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

(a) 3
(b) 9 ✓ UPSC's answer
(c) 37
(d) (X + Y + Z)

Why the answer is (b)

• Expanding the three numbers: XYZ = 100X + 10Y + Z, YZX = 100Y + 10Z + X, ZXY = 100Z + 10X + Y. Adding them gives (100X + X + 10X) + (10Y + 100Y + Y) + (Z + 10Z + 100Z) = 111X + 111Y + 111Z = 111(X + Y + Z).

• Since 111 = 3 × 37, the sum can be written as 3 × 37 × (X + Y + Z), which is always a multiple of 3 and always a multiple of 37, regardless of the digits.

• The sum is also trivially divisible by (X + Y + Z) because it equals 111 times (X + Y + Z).

• For the sum to be divisible by 9, the factor 37 × (X + Y + Z) must be divisible by 3 (since 111 already contributes one factor of 3, and 9 = 3² requires a second factor of 3).

• Because 37 is not a multiple of 3, the second factor of 3 must come from (X + Y + Z). The problem explicitly states that (X + Y + Z) is not a multiple of 3, so the sum cannot be divisible by 9.

• Therefore, XYZ + YZX + ZXY is not divisible by 9, which is option (b).

Why the other options are wrong

(a) 3: The sum equals 111(X+Y+Z) = 3 × 37 × (X+Y+Z), which always contains a factor of 3, so it is always divisible by 3.
(c) 37: The sum equals 111(X+Y+Z) = 3 × 37 × (X+Y+Z), which always contains a factor of 37, so it is always divisible by 37.
(d) (X + Y + Z): The sum equals 111(X+Y+Z), which is 111 times (X+Y+Z), so it is always divisible by (X+Y+Z) by construction.

Asked in the CSAT Paper II of the UPSC Civil Services Preliminary Examination 2020, held on 4 October 2020. Question and answer key: Union Public Service Commission. Explanation: UPSC Answer Check.