Find plaintext from two ciphers encrypted with a one-time pad and the same key I'm studying for a cryptography test and I came across this question: 


  Assume a language has three letters: A=1111, B=0011 and C=1010. Two words in this language are encrypted with a one-time pad and by using the same binary key sequence. The first four bits for both corresponding cipher texts are, respectively: $C_1= 1110$ and $C_2=1000$. Determine the first letter of both corresponding plain texts as well as the first 4 bits of the key.


Isn't this question impossible to solve without additional information? Because:


  $C_1 = L_1 + K + P_1$
  
  $C_2 = L_2 + K + P_2$
  
  Where $C_i$ is the cipher text, $L_i$ is the encrypted letter, $P_i$ is the one time pad and $K$ is the key.


Seeing how $L_1, L_2, P_1, P_2$ and $K$ are unknown, it's not possible to solve this, right? Or am I missing something?
First, there seems to be a misunderstanding: The one-time pad is the key. So your equations simplify to
$$ C_1 = L_1 + K \text, $$ 
$$ C_2 = L_2 + K \text. $$

At this point, you would be right if the $L_i$ could contain arbitrary bit sequences. However, the question imposes a constraint: $L_i\in\{A=\mathtt{1111},B=\mathtt{0011},C=\mathtt{1010}\}$.

Let's look only at the first bit of all the variables (I'll denote these in lower-case). Since $c_1=c_2$ by observation, you know that $l_1=l_2$ from the equations above. Hence given the alphabet from the question, we know that either both $L_1$ and $L_2$ are $B=\mathtt{0011}$, or they are both not $B$. You can proceed similarly for the next bits, and at some point you should be able to reduce the set of possibilities to one unique solution.