If I have a 4 digit binary number, how come there are 16 possible "numbers", and similarly, if I have a 4 digit binary number, how come there are 8 possible "numbers?"
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$\begingroup$Still not positive what exactly you're asking, but there are $8$ variations of $3$ digit binary numbers because $2^3 = 8$, and $16$ variations of $4$ digit binary numbers because $2^4 = 16$. To calculate the highest possible number (the amount of variations) in any length binary number, you use $2^n$, $n$ being the length. This applies to base $k$, to calculate the variations or highest number possible in any base, you use $k^n$.
Note that using this formula, your answer will always be in base $10$.
A 5 digit binary number can count to $32$ because $2^5 = 32$
A 5 digit hexadecimal number can count to $1,048,576$ because $16^5 = 1048576$
This works because in base $k$, adding a digit mutliplies the possible numbers by $k$. take Base $2$ as an example, in base $2$, take the number $1001$. If I add 1 digit to the begenning (or end, or anywhere for that matter), there are $2\times$ the number of solutions as before, the number can nowbe $01001$, or $11001$.
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