1. I'd like to start off by pointing out that 30 has no repeating prime factors (i.e., it's different from 8 = 2*2*2): 30 = 2*3*5. This is why I believe that if n^2 is divisible by 30, then n will also be divisible by 30. This happens because in n^2 we basically "repeat" all the prime factors in n.
A counterexample would be for that 8 I pointed out: 4^2 will be divisible by 8 but 4 is not
So 1 is sufficient to establish that n is indeed divisible by 30.
2 is not enough. Take n = 15 for instance: 30 is a factor of 2n but you can't say that n is divisible by 30. However, if you take n = 60, then 30 is a factor of 2n and n is divisible by 30.
IMHO, the answer is A.
DS
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cubicle_bound_misfit
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The answer is A.
stmt 2 says 30 is a factor of 2n
say n=15 then 30 is not a factor of n.
stmt 1 says :
n^2 is dvided by 30.
or n^2 has factors in the form of 2^a* 3^b*5^c*x^y etc.
no matter what the value of a,b,c, n will be having factors square root of them and hence divisible by 30.
Hence A wins.
stmt 2 says 30 is a factor of 2n
say n=15 then 30 is not a factor of n.
stmt 1 says :
n^2 is dvided by 30.
or n^2 has factors in the form of 2^a* 3^b*5^c*x^y etc.
no matter what the value of a,b,c, n will be having factors square root of them and hence divisible by 30.
Hence A wins.
Cubicle Bound Misfit
