Calculation: S = XYZ + YZX + ZXY S = 100X + 10Y + Z + 100Y + 10Z + X + 100Z + 10X + Y S = 111X + 111Y + 111Z S = 111 (X + Y + Z) We can observe here: 111 is divisible by 3 and 37. Hence, S will be divisible by (X + Y + Z), 3 and 37 always. ∴ 1 and 3 are correct On Checking statement 2 take X = 1, Y = 2 & Z = 4 We get S = 111 (1 + 2 + 4) = 777 Which is not divisible by 9. Hence, S will not be always divisible by 9. ∴ 2 is incorrect. Mistake Points At first, we have to understand here that We can't take x = 0 If we take any one of them equal to zero then we will not get a three-digit number like if we take x = 0, y = 2 and z = 4 then XYZ = 024 which is a two-digit number Some cases also generate divisible by 9, but in all cases not satisfy the condition. the Question said always divisible by 9, but it is not possible for all cases. Like S = 111(1 + 2 + 4) = 777, or S = 111(2 + 3 + 0) = 555 both number is not divisible by 9. Satisfy cases, S = 111(1 + 2 + 3) = 666, S = 111(2 + 3 + 4) = 999 both number also divisible. Confusion PointsX, Y and Z are not given equal to each other, hence we cannot consider this special case to answer the question.