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If a number (N) is divisible by 33, what will be the minimum value of k, such that Nk should definitely be divisible by 27?

1 2 3 4 6

I guess it should be N^k instead of Nk.

Given that N is divisible by 33=3*11. Now, in order N^k to be divisible by 27=3^3 then k must be at least 3 in order N^k to have enough 3's.

Answer: C.

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?

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.
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Re: If a number (N) is divisible by 33, what will be the minimum [#permalink]

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01 Oct 2012, 05:57

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This post received KUDOS

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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Re: If a number (N) is divisible by 33, what will be the minimum [#permalink]

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04 Dec 2015, 12:10

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