Consider all 3-digit numbers (without repetition of digits) obtained using three non-zero digits which are multiples of 3. Let S be their sum.
Which of the following is/are correct?
- 1.S is always divisible by 74.
- 2.S is always divisible by 9.
Select the correct answer using the code given below:
Answer & explanation
Answer: (c) Both 1 and 2
With the digits 3, 6 and 9, the six numbers add up to 222 × 18 = 3996, which is 74 × 54 and 9 × 444. In general, the six arrangements of three digits add to 222 × (digit sum), and 222 = 3 × 74, so both statements hold.
- The non-zero digits that are multiples of 3 are 3, 6 and 9; without repetition they give 3! = 6 numbers.
- Each digit appears twice in each place, so S = 2 × (3 + 6 + 9) × 111 = 222 × 18 = 3996.
- 3996 = 74 × 54, so S is divisible by 74.
- 3996 = 9 × 444, so S is divisible by 9.
- Check (general case): any three distinct digits a, b, c give six numbers summing to 222(a + b + c) = 3 × 74 × (a + b + c); when the numbers are multiples of 3, a + b + c is too, so S is a multiple of 9 × 74 = 666.
- ✓ 1. S = 222 × (digit sum) and 222 = 3 × 74, so 74 always divides S.
- ✓ 2. 222 carries one factor 3 and the digit sum (a multiple of 3) carries another, so 9 always divides S.
Remember · Sum of all arrangements of 3 distinct digits = 222 × (sum of the digits).
Question and answer: UPSC's official GS Paper II (2021, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). ·