Representation of base M digits in binary digits

Question

In my textbook I came across the following sentence-

"If $M = 2^{m}$, with $m$ is a positive integer, then each base $M$ digit can be represented by a sub-sequence of $m$ binary digits"

Why is this true?

Let's say we are in base $M=16$. Then $M=2^{4}$ , i.e. $m=4$. What I'm guessing the above is saying is that if I have some base 16 digits, e.g. $1, 16, 256, 4096$ then all of these digits can be represented by 4 binary digits each? Personally, I can't see how this works. Clearly $256$ cannot be represented by 4 binary digits only.

Thanks for the help!

Answer 1

The numbers $1,16,256$ and $4096$ are not base $16$ digits, but rather they are powers of $16$, i.e. $16^0,16^1,16^2,16^3$. We need instead to consider digits.

Taking base $M=16=2^4$, then the digits are $0,1,2,\dots,E,F$, and these can be represented by equivalent binary strings $0000,0001,0010,\dots,1110,1111$, each of which contains $4$ binary digits.

So, given a base $16$ number, say $E2F_{16}$, we can immediately convert this to binary as $1110\,0010\,1111_2$.

As claimed, a similar conversion can be made in any base $M=2^m$ using $m$ binary digits for each base $M$ digit.