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 :

(a) 1 only
(b) 2 only
(c) Both 1 and 2 ✓ UPSC's answer
(d) Neither 1 nor 2

Why the answer is (c)

• A 3-digit number is a multiple of 3 if and only if the sum of its digits is a multiple of 3.

• The set of all such numbers is formed by permuting all valid combinations of three distinct non-zero digits whose sum is divisible by 3.

• For any specific set of three distinct digits {a, b, c}, there are 3! = 6 permutations. In these 6 permutations, each digit appears exactly 2 times in the hundreds, tens, and units places respectively.

• The sum of the 6 permutations for a specific digit set {a, b, c} is 2 * (a+b+c) * (100 + 10 + 1) = 222 * (a+b+c).

• Since (a+b+c) is a multiple of 3, let (a+b+c) = 3k. The sum for this set is 222 * 3k = 666k.

• The total sum S is the sum of 666k over all valid digit sets. Since 666 is divisible by both 9 (6+6+6=18) and 74 (666/74=9), S is divisible by both 9 and 74.

Why the other options are wrong

(a) 1 only: Statement 2 is also correct because the sum of digits for each group is a multiple of 3, making the total sum divisible by 9.
(b) 2 only: Statement 1 is also correct because the contribution of each digit set to the total sum is a multiple of 666, which is divisible by 74.
(d) Neither 1 nor 2: Both statements are mathematically correct as derived from the structure of permutations and divisibility rules.

Asked in the CSAT Paper II of the UPSC Civil Services Preliminary Examination 2021, held on 10 October 2021. Question and answer key: Union Public Service Commission. Explanation: UPSC Answer Check.