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
Answer & explanation
Answer: (b) 9
Adding the three rotations puts each digit once in the hundreds, tens and units places, so the sum is 111(X + Y + Z) = 3 × 37 × (X + Y + Z). It is divisible by 3, 37 and (X + Y + Z), but by 9 only if (X + Y + Z) were a multiple of 3 — which it is not.
- XYZ + YZX + ZXY = (100X + 10Y + Z) + (100Y + 10Z + X) + (100Z + 10X + Y) = 111(X + Y + Z).
- 111 = 3 × 37, so the sum = 3 × 37 × (X + Y + Z): divisible by 3, 37 and (X + Y + Z).
- For 9, a second factor 3 must come from (X + Y + Z), which is not a multiple of 3.
- So the sum is not divisible by 9.
- Check: 124 + 241 + 412 = 777 = 111 × 7; 777 ÷ 9 is not whole.
Remember · Cyclic rotations of a 3-digit number always add up to 111 × (digit sum) = 3 × 37 × (digit sum).
Question and answer: UPSC's official GS Paper II (2020, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). ·