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

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

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08 Jun 2012, 09:20
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If a number (N) is divisible by 33, what will be the minimum value of k, such that N^k should definitely be divisible by 27?

A. 1
B. 2
C. 3
D. 4
E. 6

Last edited by Bunuel on 08 Jun 2012, 10:21, edited 1 time in total.
EDITED THE QUESTION.

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Re: If a number (N) is divisible by 33 [#permalink]

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08 Jun 2012, 10:25
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Expert's post
sarb wrote:
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.

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

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08 Jun 2012, 11:24
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were it to be nk, the minimum value of k is 9

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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, 06:57
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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 ...

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

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10 Jun 2012, 23:54
n = 3*11*m

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

27 = 3^3

so k must be at least 3

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Re: If a number (N) is divisible by 33 [#permalink]

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11 Jun 2012, 00:19
Bunuel wrote:
sarb wrote:
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.

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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Re: If a number (N) is divisible by 33 [#permalink]

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11 Jun 2012, 06:33
Expert's post
1
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BOOKMARKED
Bunuel wrote:
sarb wrote:
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.

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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13 Jun 2012, 13:24
Yup its 3.....Yupppppiee..
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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, 13:10
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Re: If a number (N) is divisible by 33, what will be the minimum   [#permalink] 04 Dec 2015, 13:10
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