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

Correct Option

Let the three-digit number be XYZ, which can be represented as 100X + 10Y + Z.

Similarly, the permutations YZX and ZXY can be written as:

• YZX = 100Y + 10Z + X
• ZXY = 100Z + 10X + Y

Summing these three numbers:

XYZ + YZX + ZXY = (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)

Factoring the constant 111, we find that 111 = 3 × 37.

Thus, the sum is 3 × 37 × (X + Y + Z).

For the sum to be divisible by 9, the term 37 × (X + Y + Z) must be divisible by 3. Since 37 is not divisible by 3, it necessitates that (X + Y + Z) must be divisible by 3.

However, the problem statement specifies that (X + Y + Z) is not a multiple of 3. Consequently, the expression 3 × 37 × (X + Y + Z) cannot be divisible by 9.

Incorrect Options

Based on the derived expression, (XYZ + YZX + ZXY) = 3 × 37 × (X + Y + Z):

• Option 1 (3): The expression contains a factor of 3. Therefore, it is always divisible by 3.
• Option 3 (37): The expression contains a factor of 37. Therefore, it is always divisible by 37.
• Option 4 ((X + Y + Z)): The expression is a direct multiple of (X + Y + Z). Therefore, it is always divisible by (X + Y + Z).