cybersecurity basics

tryhackme capture the flag machine!!

T R A N S L A T I O N S A N D S H I F T I N G

Question 1: c4n y0u c4p7u23 7h3 f149?

13375P34K IS A SYSTEM OF MODIFIED SPELLINGS , WHERE SOME LETTERS ARE REPLACED BY NUMBERS AND OTHER SPELLINGS.

Looking at the question, the first word is clearly “can” and “a” is encrypted as “4” and then “o” as “0” therefore randomly some letters are replaced by numbers and some other alphabets.

QUESTION 2: 01101100 01100101 01110100 01110011 00100000 01110100 01110010 01111001 00100000 01110011 01101111 01101101 01100101 00100000 01100010 01101001 01101110 01100001 01110010 01111001 00100000 01101111 01110101 01110100 00100001

Series of 1’s and 0’s clearly indicates simple binary to text convertor. We can easily do this on online encryption tool called CYBERCHEF.

Question 3: MJQXGZJTGIQGS4ZAON2XAZLSEBRW63LNN5XCA2LOEBBVIRRHOM======

It is base 32 and base 64 that have == in last and it is base 32 that uses all the capital letters therefore it is base

32. base 32 is a number system that is made up of the 26 letters of the English alphabet (A-Z) AS WELL AS the numbers 2–7, allowing for a total of 32 usable characters in each position.

Question 4: RWFjaCBCYXNlNjQgZGlnaXQgcmVwcmVzZW50cyBleGFjdGx5IDYgYml0cyBvZiBkYXRhLg==

This has to be base 64 since == are there in last and it includes both capital and small letters.

NOTE- base 32 and base 64 are online tools for this type of decryption.

Question 5: 68 65 78 61 64 65 63 69 6d 61 6c 20 6f 72 20 62 61 73 65 31 36 3f

This can be hexadecimcal also called base 16 because of the fact that each character is always 1 byte, or 8 bits long, helpfully each hexadecimal number is 4 bits, meaning 2 hex digits can ALWAYS represent a character in ASCII.The way we can tell it’s in hex is the spacing between each couplet and the character set, that is 0–9 then a-f, adding to 16.

Question 6: Ebgngr zr 13 cynprf!

ROT13 is an extremely simple cipher where each letter in a phrase is “rotated” 13 times, so that a letter becomes the letter 13 places after it.

Question 7: *@F DA:? >6 C:89E C@F?5 323J C:89E C@F?5 Wcf E:>6DX

This can be Rot47 cipher because it uses all characters in the ASCII encoding table, A-Z, 0–9, punctuation, and stuff and it’s the same basic principle as the Rot13 but instead of using just the letters A-Z.

Question 8: — . .-.. . -.-. — — — — ..- -. .. -.-. .- — .. — — -. . -. -.-. — — -.. .. -. — .

It has to be Morse code bacuse of Dots and dashes.

Question 9: 85 110 112 97 99 107 32 116 104 105 115 32 66 67 68

In this I can spot only numbers which cleraly indicates it to be a base 10 case. where each number represents its value in binary, which is then converted to ASCII and spat out as a series of letters and numbers.

Question 10: LS0tLS0gLi0tLS0gLi0tLS0gLS0tLS0gLS0tLS0gLi0tLS0gLi0tLS0gLS0tLS0KLS0t……….

And man! This was something really amazing. By looking at small and capital letters mixed we can clearly see that this has to be base 64. Converting it to base 64 gives us

— — — . — — . — — — — — — — — . — — . — — — — -

— -……

Dashes and dots reminds us of morse, right! So lets try that. Decrypting with morse gives us

01100110 01100101 00100000 01100000 01011111 01100000 00100000 01100000 01100000 01100101 00100000 01100010…

This surely has to be dycrpted to by binary decryption which further gives us fe `_` ``e b… ascii codes indicates it to be rot 47. Which further gives us 76 101 116….

Decoding this decimal finally gives us

“Let’s make this a bit trickier…”

Task 2: spectrograms

A visual representation of the spectrum of frequencies of a signal as it varies with time. When applied to an audio signal, spectrograms are sometimes called sonographs, voiceprints, or voicegrams. When the data is represented in a 3D plot they may be called waterfalls.

In this I got a audio file and heard it and it was super funny, then looking at the hint , it said ”audacity”.

$ sudo apt-get install audacity

Once installed, I open the file given called secretaudio.wav and throwing my hands here and there I change the view from waveform to spectrogram in the menu with the downwards facing black triangle. And this what it showed then:

This is what will be displayed after following above steps.

Task 3: Steganography

Steganography is “The process of hiding a message or file within another message or file”. In this we got an image file:

putting this image on steganography decoder gives us:

Gives us: SpaghettiSteg

Task 4: Security through obscurity

Security through obscurity is the reliance in security engineering on the secrecy of the design or implementation as the main method of providing security for a system or component of a system. In this we are given a file, downloading and open it leads us to the flags simply.

The file name and the message are the flags.

— — — — — — — — -THANK YOU — — — — — — — — — — —

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Radhika Agarwal

Today, the enemy is everywhere and is complacency so I want to broaden my perspective at very young age so that I do not fear the results of 100 battles