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If a number (N) is divisible by 33, what will be the minimum
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Updated on: 13 Feb 2018, 23:07
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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
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Originally posted by sarb on 08 Jun 2012, 08:20.
Last edited by abhimahna on 13 Feb 2018, 23:07, edited 2 times in total.
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Re: If a number (N) is divisible by 33, what will be the minimum
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08 Jun 2012, 09:25



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Re: If a number (N) is divisible by 33, what will be the minimum
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08 Jun 2012, 10:24
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
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10 Jun 2012, 22:54
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, what will be the minimum
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10 Jun 2012, 23:19
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. 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?



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Re: If a number (N) is divisible by 33, what will be the minimum
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11 Jun 2012, 05:33
[youtube][/youtube] vandygrad11 wrote: 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. 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
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13 Jun 2012, 12:24
Yup its 3.. ...Yupppppiee..
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Re: If a number (N) is divisible by 33, what will be the minimum
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01 Oct 2012, 05:57
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
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02 Feb 2018, 11:22
sarb wrote: 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 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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