8
$\begingroup$

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 $1999, 2003, 4002, 6003, 2002$

$\endgroup$
10
  • $\begingroup$ @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
  • 1
    $\begingroup$ @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
  • 1
    $\begingroup$ @sil I worked it out given the multi choice 😉 $\endgroup$ Mar 15 at 18:08
4
$\begingroup$

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:

$603+1399=2002$

$\endgroup$
6
  • 8
    $\begingroup$ 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
  • 2
    $\begingroup$ "the sum of the number of digits" Well, that's just nasty..... $\endgroup$
    – fleablood
    Mar 15 at 18:16
  • 1
    $\begingroup$ @hgmath nice. I figured there was something like that although I just used the calculator ;) $\endgroup$ Mar 15 at 18:17
  • 1
    $\begingroup$ @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
16
$\begingroup$

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 :

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

$\endgroup$
2
$\begingroup$

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

$\endgroup$
1
  • $\begingroup$ 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
2
$\begingroup$

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.
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.