Computational arithmetic - how many bits exactly for 8 digit number needed

Question

how many bytes (and how many bits) you need to represent number 99999999 ? I need to know this: we have a calculator, the simplest possible, and can accommodate up to 8 digits, i.e. from 0 to 99999999 (let's forget the negatives, unless if you feel comfortable to include in your answer).

How many bits/bytes do we need to store values from 0 up to and inclusive 99999999 ?

I appreciate your help, kindly provide the theoretical background and any calculations if you can. Thank you very much!

Answer 1

Because there are 8 digits and each digit can have 10 values (0, 1, …, 9), the total number of representable numbers is 10^8. To represent this many numbers in binary, we must have a number of digits N such that assigning just one of two values (0, 1) to each position gives at least as many representable numbers as we have in decimal. That is, we have to solve

2^N >= 10^8

We can take the base-2 log of both sides to get

N >= log_2(10^8) = 8 * log_2(10)

At this point, hopefully you have a calculator handy to calculate log_2(10). Note that this is equal to log_10(10)/ log_10(2) = 1/log_10(2), if your calculator does logarithms in base 10 by default. The answer comes out to:

N >= ~26.58

The smallest integer value of N that satisfies this is 27. So, 27 digits are required.