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, $2^9=512$, 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; $1000\times10=10^4$ 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.