How to determine the number of digits needed to represent a number in different bases?

Question

I am not a mathematician, but I do some computer programming and I am trying to find a solution to a fairly simple problem.

Is there any known formula/equation/function for mathematically determining the number of display digits (aka "places") that will be required when converting one number between two different number bases?

Example: Let us use the number 65535 (base 10) set as the variable N.

N base 10 requires 5 decimal digits (N10 = 65535)
N base 2 requires 16 binary digits (N02 = 1111111111111111)
N base 16 requires 4 hexadecimal digits (N16 = FFFF)

I want to feed N base 10 (N10) and another base into a formula or subroutine of some sort which returns a single number. Using the above example it might look something like this:

function(65535,2) returns 16
function(65535,16) returns 4

I want to explore some fairly large numbers in several different number bases and I am hoping this can be done as pure math, even if it requires multiple steps, rather than some sort of indexed table that would have to be created for each base.

Example: function(1222333444555,99) = ???

PS: Feel free to add any other appropriate tags since I am not sure what else would be right for this question.

Antoine · Accepted Answer · 2021-09-29T07:31:09.430

13

Yes, there is a way: $$f(n, \text{base}) = \log_{\text{base}} (n) + 1\text{.}$$

After you compute the value of $f(n,b)$, you have to use floor function to get the right integer value.

Explanation:

for numbers $b^k$, you need $k + 1$ places (digit $1$, followed by $k$ digits $0$)
for numbers $b^k -1$, you need $k$ places ($k$ digits $(b - 1)$, e.g., $99999$ for $b = 10$)
for numbers $n$ and $m$, such that $n\geq m$, then the number of places for $n$ is greater or equal to the number of places for $m$.

Hence, for all $b^{k - 1} \leq n \leq b^{k}-1$, you need $k$ places. The formula is just a fancy way to say this.

edited Sep 29 '21 at 07:31

answered May 04 '16 at 15:43

Antoine

3,439

DOH! I've been away from basic algebra too long! Put in the exponential form the equation is base^X = n, right ? – O.M.Y. May 04 '16 at 15:47
Ok I am a little confused by the explanation. You say "if n >= m" but do not explain what m is and never use it elsewhere. The rest I understand fine. – O.M.Y. May 05 '16 at 11:51
@O.M.Y. It should be clear now :) – Antoine May 05 '16 at 13:22
Its is, Thanks Antoine. – O.M.Y. May 05 '16 at 14:18

Bernard · Answer 2 · 2016-05-04T16:56:09.787

0

Explaining without logarithms, if $b^r\le n < b^{r+1}$, $n $ requires exactly $r$ digits in base $b$.

edited May 04 '16 at 16:56

answered May 04 '16 at 15:50

Bernard

175,478

@Antoine: Oops! Thanks for pointing the typo! – Bernard May 04 '16 at 16:56
If $b^r\le n < b^{r+1}$ then $n $ requires exactly $r+1$ digits in base $b$. – Christian Blatter May 05 '16 at 14:17

How to determine the number of digits needed to represent a number in different bases?

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