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

The largest number that can be represented by an n digit number in base b is bⁿ - 1. Hence, the largest number that can be represented in N binary digits is 2^N - 1. We need the smallest integer N such that:

2^N - 1 ≥ bⁿ - 1
⇒ 2^N ≥ bⁿ

Taking the base 2 logarithm of both sides of the last expression gives:

log₂ 2^N ≥ log₂ bⁿ
⇒ N ≥ log₂ bⁿ
⇒ N ≥ log bⁿ / log 2

Since we want the smallest integer N that satisfies the last relation, to find N, find log bⁿ / log 2 and take the ceiling.

In the last expression, any base is fine for the logarithms, so long as both bases are the same. It is convenient here, since we are interested in the case where b = 10, to use base 10 logarithms taking advantage of log₁₀10ⁿ == n.

For n = 3:

N = ⌈3 / log₁₀ 2⌉ = 10

For n = 4:

N = ⌈4 / log₁₀ 2⌉ = 14

For n = 6:

N = ⌈6 / log₁₀ 2⌉ = 20

And in general, for n decimal digits:

N = ⌈n / log₁₀ 2⌉