What is the least four-digit number when divided by 3, 4, 5 and 6 leaves a remainder 2 in each case?
- (a) 1012
- (b) 1022 ✓ UPSC's answer
- (c) 1122
- (d) 1222
Why the answer is (b)
• Find the Least Common Multiple (LCM) of the divisors 3, 4, 5, and 6, which is 60.
• Since the number leaves a remainder of 2 in each case, the number must be of the form 60k + 2.
• Identify the smallest four-digit number by testing values of k: for k=16, 60(16) + 2 = 962 (three digits); for k=17, 60(17) + 2 = 1022 (four digits).
• Verify that 1022 is the least four-digit number satisfying the condition, as the previous multiple (962) is a three-digit number.
• Therefore, 1022 is the correct answer.
Why the other options are wrong
- (a) 1012
- 1012 is not divisible by 3 (1012 - 2 = 1010, which is not divisible by 3).
- (c) 1122
- 1122 is a valid number of the form 60k + 2 but is larger than 1022, so it is not the least.
- (d) 1222
- 1222 is a valid number of the form 60k + 2 but is larger than 1022, so it is not the least.
Asked in the CSAT Paper II of the UPSC Civil Services Preliminary Examination 2020, held on 4 October 2020. Question and answer key: Union Public Service Commission. Explanation: UPSC Answer Check.