For any choices of values of X, Y and Z, the 6-digit number of the form XYZXYZ is divisible by:

Answer: (d) 7, 11 and 13

XYZXYZ = XYZ × 1000 + XYZ = XYZ × 1001, and 1001 = 7 × 11 × 13. So every such number is divisible by 7, 11 and 13.

1. XYZXYZ = XYZ × 1000 + XYZ = XYZ × 1001.
2. 1001 = 7 × 11 × 13.
3. So the number is always divisible by 7, 11 and 13.
4. Check: 123123 = 123 × 1001 = 123 × 7 × 11 × 13.

Remember · A repeated 3-digit block (abcabc) is a multiple of 1001 = 7 × 11 × 13; abab is a multiple of 101.

Question and answer: UPSC's official GS Paper II (2023, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). · Report an error