Where Is 1 From 2016?

We have that either one of the prime numbers are 2, or none of them are. If none of them are 2, we have that $p+q$ is even, so $p+q$ , $p(p+q)$ and $q(p+q)$ are even. The rest of the numbers are odd. But then this contradicts that we have exactly 4 even numbers. Thus, one of the primes is 2.

WLOG $q=2$ . Since we have unordered solutions, we can do this assumption.

Thus, $p$ is odd. From this, we get $q$ , $pq$ , $q(p+q)$ and $q^p$ are even, which satisfy. We must have 206 equal to one of $pq$ , $q(p+q)$ or $q^p$ . The last one can be ruled out since 206 isn't a power of 2. If $pq=206$ , we have $p=103$ , which satisfies. Thus, $|p-q|=101$ . If $q(p+q)=206$ , $p=101$ , which satisfies. Thus, $|p-q|=99$ . Therefore, the sum of all values is $\boxed{200}$ .

One of the eight given numbers is $206$ .

• $p$ or $q$ cannot be equal to 206, because it is not prime.

• $p+q$ cannot be equal to 206. If it were, both $p$ and $q$ would have to be odd, making $p, q, p^q, q^p$ , and $pq$ odd. But we are told that four of the numbers are even.

• $p^q, q^p$ cannot be equal to 206. It would require the exponent to be equal to 1, which is not prime.

• $pq = 206$ . This is possible with $p = 103$ and $q = 2$ (or v.v.). The even numbers in the list are $q, pq, q^p$ , and $(p+q)q$ . Thus all conditions are fulfilled, with $|p - q| = 101.$

• $(p+q)q = 206$ . This is possible with $p = 101$ and $q = 2$ . This situation is similar to the previous one. We find $|p - q| = 99.$ Swapping $p$ and $q$ does not result in any new solutions.

Thus the possible values of $|p - q|$ are $101$ and $99$ , with sum $\boxed{200}$ .