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 :

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 :
[A]3
[B]4
[C]5
[D]6

5
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.

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Comments

  • Nakul
    Reply

    Why we will delt 40

    • Tanmoy Das

      if you divide the six digit number with 60 then the 40 comes as remainder …