Let's get funky

GET REKT FGT

def a(c): return "%02x" % ord(c)

for c in "GET REKT FGT": print c, "=", a(c)

G = 47
E = 45
T = 54
  = 20
R = 52
E = 45
K = 4b
T = 54
  = 20
F = 46
G = 47
T = 54

We have 4745542052454b5420464754, which expands to:

47 = 0100 0111
45 = 0100 0101
54 = 0101 0100
20 = 0010 0000
52 = 0101 0010
45 = 0100 0101
4b = 0100 1011
54 = 0101 0100
20 = 0010 0000
46 = 0100 0110
47 = 0100 0111
54 = 0101 0100


010001110100010101010100001000000101001001000101010010110101010000100000010001100100011101010100

Regrouped to groups of 5: (2^5 = 32) (tip: always start from the right and pad 0's to the left)

(0000)0
10001
11010
00101
01010
10000
10000
00101
00100
10001
01010
01011
01010
10000
10000
00100
01100
10001
11010
10100

Now we need to make an "extended hexadecimal" to compensate the extra digit.

00000 = 0
00001 = 1
00010 = 2
00011 = 3
00100 = 4
00101 = 5
00110 = 6
00111 = 7
01000 = 8
01001 = 9
01010 = A (10)
01011 = B (11)
01100 = C (12)
01101 = D (13)
01110 = E (14)
01111 = F (15)
10000 = G (16)
10001 = H (17)
10010 = I (18)
10011 = J (19)
10100 = K (20)
10101 = L (21)
10110 = M (22)
10111 = N (23)
11000 = O (24)
11001 = P (25)
11010 = Q (26)
11011 = R (27)
11100 = S (28)
11101 = T (29)
11110 = U (30)
11111 = V (31)


00000 = 0
10001 = H
11010 = Q
00101 = 5
01010 = A
10000 = G
10000 = G
00101 = 5
00100 = 4
10001 = H
01010 = A
01011 = B
01010 = A
10000 = G
10000 = G
00100 = 4
01100 = C
10001 = H
11010 = Q
10100 = K


0HQ5AGG54HABAGG4CHQK (simplifies to HQ5AGG54HABAGG4CHQK)

Going in reverse, you could simply convert the number '0HQ5AGG54HABAGG4CHQK' to base 16 through arithmetic and then interpret that as ASCII, or you could do all of this in reverse, which is more realistic if you're doing it on paper. Of course, going from right-to-left, you'll end up with "100" on the far left, and you'll know to pad it to make it "0100".

Either way, if done correctly, you'll get "GET REKT FGT" back.

No, I did not expect anyone to get this. This is SUPPOSED to be extremely difficult. That's the point.