Question

for every integer \(x\) defined a function \(f\) with \(f(x)\) is the number of digits of \(x\). e.g : \(f(125) = 3\) and \(f(2012)=4\). the value of \(f\left( 2^{2012} \right) + f\left(5^{2012}\right) \) is

Answer 1

How many digits are there in \(\large 10^5\) ?

Answer 2

6 digits ?

Answer 3

is \( f(2^{2012}) + f(5^{2012}) = f( 2^{2012} \times 5^{2012})\) ?
I know the number of digits of the right side, but I don't know whether they're equal

Answer 4

they are not equal

Answer 5

use this formula

number of digits in \(x\) is given by \(\lfloor \log_{10} x \rfloor + 1\)

Answer 6

so we have
use this formula

\[f(x) = \lfloor \log_{10} x \rfloor + 1\]

Answer 7

actually you might be right! let me think a bit more..

Answer 8

could you explain about the formula? I mean, how could the function equal to logarithm ?

Answer 9

how did you say the number of digits in 10^5 is 6 ?

Answer 10

you figured out the number of digits as 6 because the "exponent" is 5, so there will be 5 zeroes in the number... plus accounting for 1gives you 6, yes ?

Answer 11

logarithm = exponent

Answer 12

yes..,

ahh..., I see...

Answer 13

\[\large \log_{10} 10^{\color{Red}{5}} = \color{Red}{5}\]

Answer 14

\[\large \log_{10} 10^{\color{Red}{3}} = \color{Red}{3}\]

Answer 15

\[\large \log_{10} 10^{\color{Red}{k}} = \color{Red}{k}\]

Answer 16

adding +1 to \(\large \color{red}{k}\) to get the number of digits

Answer 17

when \(\large \color{red}{k}\) is not an integer, you just take the integer part
\[\large\]

Answer 18

so,
\[f \left( 2^{2012} \right)=\lfloor \log _{10}2^{2012} \rfloor + 1\]

how to calculate \(\lfloor \log _{10}2^{2012} \rfloor\) ?

Answer 19

http://www.wolframalpha.com/input/?i=floor%28log_%2810%29+%282%5E2012%29%29

Answer 20

the problem should be solved without calculator :(

Answer 21

we can find it w/o calculator

Answer 22

\[\begin{align}f \left( 2^{2012} \right) +\color{blue}{f \left( 5^{2012} \right)} &=\lfloor \log _{10}2^{2012} \rfloor + 1 + \color{blue}{\lfloor \log _{10}5^{2012} \rfloor + 1}\\~\\
&=\lfloor \log _{10}(10/5)^{2012} \rfloor + 1 + \color{blue}{\lfloor \log _{10}5^{2012} \rfloor + 1}\\~\\
&=\lfloor \log _{10}10^{2012} - \log _{10}5^{2012} \rfloor + 1 + \color{blue}{\lfloor \log _{10}5^{2012} \rfloor + 1}\\~\\
&=\lfloor 2012 - \log _{10}5^{2012} \rfloor + 1 + \color{blue}{\lfloor \log _{10}5^{2012} \rfloor + 1}\\~\\
&=2012 + \lfloor - \log _{10}5^{2012} \rfloor + 1 + \color{blue}{\lfloor \log _{10}5^{2012} \rfloor + 1}\\~\\
&=\cdots
\end{align}\]

see if you can finish it off

Answer 23

Notice you can pull out integers out of greatest integer function because, for any integer \(n\) and real number \(x\) we have :
\[\lfloor n+x \rfloor = n + \lfloor x\rfloor\]

Answer 24

Here's an attempt at approximating... Hopefully you might recall that \(2^{10}=1024\approx1000=10^3\). So, we have
\[10\log_{10}2\approx\log_{10}10^3=3~~\implies~~\log_{10}2\approx0.3\]
Multiplying by \(2012\) gives \(\log_{10}2^{2012}\approx 603.6\), and so \(\lfloor\log_{10}2^{2012}\rfloor\approx603\), which means \(2^{2012}\) has somewhere around \(604\) digits (the actual answer is \(606\), but I'm not sure how to fix this approach's accuracy...)

Answer 25

Might have something to do with the fact that I'm rounding down \(1024\) to \(1000\)

Answer 26

thank you very much @ganeshie8 :))