text: # Thread: Number of possible digits in a binary representation of a number
  
  1. ## Number of possible digits in a binary representation of a number
  
  Hi all,
  
  Suppose that when n is written in decimal notation it has 100 digits. Which
  possibilities are there for the number of digits if n is written in binary notation?
  
  Any ideas?
  
  Thanks
  
  2. Hello, sirellwood!
  
  Suppose that when $\,n$ is written in decimal notation it has 100 digits.
  What possibilities are there for the number of digits if $\,n$ is written in binary notation?
  
  $\text{If }n\text{ has 100 digits: }\:10^{99}\:\le\: n\:<\:10^{100}$
  
  $\text{Suppose }2^k\text{ is the largest power-of-2 contained in }n.$
  
  . . $\text{Then: }\;10^{99}\:\le\: 2^k \:<\:10^{100}$
  
  $\text{Take logs: }\;\log(10^{99}) \:\le\;\log(2^k) \:<\:\log(10^{100})$
  
  . . . . . . . . . . . . . $99 \:\le \:k\log 2 \:<\:100$
  
  . . . . . . . . . . . . . $\dfrac{99}{\log 2} \:\le \: k \:<\:\frac{100}{\log2}$
  
  . . . . . . . . . . . . $328.87 \:\le\:k \:<\:332.19$
  
  . . . . . . . . . . . . . . $329 \:\le\:k\:\le\:332$
  
  $\text{Therefore, }n\text{, written in binary, has from 329 to 332 digits.}$
  
  3. Originally Posted by Soroban
  Hello, sirellwood!
  
  $\text{If }n\text{ has 100 digits: }\:10^{99}\:\le\: n\:<\:10^{100}$
  
  $\text{Suppose }2^k\text{ is the largest power-of-2 contained in }n.$
  
  . . $\text{Then: }\;10^{99}\:\le\: 2^k \:<\:10^{100}$
  
  $\text{Take logs: }\;\log(10^{99}) \:\le\;\log(2^k) \:<\:\log(10^{100})$
  
  . . . . . . . . . . . . . $99 \:\le \:k\log 2 \:<\:100$
  
  . . . . . . . . . . . . . $\dfrac{99}{\log 2} \:\le \: k \:<\:\frac{100}{\log2}$
  
  . . . . . . . . . . . . $328.87 \:\le\:k \:<\:332.19$
  
  . . . . . . . . . . . . . . $329 \:\le\:k\:\le\:332$
  
  $\text{Therefore, }n\text{, written in binary, has from 329 to 332 digits.}$