What is the smallest number greater than 1000 that when divided by any one of the numbers 6, 9, 12, 15, 18 leaves a remainder of 3?
- (a) 1063
- (b) 1073
- (c) 1083 ✓ UPSC's answer
- (d) 1183
Why the answer is (c)
• First, determine the Least Common Multiple (LCM) of the divisors 6, 9, 12, 15, and 18.
• The prime factorizations are: 6 = 2×3, 9 = 3², 12 = 2²×3, 15 = 3×5, 18 = 2×3².
• The LCM is the product of the highest powers of all primes involved: 2² × 3² × 5 = 4 × 9 × 5 = 180.
• The number must be of the form 180k + 3, where k is an integer.
• We need the smallest number greater than 1000. Dividing 1000 by 180 gives approximately 5.55, so we test k = 6.
• Calculating 180 × 6 + 3 gives 1080 + 3 = 1083, which matches option (c).
Why the other options are wrong
- (a) 1063
- 1063 is not divisible by 180 with a remainder of 3, as 1063 - 3 = 1060, which is not a multiple of 180.
- (b) 1073
- 1073 is not divisible by 180 with a remainder of 3, as 1073 - 3 = 1070, which is not a multiple of 180.
- (d) 1183
- 1183 is not divisible by 180 with a remainder of 3, as 1183 - 3 = 1180, which is not a multiple of 180.
Asked in the CSAT Paper II of the UPSC Civil Services Preliminary Examination 2022, held on 5 June 2022. Question and answer key: Union Public Service Commission. Explanation: UPSC Answer Check.