What is the least number of 4 digits which when divided by 5 , 8 , 10 and 15 leaves 2 as remainder in each case?

Find LCM of 5, 8, 10, 15: LCM = 120. We need the smallest 4-digit number that when divided by each divisor leaves remainder 2. This means the number is 2 more than a common multiple of 120. Smallest 4-digit number is 1000. 1000 ÷ 120 = 8 remainder 40, so 9 × 120 = 1080 is the smallest 4-digit multiple of 120. Required number = 1080 + 2 = 1082. Checking: 1082 ÷ 5 = 216 remainder 2, 1082 ÷ 8 = 135 remainder 2, 1082 ÷ 10 = 108 remainder 2, 1082 ÷ 15 = 72 remainder 2. Since 1082 is not among options A, B, or C, the correct answer is D: None of these.