\thinspace=2000+2 \\\qquad \qquad \qquad \thinspace=2002.}}}\end{align}$$
