UPSC CSE Prelims 2022
To determine the smallest number greater than 1000 that leaves a remainder of 3 when divided by 6, 9, 12, 15, and 18, the following steps are applied:
A number 'N' that leaves a remainder 'r' when divided by a set of numbers is expressed as `N = k × LCM + r`. In this case, `N = 180k + 3`, where 'k' is a positive integer.
We need to find the smallest integer 'k' such that `180k + 3 > 1000`.
The smallest integer value for 'k' that satisfies this condition is 6.
Substitute k = 6 into the expression: `N = 180 × 6 + 3 = 1080 + 3 = 1083`.
Thus, 1083 is the smallest number greater than 1000 that satisfies the given conditions. When 3 is subtracted from 1083, the result is `1083 - 3 = 1080`, which is perfectly divisible by 180.
For any number to leave a remainder of 3 when divided by 6, 9, 12, 15, and 18, subtracting 3 from that number must result in a multiple of their LCM, which is 180.
Therefore, these options do not fulfill the required condition.