question on minimum number f bits

Question

what is minimum no. of bits required to represent  //  (8)base 10              in binary  

 

formula : n >= log(8) base 2  = 3  but answer should be 4   ??  binary of 8 = 1000

Answer 1

Talking about unsigned integers, range of numbers possible using n-bits is 0 to 2ⁿ - 1

Let the largest possible number be p,

p = 2ⁿ - 1

p + 1 = 2ⁿ

So, n = log₂(p + 1)

In general for any number m, we require $\left \lceil log_{2}(m+1)\right \rceil$ bits to represent it in its binary equivalent

Comments

yes thanks ,   @ just_bhavana 

can you tell me this : 

what is minimum no. of bits required to represent  (6728)base 10   in binary    answer = 13  

6728 binary  = 1101001001000   = 13 bits   

what if question is about finding maximum  number of bits   then  n<= floor(12.7)  = 12 bits  

how if minimum is 13 bits then maximum  is 12 bits 

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here minimum would be 13 bits, maximum can be infinite as we can keep adding any number of leading 0's. Even asking question bout finding maximum number of bits is irrelevant