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Let XYZ be a 3-digit number. Let S = XYZ + YZX + ZXY. Which of the following statements is/are correct? 1. S is always divisible by 3 and (X + Y + Z) 2. S is always divisible by 9 3. S is always divisible by 37 Select the correct answer using the code given below:

Let XYZ be a 3-digit number. Let S = XYZ + YZX + ZXY. Which of the following statements is/are correct? 1. S is always divisible by 3 and (X + Y + Z) 2. S is always divisible by 9 3. S is always divisible by 37 Select the correct answer using the code given below: Correct Answer 1 and 3

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.

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