Suppose that x,y,z are positive integers satisfying x ≤ y ≤ z, and such that the product of all three numbers is twice their sum. What is the sum of all possible values of z?

xyz =2[x+y+z], solve for x

x = (2 (y + z))/(y z - 2) and yz cannot=2

Therefore x cannot be > 1, because it will violate x ≤ y ≤ z

I can only see 2 solutions as follows:

x = 1, y = 3 z = 8, and

x = 1, y = 4, z = 5

Sum of z =5 + 8 = 13

I will admit that I am tired but you lost me on the third line.

It's 17

you frogot 2,2,4