If x+2yz= xyz, where x, y and z are positive integers, what is the val

x = xyz - 2yz

Since yz is a factor of the right side of this equation, it must be a factor of the left side, so x is divisible by yz.

2yz = xyz - x

Since x is a factor of the right side of this equation, it must be a factor of the left side, so 2yz is divisible by x.

Using Statement 1, if 2yz is divisible by x, and x is odd, then yz must be divisible by x (since the "2" in "2yz" can't possibly cancel with something in "x" if x is odd). So using Statement 1, we know yz is divisible by x, but also that x is divisible by yz. That can only happen if x and yz are the same number. So x = yz, and the equation becomes: x + 2x = x*x , so 3x = x^2 and x = 3. Since x = yz, we thus know yz = 3, and since y and z are positive integers, one of them is 3, the other is 1, and y + z = 4. So Statement 1 is sufficient.

Statement 2 is not sufficient, since we can have the values found above (x = 3, y = 3 and z = 1, say), or we can have x = 4, y = 1 and z = 2. That's the only possible solution if x is even; rewriting the equation as x = xyz - 2yz, we find x = yz(x - 2), and x-2 must be a factor of x. If x is even, that will only happen if x is 4. Actually you could have used that observation to solve the entire question -- if x-2 is a factor of x, then x can only equal 3 or 4 (or I suppose x could be 1, though that's impossible here), from which the solution follows. But that relies on an observation that one won't be able to make in many questions -- the method I used above is more generally useful, so I prefer that approach.