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 $1999, 2003, 4002, 6003, 2002$

  • $\begingroup$ @samerivertwice yes base 10 $\endgroup$ Mar 15 at 17:15
  • $\begingroup$ Not hard to do with a computer program, but perhaps that's not allowed. $\endgroup$
    – paw88789
    Mar 15 at 17:18
  • $\begingroup$ @Sil I am certain that that is what the question asks $\endgroup$ Mar 15 at 17:36
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    $\begingroup$ @Sil I'll ask the author for an explanation, as soon as I get a response, I'll post it here $\endgroup$ Mar 15 at 18:05
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    $\begingroup$ @sil I worked it out given the multi choice 😉 $\endgroup$ Mar 15 at 18:08

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:


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    $\begingroup$ remark: without taking the ceiling function, $$2001\log_{10}(2)+2001\log_{10}(5) = 2001(\log_{10}(2)+\log_{10}(5)) = 2001\log_{10}(10) = 2001$$ so no calculator is needed to know the answer is in $[2001,2003)$ $\endgroup$
    – hgmath
    Mar 15 at 18:14
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    $\begingroup$ "the sum of the number of digits" Well, that's just nasty..... $\endgroup$
    – fleablood
    Mar 15 at 18:16
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    $\begingroup$ @hgmath nice. I figured there was something like that although I just used the calculator ;) $\endgroup$ Mar 15 at 18:17
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    $\begingroup$ @samerivertwice your solution is brilliant, thank you very much $\endgroup$ Mar 15 at 18:18
  • $\begingroup$ @MichaelBlane for the logarithm I just used a calculator. If needed to do without a calculator this can be done, using an infinite series for the log function but the real magic is in hgmath's comment - if you want to learn something, that comment is the thing to think about. $\endgroup$ Mar 15 at 18:20

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,




Finally, the sum of digits of $2^{2001}$ and $5^{2001}$ is equal :

$$\begin{align}\color {gold}{\boxed {\color{black}{m+1+n+1=m+n+2\\ \qquad \qquad \qquad\thinspace=2000+2 \\\qquad \qquad \qquad \thinspace=2002.}}}\end{align}$$


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, $$ \begin{align} \lfloor2001\log_{10}(2)\rfloor+1+\lfloor2001\log_{10}(5)\rfloor+1 &=\lfloor2001\log_{10}(2)+2001\log_{10}(5)\rfloor+1\\ &=2002 \end{align} $$

  • $\begingroup$ I see that I answered the question after it was edited to be correct. It was probably that edit which brought this question to the top of the front page just as I was looking. $\endgroup$
    – robjohn
    Mar 15 at 22:09

Generalization of the problem:

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

$$10^m<\underbrace {2^{N}}_{m+1 ~ \text{digits}}<10^{m+1}$$

$$10^n<\underbrace{5^{N}}_{n+1 ~ \text{digits}}<10^{n+1}$$


$$N= m+n+1$$

$$ m+n=N-1$$

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.

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