N is divisible by 33 ...

Prime factors of 33 are 3 x 11 ...

Therefore N has at the least one three and one eleven as its prime factors ..

Now N^k should be divisible by 27 i.e. 3 x 3 x 3

In order for N^K to be divisible by 27 , N^k must contain at least 3 three's .. In order for N^k to be DEFINITELY divisible by 27 it needs to have at least 3 3's as its prime factors .. we know that n has at least one 3 , therefore it needs three 3's , thus min possible value of k is 3 , and that would occur if n has exactly one three as one of its prime factor .. The max value for k would obviously be anything more than 3 uptil infinite ...

Therefore answer is 3 (C)

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