Five cards are lying on a table as shown.
\[\begin{matrix} & \qquad & \boxed{\tt{P}} & \qquad & \boxed{\tt{Q}} \\  \\ \boxed{\tt{3}} & \qquad & \boxed{\tt{4}} & \qquad & \boxed{\tt{6}} \end{matrix}\]
Each card has a letter on one side and a [whole number](https://artofproblemsolving.com/wiki/index.php/Whole_number) on the other side.  Jane said, "If a vowel is on one side of any card, then an [even number](https://artofproblemsolving.com/wiki/index.php/Even_number) is on the other side."  Mary showed Jane was wrong by turning over one card.  Which card did Mary turn over? (Each card number is the one with the number on it. For example card 4 is the one with 4 on it, not the fourth card from the left/right)
$\text{(A)}\ 3 \qquad \text{(B)}\ 4 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ \text{P} \qquad \text{(E)}\ \text{Q}$
Logically, Jane's statement is equivalent to its [contrapositive](https://artofproblemsolving.com/wiki/index.php/Contrapositive), 

If an even number is not on one side of any card, then a vowel is not on the other side.

For Mary to show Jane wrong, she must find a card with an [odd number](https://artofproblemsolving.com/wiki/index.php/Odd_number) on one side, and a vowel on the other side.  The only card that could possibly have this property is the card with $3$, which is answer choice $\boxed{\textbf{(A)}}$