Let the least number of six digit which when divided by 4, 6, 10 and 15 leaves in each case same remainder 2 be N. The sum of digit in N is :
Q. Let the least number of six digit which when divided by 4, 6, 10 and 15 leaves in each case same remainder 2 be N. The sum of digit in N is :
Answer: 5
Notes: LCM of 4, 6, 10, 15 = 60 Least number of 6 digits = 100000 The leats number of 6 digits which is exactly divisible by 60 = 100000 + (60 - 40) = 100020 ∴ Required Number = 100020 + 2 = 100022 Hence, the sum of digits = 1 + 0 + 0 + 0 + 2 + 2 = 5 Hence option [C] is correct answer.