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Director  V
Joined: 18 Feb 2019
Posts: 573
Location: India
GMAT 1: 460 Q42 V13 GPA: 3.6
If N is a positive integer, is N^3/4 an integer?  [#permalink]

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10 00:00

Difficulty:   55% (hard)

Question Stats: 63% (01:54) correct 37% (01:53) wrong based on 124 sessions

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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.
GMAT Tutor G
Joined: 24 Jun 2008
Posts: 1843
Re: If N is a positive integer, is N^3/4 an integer?  [#permalink]

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2
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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Intern  B
Joined: 07 Jun 2019
Posts: 3
Re: If N is a positive integer, is N^3/4 an integer?  [#permalink]

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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.
Manager  S
Joined: 03 Aug 2009
Posts: 58
Re: If N is a positive integer, is N^3/4 an integer?  [#permalink]

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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.
Intern  B
Joined: 03 Oct 2016
Posts: 1
Re: If N is a positive integer, is N^3/4 an integer?  [#permalink]

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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?
Intern  B
Joined: 16 Jul 2019
Posts: 1
Re: If N is a positive integer, is N^3/4 an integer?  [#permalink]

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Can anyone please explain what does second statement mean?

Posted from my mobile device
Intern  B
Joined: 11 Jun 2019
Posts: 29
Location: India
Re: If N is a positive integer, is N^3/4 an integer?  [#permalink]

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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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Manager  G
Joined: 28 Feb 2014
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Concentration: General Management, International Business
GPA: 3.97
WE: Engineering (Education)
Re: If N is a positive integer, is N^3/4 an integer?  [#permalink]

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