If a number (N) is divisible by 33, what will be the minimum

n = 3*11*m

n^k = 3^k*11^k*m^k

27 = 3^3

so k must be at least 3

Thanks as usual, Bunuel. Could you help me wrap my head around this some more?

So N is divisible by 3*11
and N^k is divisible by 3*3*3.

So why exactly does the minimum value for k only need to be 3? Could you possibly use some example values here?

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N^k to be divisible by 3^3 should have 3^3 as a factor and since N has only one 3 in its prime factorization then k must be at least 3.

Yup its 3.....Yupppppiee..

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)

Since N is divisible by 33, we see that N has, at a minimum, one prime factor of 3. In order for N^k/27 = integer, we see that N must have at least three primes of 3 (since 27 has three factors of 3), and thus the minimum value of k would have to be 3, since 33^3 = 3^3 x 11^3 = 27 x 11^3.

Answer: C

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