A 4-digit number N is such that when divided by 3, 5, 6, 9 leaves a remainder 1, 3, 4, 7 respectively. What is the smallest value of N ?

(a) 1068
(b) 1072
(c) 1078 ✓ UPSC's answer
(d) 1082

Why the answer is (c)

• The condition N ≡ 7 (mod 9) already implies N ≡ 1 (mod 3), so the divisor 3 condition is redundant.

• Combine N ≡ 4 (mod 6) and N ≡ 3 (mod 5): testing 4, 10, 16, 22, 28 modulo 30 gives N ≡ 28 (mod 30).

• Combine N ≡ 28 (mod 30) with N ≡ 7 (mod 9): write N = 28 + 30k; since 28 ≡ 1 and 30 ≡ 3 modulo 9, 1 + 3k ≡ 7, so k ≡ 2 (mod 3).

• The smallest such k is 2, giving N = 28 + 60 = 88, so all solutions are N = 88 + 90m.

• The smallest 4-digit value is for m = 11: 88 + 990 = 1078, which is option (c).

Why the other options are wrong

(a) 1068: 1068 leaves remainder 0 when divided by 3, not the required remainder 1.
(b) 1072: 1072 leaves remainder 2 when divided by 5, not the required remainder 3.
(d) 1082: 1082 leaves remainder 2 when divided by 3, not the required remainder 1.

Asked in the CSAT Paper II of the UPSC Civil Services Preliminary Examination 2025, held on 25 May 2025. Question and answer key: Union Public Service Commission. Explanation: UPSC Answer Check.