Question

How many bits are required to represent the decimal numbers in the range from 0 to 999 in straight binary code?

Answer 1

Note that powers of 2 can be written in binary as


`2^0=1_2`

`2^1=10_2`

`2^2=100_2`

and so on. Observe that
`n+1` digits are required to represent the
`n`-th power of 2 in binary.

Also observe that


`\log_2(2^n)=n\log_22=n`

so we need only add 1 to the logarithm to find the number of binary digits needed to represent powers of 2. For any other number (non-power-of-2), we would need to round down the logarithm to the nearest integer, since for example,


`2_(10)=10_2\iff\log_2(2^1)=\log_22=1`

`3_(10)=11_2\iff\log_23=1+(\text{some number between 0 and 1})`

`4_(10)=100_2\iff\log_24=2`

That is, both 2 and 3 require only two binary digits, so we don't care about the decimal part of
`\log_23`. We only need the integer part,
`\lfloor\log_23\rfloor`, then we add 1.

Now,
`2^9=512<1024=2^(10)`, and 999 falls between these consecutive powers of 2. That means


`\log_2999=9+\text{(some number between 0 and 1})`

which means 999 requires
`\lfloor\log_2999\rfloor+1=9+1=10` binary digits.

Your question seems to ask how many binary digits in total you need to represent all of the numbers 0-999. That would depend on how you encode numbers that requires less than 10 digits, like 1. Do you simply write
`1_2`? Or do you pad this number with 0s to get 10 digits, i.e.
`0000000001_2`? In the latter case, the answer is obvious;
`1000*10=10^4` total binary digits are needed.

In the latter case, there's a bit more work involved, but really it's just a matter of finding how many number lie between successive powers of 2. For instance, 0 and 1 both require one digit, 2 and 3 require two, while 4-7 require three, while 8-15 require four, and so on.