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If n^2/n yields an integer greater than 0, is n divisible by 30?
(1) n^2 is divisible by 20
(2) n^3 is divisible by 12
(1) n^2 is divisible by 20
(2) n^3 is divisible by 12
For me its B. Can someone please confirm?
28 Jan 2012, 03:25
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enigma123
If n^2/n yields an integer greater than 0, is n divisible by 30?
Given: n^2/n yields an integer greater than 0 --> n>0 --> n^2/n=n=integer. So, the question basically asks whether n is a multiple of 30.
(1) n^2 is divisible by 20 --> 20=2^2*5 --> n is divisible by at least 2 and 5, otherwise these factors could not appear in n^2 (exponentiation does not "produce" primes), though we don't know about other possible factors of n. Not sufficient.
(2) n^3 is divisible by 12 --> 12=2^2*3 --> the same way here: n is divisible by at least 2 and 3, though we don't know about other possible factors of n. Not sufficient.
(1)+(2) n is divisible by 2, 3 and 5 hence by 2*3*5=30. Sufficient.
Answer: C.
09 Oct 2013, 22:47
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rwj
Hi,
If n*n is divisible by 20 (2*2*5) then in the numerator there has to be minimum 2*2*5*5 so as to make it a perfect square and that 20 divides the numerator properly and gives us an integer. So n has atleast one 2 and one 5.
Similarly, n*n*n = n^3 is divisible by 12 (2*2*3). So numerator has to have 2*2*2*3*3*3 so as to make it a perfect cube and that 12 divides the numerator and gives us an integer value. So n has atleast one 2 and one 3
Therefore combining the 2 statements n has atleast one 2, one 3 and one 5. Hence it should be divisible by 30.
Consider Kudos if the post helps!!
