Maximum number of prime sums Zach thinks of six different positive integers, and for each pair of numbers, he adds them. This gives him $\binom{6}{2}=15$ sums. Among these sums, find the maximum number that can be prime numbers. I deducted that there has to be atleast one even number among the $6$ numbers. But will there be one even number, or two evens. Then what will the other numbers be.

The sum of two different positive integers is always at least $3$, and so in order for it to be prime, it has to be odd. So one summand has to be even, the other odd. If there are $a$ even numbers and $b = 6-a$ odd numbers, then the potential number of primes is at most $ab \leq 9$, attained when $a=b=3$. Can this bound be attained? Here is an example:
$$
2,3,4,8,9,15.
$$
All of the following are primes:
$$
2+3 = 5 \\
3+4 = 7 \\
2+9 = 3 + 8 = 11 \\
4+9 = 13 \\
2+15 = 8+9 = 17 \\
4+15 = 19 \\
8+15 = 23
$$
If you don't like the repeated sums, you can try instead:
$$
2,3,4,8,39,99.
$$
All of the following are prime:
$$
2+3 = 5 \\
3+4 = 7 \\
3+8 = 11 \\
2+39 = 41 \\
4+39 = 43 \\
8+39 = 47 \\
2+99 = 101 \\
4+99 = 103 \\
8+99 = 107
$$
