If x, y, and z are 3 positive consecutive integers such that x < y < z, what is the remainder when the product of x, y, and z is divided by 8?
(1) (xz)^2 is even
(2) 5 . y^3 is odd
Going for D here ... wait for OA though
From 1) possible cases - (x even, z even) or (x even, z odd) or (x odd, z even)
But from the question stem ...x < y < z , here x and z are both either even or odd.
so only even possible ..that means multiple of 8
From 2) y ^3 has to be odd..so y odd ..so x and z are even...divisible by 8