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If N is a positive integer, is N^3/4 an integer?
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Updated on: 10 Jun 2019, 22:50
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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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Originally posted by kiran120680 on 09 Jun 2019, 09:18.
Last edited by kiran120680 on 10 Jun 2019, 22:50, edited 1 time in total.



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Re: If N is a positive integer, is N^3/4 an integer?
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09 Jun 2019, 10:40
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?
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10 Jun 2019, 08:30
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?
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08 Aug 2019, 04:53
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?
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18 Aug 2019, 03:09
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?
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18 Aug 2019, 11:20
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?
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18 Aug 2019, 11:53
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?
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19 Aug 2019, 01:43
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?
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