Product makes Sum Parity Undecidable

For this situation to happen, we will first require that the product of the two remaining numbers be a value that can be obtained using at least two distinct pairs of numbers from the original set. This gives us two options, namely (i), $(1,6), (2,3)$ , and (ii) $(3,4), (2,6)$ . Now for Claudia to be unable to tell whether the sum of Zach's 5 numbers is even or odd, the pairs noted in each of the aforementioned options would have to be such that the sum of the elements of one pair is even and the sum of the elements of the other pair is odd. This is only the case in option (ii), as $3 + 4 = 7$ is odd and $2 + 6 = 8$ is even. Thus the product of Zach's 5 numbers must be $7! / 12 = \boxed{420}$ .