decrypted data

[Jumper]

Are you looking for:

A: A way to encrypt and decrypt your own data using PBP. If this is what you are looking for there are several ways to do it. TEA, XTEA and so on.. or a more simple XOR encryption

B: Get help to decrypt information from a different source than your own?


??

[RAYMON]

Hello

I think it's complicated, but you must know.
And you how you could have done to protect data of a eeprom 24lc512
Or perhaps the program that sends false with true and blurs the text.
thank you for your help

[RAYMON]

helo


Apparently in order to decode it must test all the values from 0 to 255 and
My case I'm only 10 digits? 0 to 9. As is always numeric characters
over a length of 40 characters, there will be several repetitions of figure?

If coding for each digit is a different value to not be able to compare
compared to its own value and its relative position in the text encrypt
so it will not have duplicates so it must be visible in the text by encrypting
report its presence (1 time only) or all the characters this one
both in the text after eliminating pressent characters several times, remains in the form code.

adqtuy / egil / pjkn / obfxzh
(1 time) / (2 times) / (3 times) / (4 5 8 9 times)

(A d t q u y) once in the text

9 9 9 9 8 8 8 8 7777 6666 = 5555 4444 3333 2222 1111
adqtuy - - - - - - - - - - -----------------------------------

Example: 1234567890123456 = 78900987654321987654
or 1111111111111111 = 11111111111111111111
or 1111777511456421 = 66880044411009922336


and the first digit has a different value of the second and so on. .... So the first 1 will be A, B will be 1 second and so on, more characters to scramble.

So with 10 digits and can not exceed 255 reduces its encryption.

Converted to decimal 0 = 48 and 30 in Hexa
255: 30 = 9 MAX 255: 48 MAX = 6

9 converted to decimal = 57 and 39 in Hexa
255: 39 = 7 MAX 255: 57 = 65 MAXI

So the division, subtraction, multiplication will not be used and still adding
that may serve, by inference we can easily find its encryption method.

Permuting rest but will have identical characters and easily detectable.
or the formula in exel and must be tested from 0 to 255.

EXEL SOUX = ((0 * C86/256)-T (0 * C86/256)) * 256

EXEL SOUX = ((255 * C86/256)-ENT (255 * C86/256)) * 256



Which are coded and which are repeated, have tested the formula in exel there are chances of
Find a match and that number is found the rest will be easy.

I clear the text and I encrypt the text, and my friend only encryption key but
do not know the plaintext

Its key it can decrypt and discover the plaintext, I guess it is not
also complicated, because the figures have the same value once encrypt

Example: A value will b 5, an R-value will be 0 ... etc ...

So they will not take the value of another figure, and will not be repeated
in the encrypt text, it will always be one (b for 5) and no other (b)?
significantly reduced its combinations.







On the forums:

Simply because this key will affect the interference characters by their ASCII value.

ASCII codes represent the letters A to Z (eg "A" is 65, "B" is 66, ...) and all the figures and the other symbols on your keyboard.

But the problem is that ASCII values range from 0 to 255.

The most important part of encryption is the way to blur the text.
ASCII does not constitute an encoding of secret is to encrypt the ASCII code
Encryption using the following:
C note on the encryption feature which, at any integer n in [0, 255] combines the remainder of the division by 256 of 7n. Let C (n) the rest. We use the formula "= MOD (7 * n, 256).