UPSC Prelims 2020·CSAT·Quantitative Aptitude·Number System

What is the least four-digit number when divided by 3, 4, 5 and 6 leaves a remainder 2 in each case?

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Last updated 23 May 2026, 3:31 pm IST
  1. A1012
  2. B1022Correct
  3. C1122
  4. D1222

Explanation

1. Find the Least Common Multiple (LCM) of the divisors 3, 4, 5, and 6. LCM(3, 4, 5, 6) = 60. 2. Any number that leaves a remainder of 2 when divided by 3, 4, 5, and 6 must be of the form (LCM * k) + 2, where 'k' is an integer. So, the numbers are of the form (60k + 2). 3. We are looking for the least four-digit number. The smallest four-digit number is 1000. We need to find the smallest integer 'k' such that (60k + 2) is greater than or equal to 1000. 60k + 2 >= 1000 60k >= 998 k >= 998 / 60 k >= 16.63... 4. Since 'k' must be an integer, the smallest integer value for 'k' that satisfies the condition is 17. 5. Substitute k = 17 into the expression (60k + 2): Number = (60 * 17) + 2 = 1020 + 2 = 1022. 6. Check the options: A) 1012: (1012 - 2) = 1010. 1010 is not divisible by 60. B) 1022: (1022 - 2) = 1020. 1020 is divisible by 60 (1020 / 60 = 17). This matches our calculated number. C) 1122: (1122 - 2) = 1120. 1120 is not divisible by 60. D) 1222: (1222 - 2) = 1220. 1220 is not divisible by 60. The least four-digit number satisfying the condition is 1022. The final answer is B
Quantitative Aptitude: What is the least four-digit number when divided by 3, 4, 5 and 6 leaves a remainder 2 in each case?

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