QUESTION

CSAT

Hard

Maths

Prelims 2020

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

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Explanation

Let XYZ represent a three-digit number, which can be expressed as: XYZ = $100X + 10Y + Z$ , where X is the hundreds digit, Y is the tens digit, and Z is the ones digit.

Now, consider the sum: XYZ + YZX + ZXY = $(100X + 10Y + Z) + (100Y + 10Z + X) + (100Z + 10X + Y)$ .

Simplifying this expression: XYZ + YZX + ZXY = $111(X + Y + Z)$ .

This shows that the sum is divisible by $(X + Y + Z)$ . Additionally, since 111 is divisible by both 3 and 37, the expression XYZ + YZX + ZXY is divisible by $(X + Y + Z)$ , but not by 9.