Solutions to the cipher puzzles - Part 2

in #newbiegames6 years ago (edited)

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This article lists the correct answers to my cipher puzzle contest.

I will update this article when each puzzle is solved - unless the 7 days editing limit goes by - then I will make a new post.

Puzzle 1 - solved by @markgritter

"756e7769656c6479"

The numbers are ASCII characters in hexadecimal numbers. You can solve the puzzle by hand, using a hexadecimal to ASCII chart or with a hexadecimal to ASCII converter.

Below is the solution and the cipher in python:

print(bytes.fromhex('756e7769656c6479').decode('ascii'))
print(binascii.b2a_hex(b'unwieldy'))

Puzzle 2 - solved by @nathanielcwm

"113 117 105 120 111 116 105 99"

The numbers are ASCII characters in decimals. You can solve the puzzle by hand, using a decimal to ASCII chart or with a decimal to ASCII converter.

Below is the solution and the cipher in python:

print(''.join(chr(int(c)) for c in "113 117 105 120 111 116 105 99".split(' ')))
print(' '.join(str(ord(c)) for c in 'quixotic'))

Puzzle 3 - solved by @nathanielcwm

"ZmFsbGFjaW91cw=="

This cipher is in base64 and can be solved with a base64 decoder.

The puzzle and it's solution in python:

print(base64.b64decode('ZmFsbGFjaW91cw=='))
print(base64.b64encode(b'fallacious'))

Puzzle 4 - solved by @markgritter

"65736F69646E617267"

This cipher is ASCII characters in hexadecimal like puzzle 1, but the word is spelled backwards: "esoidnarg". So the correct answer is "grandiose".

The puzzle and it's solution in python:

print(bytes.fromhex('65736F69646E617267')[::-1])
print(binascii.b2a_hex(b'grandiose'[::-1]))

Puzzle 5 - solved by @markgritter

"MzAgMzEgMzEgMzEgMzAgMzAgMzEgMzEgMjAgMzAgMzEgMzEgMzAgMzAgMzAgMzAgMzEgMjAgMzAgMzEgMzEgMzEgMzAgMzEgMzEgMzAgMjAgMzAgMzEgMzEgMzAgMzEgMzEgMzEgMzEgMjAgMzAgMzEgMzEgMzEgMzAgMzAgMzEgMzAgMjAgMzAgMzEgMzEgMzEgMzEgMzAgMzAgMzE="

This cipher was encoded three times and could be solved for example with these decoder tools. You needed to decode base64, then hexadecimal to ASCII and then binary to ASCII.

The answer is "savory".

Puzzle 6 - solved by @markgritter

"c8cae8cae4dad2dccac8"

This cipher is ASCII codes multiplied by 2 (or bit-shifted by 1) and given in hexadecimal.

The puzzle and it's solution in python:

print(''.join(chr(int(b/2)) for b in bytes.fromhex('c8cae8cae4dad2dccac8')))
print(''.join('{:02x}'.format(ord(c)*2) for c in "determined"))

Congratulations to the solvers!

Puzzles 7 and 8 are still unsolved - Go and solve them to earn a little STEEM!