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If N is a positive integer, is N^3/4 an integer?

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If N is a positive integer, is N^3/4 an integer?  [#permalink]

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New post Updated on: 10 Jun 2019, 23:50
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If N is a positive integer, is (N^3)/4 an integer?

I. N^2 + 3 is a prime number.
II. N is the number of odd factors of 6.

Originally posted by kiran120680 on 09 Jun 2019, 10:18.
Last edited by kiran120680 on 10 Jun 2019, 23:50, edited 1 time in total.
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Re: If N is a positive integer, is N^3/4 an integer?  [#permalink]

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New post 09 Jun 2019, 11:40
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Strange question. If n is even, n^3 will be divisible by 2^3, so it will definitely be divisible by 2^2, and n^3/4 will be an integer. If n is odd, then n^3 won't even be divisible by 2, so n^3/4 certainly won't be an integer. So the question is just asking "is n even?"

Using Statement 1, n^2 + 3 can never equal 2 (since then n^2 would be negative, which is impossible). So if n^2 + 3 is a prime, n^2 + 3 must be an odd prime, and n^2 must be even, so n is even, and Statement 1 is sufficient.

Statement 2 unambiguously tells us the numerical value of n, so it has to be sufficient; there's no reason to spend any time actually working out what n is. So the answer is D.
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Re: If N is a positive integer, is N^3/4 an integer?  [#permalink]

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New post 10 Jun 2019, 09:30
The 2nd condition tells me N = 2 and it is sufficient clearly.
But I don't get why the 1st condition is sufficient.
I can see that N must be even but I don't understand how it helps to get an answer.
I would appreciate if anyone gives me any explanation.
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Re: If N is a positive integer, is N^3/4 an integer?  [#permalink]

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New post 08 Aug 2019, 05:53
Lilyj, let me try to help you

\(\frac{N^{3}}{4}\) to be an integer, N has to even. Why? because \(Odd*Odd*Odd = Odd\). If n is even \(=> N^{2}\) is even.

Statement 1 says that \(N^{2} + 3\) is a prime number. All prime numbers, except 2, are odd.
Let \(N^{2} + 3 = X\). If \(X=2 => N^{2} + 3 = 2=> N^{2} = -1\), which is impossible as original statement says that N is a positive integer. Therefore, X has to be odd.
3 is Odd => \(N^{2}+Odd = Odd\). The only way previous equation to hold true is for \(N^{2}\) to be Even (Even+Odd=Odd).

As we needed to prove that \(N^{2}\) is even, statement 1 is sufficient.

Hope this helps.
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Re: If N is a positive integer, is N^3/4 an integer?  [#permalink]

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New post 18 Aug 2019, 04:09
kiran120680 wrote:
If N is a positive integer, is (N^3)/4 an integer?

I. N^2 + 3 is a prime number.
II. N is the number of odd factors of 6.


Pl explain what does statement II means?
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Re: If N is a positive integer, is N^3/4 an integer?  [#permalink]

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New post 18 Aug 2019, 12:20
Can anyone please explain what does second statement mean?

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Re: If N is a positive integer, is N^3/4 an integer?  [#permalink]

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New post 18 Aug 2019, 12:53
Hi,
2nd statement says N= number of odd factors of 6. Now we know that 6 has 4 factors (1,2,3,6) out of which 2 are odd. Hence N=2.which is sufficient to answer the question

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Re: If N is a positive integer, is N^3/4 an integer?  [#permalink]

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New post 19 Aug 2019, 02:43
kiran120680 wrote:
If N is a positive integer, is (N^3)/4 an integer?

I. N^2 + 3 is a prime number.
II. N is the number of odd factors of 6.


is (N^3)/4 an integer OR Is N divisible by 2?

(1) N^2 + 3 is a prime number, only possible when N=2. Sufficient.

(2) N is the number of odd factors of 6, there are 2 factors of 6 which are odd. Sufficient.

D is correct.
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Re: If N is a positive integer, is N^3/4 an integer?   [#permalink] 19 Aug 2019, 02:43
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