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 $1999, 2003, 4002, 6003, 2002$
Best Answer
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 :
$$\begin{align}\color {gold}{\boxed {\color{black}{m+1+n+1=m+n+2\\ \qquad \qquad \qquad\thinspace=2000+2 \\\qquad \qquad \qquad \thinspace=2002.}}}\end{align}$$