# Converting Base 10 numbers to an Alphabetical Number System(Similar to the excel column labelling system)

I am interested in converting a base 10 number into an Alphabetical Number System (like the one used to label columns in excel.) For example,

$55 = BC$ in this system because $2*26^1 + 3*26^0 = 55$.

(The 2 and 3 shown above are because B and C are the second and third letter in the alphabet, respectively.)

Is there some sort of formula which can be used to derive the correct numbers correlating to the Alphabet?

EDIT: I am trying to go from 55 -> BC

Other Examples:

$AAA = 703 = 1*26^2 + 1*26^1 + 1*26^0$

$ZZ = 702 = 26*26^1 + 26*26^0 = 27 * 26$

$AA = 27 = 1*26^1 + 1*26^0$

• It seems you already have all you need using this base-$26$ formulas. What do you want exactly, find the letters or find the numbers before $26^k$? – zwim Dec 30 '17 at 1:12
• @zwim I am trying to go from 55 -> BC – Arvind Ganesh Dec 30 '17 at 1:24

To express a number $n$ in base $b$ ($26$ in your example), you do division with remainder. Write $n=qb+r$ with $0 \le r \lt b$. The units digit is $r$. Now do the same with $q$ and the remainder is the next digit. Keep going until you don't get a quotient. This is the standard approach when you allow $0$ digits and not $b$ digits. In your system you do not allow $0$ and do allow $26$, so the condition on $r$ should be $0 \lt r \le 26$
• It's not quite that because his system doesn't have a $0$ digit, but does have a $26$ digit, so the condition should be $0<r\le b$. – Henning Makholm Dec 30 '17 at 1:20
• In fact you should better keep the $q,r$ given by Euclidean division. In ASCII code, the letters are consecutive and you are assured that $\operatorname{chr}(\operatorname{ord}('A')+i)$ with $0\le i\le 25$ represents all the letters, thus making math basis compatible with alphabetic system. I'm pretty sure that keeping $1,...,26$ will call for extra unnecessary conversions. – zwim Dec 30 '17 at 1:56