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

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kiran120680
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If N is a positive integer, is N^3/4 an integer? [#permalink] 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.
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D

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] 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] New post   18 May 2022, 00:42
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QuantMadeEasy wrote:
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.





If n^2+3 is taken as 19 which is also prime then we have 2 nos 2&4. How is statement 1 sufficient. Can you please explain?

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Re: If N is a positive integer, is N^3/4 an integer? [#permalink] New post   18 May 2022, 03:32
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Knightwing wrote:
QuantMadeEasy wrote:
(1) N^2 + 3 is a prime number, only possible when N=2. Sufficient.


If n^2+3 is taken as 19 which is also prime then we have 2 nos 2&4. How is statement 1 sufficient. Can you please explain?


The solution you've quoted isn't correct -- Statement 1 here is true for an infinite number of values of N (as you point out, N can be 2 or 4, and it can also be 8, 10, 14, 22 and an infinite number of other values). The statement is still sufficient, though, because while we don't know the value of N, we can still answer the question.
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Re: If N is a positive integer, is N^3/4 an integer? [#permalink] 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] 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] 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] 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] 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] 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] New post   23 May 2023, 06:56
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Re: If N is a positive integer, is N^3/4 an integer? [#permalink]
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