Both roots of the quadratic equation $x^2 - 63x + k = 0$ are prime numbers. The number of possible values of $k$ is 
$\text{(A)}\ 0 \qquad \text{(B)}\ 1 \qquad \text{(C)}\ 2 \qquad \text{(D)}\ 4 \qquad \text{(E) more than 4}$
Consider a general quadratic with the coefficient of $x^2$ being $1$ and the roots being $r$ and $s$. It can be factored as $(x-r)(x-s)$ which is just $x^2-(r+s)x+rs$. Thus, the sum of the roots is the negative of the coefficient of $x$ and the product is the constant term. (In general, this leads to [Vieta's Formulas](https://artofproblemsolving.com/wiki/index.php/Vieta%27s_Formulas)).
We now have that the sum of the two roots is $63$ while the product is $k$. Since both roots are primes, one must be $2$, otherwise, the sum would be even. That means the other root is $61$ and the product must be $122$. Hence, our answer is $\boxed{\text{(B)}\ 1 }$.