Calculating bits required to store decimal number

Question

This is a homework question that I am stuck with.

Consider unsigned integer representation. How many bits will be required to store a decimal number

Answer 1

For a binary number of n digits the maximum decimal value it can hold will be

2^n - 1, and 2^n is the total permutations that can be generated using these many digits.

Taking a case where you only want three digits, ie your case 1. We see that the requirements is,

2^n - 1 >= 999

Applying log to both sides,

log(2^n - 1) >= log(999)

log(2^n) - log(1) >= log(999)

n = 9.964 (approx).

Taking the ceil value of n since 9.964 can't be a valid number of digits, we get n = 10.