# Find the code

• Jun 12th 2009, 05:19 AM
Raoh
Find the code
hi(Happy)
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$\displaystyle MGE$ (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(Wondering)
thanks a lot.
• Jun 12th 2009, 05:46 AM
Soroban
Hello, Raoh!

Quote:

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

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

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

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": .$\displaystyle A,B,C, \hdots Z$

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

• Jun 12th 2009, 06:01 AM
Raoh
thanks a lot(Happy) ($\displaystyle 3273^{rd}$ (Wink))
• Jun 13th 2009, 04:44 AM
aidan
Quote:

Originally Posted by Soroban
Hello, Raoh!

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

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

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

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

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

12 = M
6 = G
4 = E

If I were doing the coding,
the 8273rd book would have code: MGE
• Jun 13th 2009, 05:32 AM
Raoh
thanks a lot(Happy),i was searching for an explanation like that,very well indeed(Wink)