Let XYZ be a three-digit number, where (x + y + z) is not a multiple of 3. Then (XYZ + YZX + ZXY) is not divisible by

Let the three-digit number XYZ be represented as 100x + 10y + z. The sum (XYZ + YZX + ZXY) can be written as: (100x + 10y + z) + (100y + 10z + x) + (100z + 10x + y) = (100x + x + 10x) + (10y + 100y + y) + (z + 10z + 100z) = 111x + 111y + 111z = 111(x + y + z)

We know that 111 can be factored as 3 * 37. So, the sum = 3 * 37 * (x + y + z).

Now let's analyze the options based on the given condition that (x + y + z) is not a multiple of 3:

A) Divisible by 3: The sum is 3 * 37 * (x + y + z). Since 3 is a factor, the sum is always divisible by 3. So, A is incorrect.

C) Divisible by 37: The sum is 3 * 37 * (x + y + z). Since 37 is a factor, the sum is always divisible by 37. So, C is incorrect.

D) Divisible by x + y + z: The sum is 111 * (x + y + z). Since (x + y + z) is a factor, the sum is always divisible by (x + y + z). So, D is incorrect.

B) Divisible by 9: For the sum (3 * 37 * (x + y + z)) to be divisible by 9, (37 * (x + y + z)) must be divisible by 3. Since 37 is not divisible by 3, it implies that (x + y + z) must be divisible by 3. However, the problem explicitly states that (x + y + z) is NOT a multiple of 3. Therefore, the sum is not divisible by 9. So, B is the correct answer.

The final answer is B