Winning Number

• Only one of statements a) and b) can be true, and at most one of c) and d) can be true.
• Then (since there are four true statements) it follows that e) and f) must be true, and that exactly one of c) and d) is true.
• From e) and f) are choices are limited to 10, 11 and 12; and only 11 satisfies one of c) and d).

$\text{possible elements for (It was even)={2,4,6,8,10,12,14,16}}$

$\text{possible elements for (It was odd)={1,3,5,7,9,11,13,15,}}$ $\implies$ $\boxed{11}$

$\text{possible elements for (It was prime)={2,3,5,7,11,13}}$ $\implies$ $\boxed{11}$

$\text{possible elements for (It was a square number)={4,9,16}}$

$\text{possible elements for (It had two digits)={10,11,12,13,14,15,16}}$ $\implies$ $\boxed{11}$

$\text{possible elements for (It was between 6 and 12)={7,8,9,10,11}}$ $\implies$ $\boxed{11}$

Hence the answer is $\color{#20A900}{\boxed{11}}$

Even <==> Odd
Prime <==> Perfect square

Since a) & b) are mutually exclusive and c) & d) are also mutually exclusive, we can only choose one out of each pair, and that means that e) & f) are both the truth that describes the winning number. Two digit numbers within the limits of 6 ≤ x ≤ 12 is just 10, 11 and 12. Because none of these are perfect squares, our number must be a prime.

Answer = 11, a two-digit odd prime number between 6 and 12.