Question

fine the least number between 200 and 500 which leaves a remainder of 3 in each case when divided by 8, 10, 12 and 30

Answer 1

If a number (n) leaves a remainder of 3,
n-3 must be divisible by each number (8,10,12, and 30).
Their minimum common multiple is 120
the smallest number that works with n-3=240 is 243.
so the answer is 243.