Q. Let the least number of six digits which when divided by 4, 6, 10, 15 leaves in each case same remainder 2 be N. The sum of digits in N is :
✅ Correct Answer: (C)
5
Explanation: LCM of 4, 6, 10, 15 = 60
Least number of 6 digits = 100000
The least number of 6 digits which is exactly divisible by 60 = 100000 + (60 – 40) = 100020
∴ Required number (N) = 100020 + 2 = 100022
Hence, the sum of digits = 1 + 0 + 0 + 0 + 2 + 2 = 5
Least number of 6 digits = 100000
The least number of 6 digits which is exactly divisible by 60 = 100000 + (60 – 40) = 100020
∴ Required number (N) = 100020 + 2 = 100022
Hence, the sum of digits = 1 + 0 + 0 + 0 + 2 + 2 = 5
Explanation by: Mr. Dubey
LCM of 4, 6, 10, 15 = 60
Least number of 6 digits = 100000
The least number of 6 digits which is exactly divisible by 60 = 100000 + (60 – 40) = 100020
∴ Required number (N) = 100020 + 2 = 100022
Hence, the sum of digits = 1 + 0 + 0 + 0 + 2 + 2 = 5
Least number of 6 digits = 100000
The least number of 6 digits which is exactly divisible by 60 = 100000 + (60 – 40) = 100020
∴ Required number (N) = 100020 + 2 = 100022
Hence, the sum of digits = 1 + 0 + 0 + 0 + 2 + 2 = 5