Factoring a number to get an encoded string Say we encode the string $BCA$ in the following way:
\begin{align*}BCA &= 2 \times 26^2 + 3 \times 26^1 + 1 \times 26^0 \\ &= 1456 \end{align*}
Is it possible to get back to the string $BCA$, given only the number $1456$?
 A: Yes. (But note that 1456 is not correct; $2\cdot26^2 + 3\cdot26 + 1 = 1431$, not $1456$.)
First, divide your number, 1431, by 26, giving  quotient, 55, and a remainder, 1.  The remainder of 1 means that the last letter is A.  
Now divide the quotient 55 by 26, giving a quotient of 2 and a remainder of 3.  The remainder, 3, says that the next-to-last letter is C.
Finally, divide the quotient 2 by 26, giving a quotient of 0 and a remainder of 2.  This says that the next previous letter is B.
Now the quotient is 0, so we are done.
A: Yes, it is possible as Mark has shown you one way to get the string digits from the least significant (that is getting $A$ first then $C$ and then $B$ finally).
But you have to be careful in using this method of encoding, because you are using base $26$, the number could be large if you are dealing with say strings of length $10$.
Another way to get back the strings is by calculating
$$  \left \lfloor log_{26} 1432 \right \rfloor  = 2 $$ 
This tells you that the highest power of $26$ is $2$. 
Start dividing by $26^2$ to get the most significant digit, i.e.
$$  \left \lfloor \frac{1431}{26^2} \right \rfloor = 2 \tag{1}$$
and from $(1)$ you know that most significant digit is $B$.  The remaining number is 
$$ 1431 - 2 \times 26^2 = 1431 - 1352 = 79$$
To find the next significant digit 
$$  \left \lfloor log_{26} 79 \right \rfloor  = 1 $$ 
and
$$ 79 = \bf{3} \times 26 + \bf{1}$$
Therefore you get the digits from the most significant to least as $B$, $C$ and $A$.  
