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kiran120680
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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.
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kiran120680
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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Can anyone please explain what does second statement mean?

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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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kiran120680
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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QuantMadeEasy
kiran120680
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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QuantMadeEasy

(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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Bunuel can you please help with explanation here. Explanations above didn't make sense to me. Even chatgpt said only statement 2 is sufficient 😂
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Addu.23

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

(1) N^2 + 3 is a prime number.
(2) N is the number of odd factors of 6.

Bunuel can you please help with explanation here. Explanations above didn't make sense to me. Even chatgpt said only statement 2 is sufficient 😂

ChatGPT is not good with math.

(1) N^2 + 3 is a prime number.

N^2 + 3 > 3, so N^2 + 3 can only be an odd prime. N^2 + 3 = odd implies N^2 = even, which further implies that N is even. If N is even, then N^3/4 will for sure be an integer. So, (1) is sufficient.

Hope it helps.
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