Question

How many different values can a 16-bit binary number store? How many digits are needed fora hexadecimal number to do the same? How about a base-5 number? Show your work.

Answer 1

A very simple way to know the number of possible combinations in a binary number is given by the next formula:

`2^n=2^(16)=65536^{}`

Now, we should remember that an hexadecimal number contain 16 different digits, from 0 to F. Therefore

`\begin{gathered} 16^n=65536 \\ n\ln 16=\ln 65536 \\ n=(\ln 65536)/(\ln 16) \\ n=4 \\ 16^4=65536 \end{gathered}`

Thus, we will need 4 hexadecimal numbers to achieve the same values as a 16-bit number.

For the base-5 number we can do a similar procedure:

`\begin{gathered} 5^m=65536 \\ m\ln 5=\ln 65536 \\ m=(\ln 65536)/(\ln 5) \\ m=6.89\approx7 \\ 5^7^{}=78125 \end{gathered}`

Therefore, we would need at least 7 numbers base-5 to achieve something similar to 16-Bit number