1. Find the code

hi
In a library,books are being coded with this method AAA, AAB,...,AAZ, ABA, ABB, ..., ABZ, ACA, ...
What is the code of the 8273(th) book ?
i followed some hints,and i found the code $MGE$ (Euclid's division)
is it correct ?
is there any other way to solve this problem ? because i don't understand how the Euclid's division helps find the solution
thanks a lot.

2. Hello, Raoh!

In a library, books are being coded with this method:
. . $AAA, AAB \hdots AAZ,\;ABA, ABB \hdots ABZ,\;ACA,ACB \hdots ACZ,\;\hdots ZZZ$

What is the code of the $8273^{rd}$ book?

i followed some hints,and i found the code $MGE$ (Euclid's division)
Is it correct? . . I got ${\color{blue}MFE}$

Is there any other way to solve this problem?
Because i don't understand how the Euclid's division helps find the solution.

We are working with three-digit "numbers", written in base-26,
. . using the "digits": . $A,B,C, \hdots Z$

With the Euclidean Algorithm, we have: . $8273 \;=\;(12)(6)(5)_{26} \;=\;MFE_{26}$

3. thanks a lot ( $3273^{rd}$ )

4. Originally Posted by Soroban
Hello, Raoh!

We are working with three-digit "numbers", written in base-26,
. . using the "digits": . $A,B,C, \hdots Z$

With the Euclidean Algorithm, we have: . $8273 \;=\;(12)(6)(5)_{26} \;=\;MFE_{26}$
In base 26 notation it appears that
A=0, B=1, C=2, D=3, E=4, F=5, G=6, ... ,M=12

IF AAA identifies the first book:
$A \times 26^2 + A \times 26^1 + A \times 26^0$ = 0
The first book has serial number equivalent to zero.

The 8273rd book should therefore should have a serial number equivalent to 8272.

$12 \times 26^2 + 6 \times 26^1 + 4 \times 26^0$ = 8272

12 = M
6 = G
4 = E

If I were doing the coding,
the 8273rd book would have code: MGE

5. thanks a lot,i was searching for an explanation like that,very well indeed