How many three digit number which are divisible by 6 and 3 both?
Explanation: Divisibility rule by 6: the number must be an even number and sum of digit must be divisible by 3.
The first number which is divisible by 6 is 102 and the last number is 996. Applying arithmetic formula for finding number of terms with common difference (d) 6, first term(A1) 102 and last term(An) 996:
An = A1 + (n - 1)*d 996 = 102 + (n - 1)*6 n-1 = 894 / 6 = 149 n = 150
So, option (C) is correct.
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