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

What is the least four-digit number when divided by 3, 4, 5 and 6 leaves a remainder 2 in each case?
  1. 1012
  2. 1022
  3. 1122
  4. 1222

Solution

LCM of 3, 4, 5, 6 is 60. The number is of the form $60k + 2$. For the least four-digit value, $60k + 2 \ge 1000$ gives $k \ge 16.63$, so $k = 17$, giving $60 \times 17 + 2 = 1020 + 2 = 1022$.

Asked in: CSAT 2020

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