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Bunuel
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Bunuel
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I think this is a high-quality question and I agree with explanation.
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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Bunuel
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(1) 18 is a factor of \(n^2\).
Therefore, \(n = 3\sqrt{2p}\) can take values that are multiples of \(3*2 = 6\), such as 6, 12, 18, etc. Based on this, \(n\) might or might not be a multiple of 9. Not sufficient.

I am having some difficulty understanding this part. I can see if p=2, we can get n = 6, and if p = 8, we get 12, but I am not sure, how we can get p to be 18? My understanding is that for 3root2p to be an integer, p has to contribute odd number of 2's to make root2p an inteher, and hence p can be 2, 8, 32, etc. In these cases, we get the value of n as 6, 12, 24.

Bunuel - would love to know where I am going wrong here.

Thanks
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TargetMBA007
Bunuel
Official Solution:

(1) 18 is a factor of \(n^2\).
Therefore, \(n = 3\sqrt{2p}\) can take values that are multiples of \(3*2 = 6\), such as 6, 12, 18, etc. Based on this, \(n\) might or might not be a multiple of 9. Not sufficient.

I am having some difficulty understanding this part. I can see if p=2, we can get n = 6, and if p = 8, we get 12, but I am not sure, how we can get p to be 18? My understanding is that for 3root2p to be an integer, p has to contribute odd number of 2's to make root2p an inteher, and hence p can be 2, 8, 32, etc. In these cases, we get the value of n as 6, 12, 24.

Bunuel - would love to know where I am going wrong here.

Thanks

Essentially, p must contain an odd number of 2s and an even number of other primes. To get n = 18, consider p = 2*3^2. In this case \(n = 3\sqrt{2p}=3\sqrt{2^2*3^2}=3*2*3=18\).

Hope it helps.
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