Find the sum of a least six digits number, which when divided by 4, 6, 10 and 15 leaves the remainder of 2 in each case.

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Solution

The correct option is C 5
Least number of 6 digits is 100000.
L.C.M. of 4, 6, 10 and 15 is 60.
On dividing 100000 by 60, the remainder obtained is 40.
Least number of 6 digits divisible by 4, 6, 10 and 15 = 100000 + (60 - 40) = 100020
N = (100020 + 2) = 100022
Sum of digits in N = (1 + 2 + 2) = 5.

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Find the sum of a least six digits number, which when divided by 4, 6, 10 and 15 leaves the remainder of 2 in each case.

The correct option is C 5Least number of 6 digits is 100000. L.C.M. of 4, 6, 10 and 15 is 60. On dividing 100000 by 60, the remainder obtained is 40. Least number of 6 digits divisible by 4, 6, 10 and 15 = 100000 + (60 - 40) = 100020 N = (100020 + 2) = 100022 Sum of digits in N = (1 + 2 + 2) = 5.

The correct option is C 5
Least number of 6 digits is 100000.
L.C.M. of 4, 6, 10 and 15 is 60.
On dividing 100000 by 60, the remainder obtained is 40.
Least number of 6 digits divisible by 4, 6, 10 and 15 = 100000 + (60 - 40) = 100020
N = (100020 + 2) = 100022
Sum of digits in N = (1 + 2 + 2) = 5.