Mix
Description
At the BASEment no. 65536, A man is irritated with SHIFT key in his KEYBOARD as it's a sticky key, A kid is having chocolate icecream with a SPOON.
Author : Whit3_D3vi1
Challenge Files
flag.txt reads
If you opened this then you are a n00b
HAHA VERY FUNNY
chall_encrypted.txt contains some seemingly chinese.
since BASE65536 is a BIG hint to base 65536, We do stuff.
import base65536
with open('chall_encrypted.txt','r') as chall_enc:
data = chall_enc.read()
data1 = "".join( chr(ord(i)%256)+chr(ord(i)//256) for i in data )
d = base65536.decode(data)
print(d.decode())
Which prints out some neat stuff
flag 1:
Yjod od s lrunpstf djogy vo[jrtyrcy jrtr od upi g;sh xj4t-}U-i+dit4+
flag 2:
3030313130313030203031313130303130203030313130303131203031303131313131203030313130313030203031313130313131203030313130303131203031313130303131203030313130303030203031313031313031203030313130303131203031303131313131203031313130313131203031313031303031203030313130313131203031313031303030
flag 3:
MTExMTExMTExMTAwMTAwMDEwMTAxMDExMTAxMDExMTExMTEwMTAxMTExMTExMTExMDExMDExMDExMDExMDAwMDAxMTAxMDAxMDAxMDAxMDAwMDAwMDAwMDAwMDAwMDAwMTAxMDAxMTAwMDAwMDAwMDAwMDAwMDAwMDAwMDAwMDAwMDAwMDAwMDAwMDAwMDAwMDAwMDAwMDAwMDAwMDAwMTAxMDAwMDAwMDAwMDAwMTAxMDAwMTAxMDAxMDAwMTAxMDAxMTExMTExMTAwMTAxMDAxMDExMTExMTExMTAwMTAxMDAxMTAwMDAwMDAwMDAwMDAwMTAxMDAxMDAwMDAwMDAwMDAwMDAwMDAwMDAwMDAwMDAwMDAwMTAxMDExMTExMTExMTExMTExMTExMTExMDAxMDEwMDAwMDAwMDAwMDAwMDAwMDAwMDAwMDAwMDAxMDEwMDAwMDAwMDAxMDEwMDExMDAwMDAwMDAxMDEwMDAxMDEwMTExMTAwMTAxMDAxMDExMTExMTExMTExMTExMTExMTExMDAxMDEwMDExMDExMDExMTExMTExMTExMTAwMTAxMA==
Now we have 3 parts, lets work with hints as usual.
So the first part must be KEYBOARD SHIFT cipher
At this point, I am wondering why is dcode.fr S0 GOOD.
And we get the first part as This is a keyboard shift ciphertext here is you flag zh3r0{Y0u_sur3_
Now moving to second part, we dont even need a hint,
f2 = "3030313130313030203031313130303130203030313130303131203031303131313131203030313130313030203031313130313131203030313130303131203031313130303131203030313130303030203031313031313031203030313130303131203031303131313131203031313130313131203031313031303031203030313130313131203031313031303030"
print(bytes.fromhex(f2))
# b'00110100 01110010 00110011 01011111 00110100 01110111 00110011 01110011 00110000 01101101 00110011 01011111 01110111 01101001 00110111 01101000'
flag2 = "".join(chr(int(i,2)) for i in bytes.fromhex(f2).split())
print(flag2)
And just like that, we have 4r3_4w3s0m3_wi7h as second part of flag
Coming to third part, it seemingly is base64 encoding,
'MTExMTExMTExMTAwMTAwMDEwMTAxMDExMTAxMDExMTExMTEwMTAxMTExMTExMTExMDExMDExMDExMDExMDAwMDAxMTAxMDAxMDAxMDAxMDAwMDAwMDAwMDAwMDAwMDAwMTAxMDAxMTAwMDAwMDAwMDAwMDAwMDAwMDAwMDAwMDAwMDAwMDAwMDAwMDAwMDAwMDAwMDAwMDAwMDAwMDAwMTAxMDAwMDAwMDAwMDAwMTAxMDAwMTAxMDAxMDAwMTAxMDAxMTExMTExMTAwMTAxMDAxMDExMTExMTExMTAwMTAxMDAxMTAwMDAwMDAwMDAwMDAwMTAxMDAxMDAwMDAwMDAwMDAwMDAwMDAwMDAwMDAwMDAwMDAwMTAxMDExMTExMTExMTExMTExMTExMTExMDAxMDEwMDAwMDAwMDAwMDAwMDAwMDAwMDAwMDAwMDAxMDEwMDAwMDAwMDAxMDEwMDExMDAwMDAwMDAxMDEwMDAxMDEwMTExMTAwMTAxMDAxMDExMTExMTExMTExMTExMTExMTExMDAxMDEwMDExMDExMDExMTExMTExMTExMTAwMTAxMA=='
print(base64.b64decode(f3))
# b'1111111111001000101010111010111111101011111111110110110110110000011010010010010000000000000000001010011000000000000000000000000000000000000000000000000000000001010000000000001010001010010001010011111111001010010111111111001010011000000000000001010010000000000000000000000000000001010111111111111111111110010100000000000000000000000000010100000000010100110000000010100010101111001010010111111111111111111110010100110110111111111111001010'
Which turns out to be some sort of binary, lets take third hint SPOON to our help and hey, never to forget dcode.fr
We get the third part of flag _411_7h3_ski115}
zh3r0{Y0u_sur3_4r3_4w3s0m3_wi7h_411_7h3_ski115}
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