Three primes $p,q$, and $r$ satisfy $p+q = r$ and $1 < p < q$. Then $p$ equals
$\textbf{(A)}\ 2\qquad \textbf{(B)}\ 3\qquad \textbf{(C)}\ 7\qquad \textbf{(D)}\ 13\qquad \textbf{(E)}\ 17$
We are given that $p,q$ and $r$ are primes. In order for $p$ and $q$ to sum to another prime, either $p$ or $q$ has to be even, because the sum of two odd numbers would be even, and the only even prime is $2$ (but $p + q = 2$ would have, as the only solution in positive integers, $p = q = 1$, and $1$ is not prime). Thus, with one of either $p$ or $q$ being even, either $p$ or $q$ must be $2$, and as $p < q$, we deduce $p = 2$ (as $2$ is the smallest prime). This means the answer is $\boxed{\textbf{(A)}\ 2}$.