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

The correct option is 9.

Let us analyze the three-digit numbers given in the problem:

A three-digit number $XYZ$ can be written in expanded form as:

$XYZ=100X+10Y+Z$

Similarly, the cyclic permutations $YZX$ and $ZXY$ can be expanded as:

$YZX=100Y+10Z+X$

$ZXY=100Z+10X+Y$

Now, let us find the sum of these three numbers, $(XYZ+YZX+ZXY)$:

$XYZ+YZX+ZXY=(100X+10Y+Z)+(100Y+10Z+X)+(100Z+10X+Y)$

Grouping the terms containing $X$, $Y$, and $Z$:

$=(100+1+10)X+(10+100+1)Y+(1+10+100)Z$

$=111X+111Y+111Z$

$=111(X+Y+Z)$

We know that $111$ can be prime factorized as:

$111=3\×37$

Therefore, the sum becomes:

$XYZ+YZX+ZXY=3\×37\×(X+Y+Z)$

From this expression, we can check the divisibility by each option:

1. It is clearly divisible by 3 because 3 is a factor in the expression.

2. It is clearly divisible by 37 because 37 is a factor in the expression.

3. It is clearly divisible by (X + Y + Z) because $(X+Y+Z)$ is a factor in the expression.

Now, let us consider divisibility by 9:

For the expression $3\×37\×(X+Y+Z)$ to be divisible by 9, the factor $(X+Y+Z)$ must contribute another factor of 3 (since 9 = 3 × 3).

However, the question explicitly states that (X + Y + Z) is not a multiple of 3. This means $(X+Y+Z)$ cannot supply the additional factor of 3 needed to make the sum divisible by 9.

Hence, $(XYZ+YZX+ZXY)$ is not divisible by 9.