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09 Mar 2016, 12:48
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D
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Difficulty:
65%
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Question Stats:
53% (02:08) correct
47%
(02:01)
wrong
based on 124
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If n is an integer, is n^3 divisible by 54?
(1) n^2 is divisible by 6.
(2) n^3 is divisible by 36.
(1) n^2 is divisible by 6.
(2) n^3 is divisible by 36.
stonecold
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09 Mar 2016, 21:26
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Here is my approach ]
here we need to check if N^3 is divisible by 54 so => N^3 must contain a 2 and three 3's
statement 1 => N^2 is divisible by 6 => 2,3 must be the primes of N^2
now as N^anypower and N have same prime factors => N must have 2 and three as Prime factors which in turn means N^3 will have atleast 3 two's and 3 three's Hence suffficient
Statement 2 => hence N^3 is divisible by 36 => 2,3 must be its prime factors => 2,3 must be the prime factors of N too => sufficient
Hence D is sufficient.
Here the clear trap was statement 2 as N^3 was divisible by 36 ..
I hope i am right...
here we need to check if N^3 is divisible by 54 so => N^3 must contain a 2 and three 3's
statement 1 => N^2 is divisible by 6 => 2,3 must be the primes of N^2
now as N^anypower and N have same prime factors => N must have 2 and three as Prime factors which in turn means N^3 will have atleast 3 two's and 3 three's Hence suffficient
Statement 2 => hence N^3 is divisible by 36 => 2,3 must be its prime factors => 2,3 must be the prime factors of N too => sufficient
Hence D is sufficient.
Here the clear trap was statement 2 as N^3 was divisible by 36 ..
I hope i am right...
09 Mar 2016, 21:49
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Bunuel
Hi,
A good Q..
Info from Q..
1) VERY IMP- n is an integer..
2) Is n^3 div by 54
Inference--
If n is an integer and n^3 has to be div by 54, or 2*3^3..
so n has to be div by 2*3 as n^3 by 3^3 means n by 3..
Lets see the statements--
(1) n^2 is divisible by 6.
since n is an integer, n^2 by 2*3 will mean even n by 2*3..
meets our INFERENCE
Suff
(2) n^3 is divisible by 36.
n^3 by 2^2*3^2..
again as n is an integer n will be div by 2 and 3..
Again meets our INFERENCE
Suff
D
