Consider an n bit binary number. If the number is to be encoded in Octal, how many Octal digits would be required approximately? if u say n/3 then u r r* bt will get half marks only. Right answer is: Ceil(n/3) tell me why it is so.

Please don't jump into any thread with your questions. If you don't have the number of binary digits divisible by 3 then you may need an extra octal digit to convert the binary number into octal

ceil function is nothing but it will round of to next high digit for example if u consider 4 bit number as 1001 then 4/3 results in 1.3 but ceil(n/3) =ceil(1.3)=2 so oct representation of 1001=21(which has 2 digits)

namune ji when u take it as integer u need not to worry about decimals and u no need to use ceil function.got it