# Cryptography

## Tools Used
- RapidTables
    - https://www.rapidtables.com/web/tools/index.html
        - Base64
        - Binary
        - Hex String
- Cryptii
    - https://cryptii.com/
    - Decode
- Decode
    - https://www.dcode.fr/
    - RSA Cipher
- Kali Linux
    - Strings

## Challange 1
Using rapidtables I was able to decode the encoded strings. The first string was a hexadecimal string. I ommited the 0x and pasted the remaining string into the deocoder.
`0x73636f7270696f6e` `scorpion`
The decoded string is **scorpion**.

I then proceeded to the next string. The string is a base64 string. I then pased that into the decoder. 
`c2NyaWJibGU` `scribble`
The decoded string is **scribble**.

The next string was a binary string. 
`01110011 01100101 01100011 01110101 01110010 01100101 01101100 01111001` `securely`
The decoded string is **securely**.

The last string was a double encoded string. Original word was first encoded into a base64 and then that base64 was encoded into a binary string. I first decoded the binary to reveal the base64 and then decoded the base64 to reveal the original string. 
`01100010 01000111 00111001 01110011 01100010 01000111 01101100 01110111 01100010 00110011 01000001 00111101` `bG9sbGlwb3A=`
`lollipop`
The decoded string is **lollipop**.


## Challange 2
Using crpytii I took the encrpyred string and decoded it using a ceaser cipher with a shift of 13. I used the Large Language Model Gemma to aid in my decription. The model was able to tell me the most logical cipher based on the encoding.
`iveghny ynxr` `virtual lake`
The decoded string is **virtual lake**.

## Challange 3
Using cryptii and gima I was able to determine that the cipher is a alphabetical substituion using the alphabet in reverse. 
`hzuvob lyerlfh xzev` `safely obvious cave`
The decoded string is **safely obvious cave**.


## Challange 4
Based on the ... and the -- I was able to determine this was morse code. I used crpytii to convert the string. 
`.... . / ... . -.-. .-. . - / --- ..-. / --. . - - .. -. --. / .- .... . .- -.. / .. ... / --. . - - .. -. --. / ... - .- .-. - . -.. / ... -.- -.-- / -.. -.- ...- -... / ----. ---.. .---- -....`  `the secret of getting ahead is getting started sky dkvb 9816`
The decoded string is **the secret of getting ahead is getter strated sky dkvb 9816**.

## Challange 5
The notes for this challange is that there were handwritten notes with 3 & 5. Using this information I looked for ciphers on cryptii where the numbers could be insterted and keys. I was able to figure out this was a Rail Cipher. 3 correlated to the first message and 5 the second. 
`Cair eruSA-0org sgaeudrpesr K-II98.ue cn seYQ3` `Courage is grace under pressure SKY-AIQI-9380.`
The deconded first string is **Courage is grace under pressure SKY-AIQI-9380.**
`F daS-eefn  n KZ3eheadty.YI8lta oiwy-Q0. r aI2` `Feel the fear and do it anyway. SKY-IQIZ-3802.`
The decoded second string is **Feel the fear and do it anyway. SKY-IQIZ-3802.**

## Challange 6
The notes for this challange was that the keyword is **qizkwcgqbs**. The let me to a vigenere cipher because it uses a key. Using cryptii and the keyword I was able to decipher the phrase. 
`Y ln xkv lubj swlzqvkht, A vmzb pjk bbua we ddgs ILQ-GQYU-8026` `I do not fear computers, I fear the lack of them SKY-QIZK-8026`
The decrypted string is **I do not fear computers, I fear the lack of them SKY-QIZK-8026**.

## Challange 7
For this challange I used Kali Linux. The provided file had a string encoded into it. I first tried looking through the metadata and through steghide. The strign was hidden in the strings metadata of the file. Using the following command `strings Step1.jpg | grep SKY` I was able to pull tjust the string I needed out of the file. `SKY-TVJI-2063`
The decoded string is **SKY-TVJI-2063**.

## Challage 8
This challange is decoding a string using RSA. I used dcode.fr and their RSA decrypter to decipher the text. The provided values were N, C, and E. I was then able to use that to pull P adn Q from dcode. Then With that I was able to decode the string. C is the encoded string.
`n = 1079`
`e = 43`
`c = 996 894 379 631 894 82 379 852 631 677 677 194 893`
`p = 13`
`q = 83`
`SKY-KRYG-5530`
The decoded string is **SKY-KRYG-5530**.