This time the flag is padded to a random length between 250 and 550 with random chars (both leading and trailing).
```php
//For these 3 problems I'll only change these 2 lines:
$flag = trim(file_get_contents("../../flag3.txt"));
$key = pad_string($flag, mt_rand(250,550));
```

Start with the same script as OTP2, we try to find the key length and position of the flag.
```python
for klen in range(250, 550):
find_kpos(klen)
```

Output:
```
354 245 LDBEAS FICEAN
355 245 LDBEAS FICEAN
356 245 LDBEAS FICEAN
...
547 245 LDBEAS FICEAN
548 245 LDBEAS FICEAN
549 245 LDBEAS FICEAN
```

Looks like the key length doesn't matter, we really have an one time pad now. Oh wait, the same key is used to encrypt 2 plaintexts, so two time pad actually. We also got 1 important piece of information: flag position is 245.

Now continue using the same strategy as OTP2
```python
klen = 549
kpos = 245
find_nextchar(key, kpos, klen)
```

Output:
```
[121, 122, 123, 124, 125, 126, 127, 112, 113, 114, 115, 116, 117, 118, 119, 104, 105, 106, 107, 108, 109, 110, 111, 96, 97, 98]
y LDBEASA FICEANJ
z LDBEASB FICEANI
{ LDBEASC FICEANH
| LDBEASD FICEANO
} LDBEASE FICEANN
~ LDBEASF FICEANM
 LDBEASG FICEANL
p LDBEASH FICEANC
q LDBEASI FICEANB
r LDBEASJ FICEANA
t LDBEASL FICEANG
u LDBEASM FICEANF
v LDBEASN FICEANE
w LDBEASO FICEAND
i LDBEASQ FICEANZ
j LDBEASR FICEANY
k LDBEASS FICEANX
` LDBEASX FICEANS
a LDBEASY FICEANR
b LDBEASZ FICEANQ
```

With some trials and errors, eventually we recover the flag: `UDCTF{wh3n_th3y_s4y_1_t1m3_th3y_mean_1t}`

Full script: [otp3_sol.py](https://github.com/CTF-STeam/ctf-writeups/blob/master/2021/BlueHensCTF/OTP3/otp3_sol.py)

Original writeup (https://github.com/CTF-STeam/ctf-writeups/tree/master/2021/BlueHensCTF#otp3-crypto---482-pts).