Logic - Prove the following Here's the Problem:

Which one of these is true?  
  A) All of the below 
  B) None of the below 
  C) Some of the above 
  D )None of the above 
  E )None of the above

My attempt:

  
*
  
*Suppose A is true, then B,C,D,E are true. But, D says A,B,C are false. So, A is false.
  
*Suppose B is true, then C,D,E are false. But, C says either A or B is true, which is true. So, B is false 
  
*Suppose C is true, then either A or B is true. But, we know A and B are false from the previous assumptions. So, C is false
  
  
*Suppose D is true, then A,B,C are false. <----ANS 
  
*Suppose E is true, then A,B,C,D are false. But, D is true. So, E is false
  

Is my answer correct? If so, how can I prove this using mathematics? 
 A: I can't see a problem with your reasoning, and assuming there isn't a problem I can't see, you just did prove it using mathematics. Specifically, you used a series of arguments by contradiction.
Edit: After reading your comment: I'm pretty sure there's no way to prove this directly, considering we don't have any actual information to show a statement is true without assuming one of them. Proofs by contradiction aren't as satisfying as direct proofs, but if it's all we have, it's what we use.
A: Here is a more formal way to handle this question. I'm adding this answer just for the fun of it-- your approach is fine.
But first, I see one issue with your attempt.  You correctly say "Suppose D is true, then A,B,C, are false."  And you've correctly established that A,B,C are false.  But that does not allow us to conclude that "D is true"!  In fact, we're only allowed to conclude that D is true, because of what you correctly say in a comment: "~D is logically equivalent to A,B,C are true."
Now for the more formal approach, where we will actually use the shape of the formulas to guide the order in which we use the given information.

$
\newcommand{\calc}{\begin{align} \quad &}
\newcommand{\calcop}[2]{\\ #1 \quad & \quad \unicode{x201c}\text{#2}\unicode{x201d} \\ \quad & }
\newcommand{\endcalc}{\end{align}}
\newcommand{\ref}[1]{\text{(#1)}}
\newcommand{\true}{\text{true}}
\newcommand{\false}{\text{false}}
$Translating the given problem into formal logic, letting $\;A\;$ stand for "A is true", etc., you are given that
\begin{align}
\tag a A &\;\equiv\; B \land C \land D \land E \\
\tag b B &\;\equiv\; \lnot C \land \lnot D \land \lnot E \\
\tag c C &\;\equiv\; A \lor B \\
\tag d D &\;\equiv\; \lnot A \land \lnot B \land \lnot C \\
\tag e E &\;\equiv\; \lnot A \land \lnot B \land \lnot C \land \lnot D \\
\end{align}
and you want to determine the truth values of $\;A,B,C,D,E\;$.
(Note: The original formulation for C could also be interpreted as "Some but not all of the above".  That could have led to $\;C \;\equiv\; A \not\equiv B\;$, but the end result would have been the same.)

Looking at the shape of these formulas, it is clear that we can substitute $\ref b$ in $\ref a$:
$$\calc
A
\calcop={by $\ref a$}
B \land C \land D \land E
\calcop={by $\ref b$}
\lnot C \land \lnot D \land \lnot E \land C \land D \land E
\calcop={simplify using contradition}
\false
\endcalc$$
In other words, we're proven $\;\lnot A\;$. Similarly, we can substitute $\ref d$ in $\ref e$, resulting in $\;\lnot E\;$.
Now we see a similarity between the right hand sides of $\ref c$ and $\ref d$:
$$\calc
D
\calcop={by $\ref d$}
\lnot A \land \lnot B \land \lnot C
\calcop={DeMorgan -- to prepare for the next step}
\lnot (A \lor B) \land \lnot C
\calcop={by $\ref c$}
\lnot C \land \lnot C
\calcop={simplify}
\lnot C
\endcalc$$
So $$\tag 1 D \;\equiv\; \lnot C$$  Substituting $\ref 1$ in $\ref b$ quickly (by contradiction) results in $\;\lnot B\;$. And substituting that together with our initial $\;\lnot A\;$ in $\ref c$ results in $\;\lnot C\;$. Finally, that with $\ref 1$ gives us $\;D\;$.
So our conclusion is that D is true, and all others false.
