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If N is a positive integer

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If N is a positive integer  [#permalink]

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New post Updated on: 13 Aug 2018, 02:27
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Question Stats:

68% (01:58) correct 32% (01:45) wrong based on 189 sessions

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e-GMAT Question:



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 \(6\).

    A) Statement (1) ALONE is sufficient, but statement (2) ALONE is not sufficient.
    B) Statement (2) ALONE is sufficient, but statement (1) ALONE is not sufficient.
    C) Both statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
    D) EACH statement ALONE is sufficient.
    E) Statement (1) and (2) TOGETHER are NOT sufficient.

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Question 10 of The e-GMAT Number Properties Marathon




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Question 11 of the Marathon


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Originally posted by EgmatQuantExpert on 28 Feb 2018, 03:04.
Last edited by EgmatQuantExpert on 13 Aug 2018, 02:27, edited 2 times in total.
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Re: If N is a positive integer  [#permalink]

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New post 28 Feb 2018, 03:36
EgmatQuantExpert wrote:
Question:

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 \(6\).

    A) Statement (1) ALONE is sufficient, but statement (2) ALONE is not sufficient.
    B) Statement (2) ALONE is sufficient, but statement (1) ALONE is not sufficient.
    C) Both statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
    D) EACH statement ALONE is sufficient.
    E) Statement (1) and (2) TOGETHER are NOT sufficient.



we need to know if n is a multiple of 2

1) n^2 + 3 is prime

if n =1
n=2

the statement is true for both

but n^3/4 is not a integer if n=1 and an integer when n=2

2) n= number of odd factors of 6

6 = 3x2 i.e factors = 1,2,3,6

1 is odd and 3 is odd

n=2

2^3/4= 2(integer)

sufficient

(B) imo
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Re: If N is a positive integer  [#permalink]

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New post 28 Feb 2018, 07:19
1
Hatakekakashi wrote:
EgmatQuantExpert wrote:
Question:

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 \(6\).

    A) Statement (1) ALONE is sufficient, but statement (2) ALONE is not sufficient.
    B) Statement (2) ALONE is sufficient, but statement (1) ALONE is not sufficient.
    C) Both statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
    D) EACH statement ALONE is sufficient.
    E) Statement (1) and (2) TOGETHER are NOT sufficient.



we need to know if n is a multiple of 2

1) n^2 + 3 is prime

if n =1
n=2

the statement is true for both

but n^3/4 is not a integer if n=1 and an integer when n=2

2) n= number of odd factors of 6

6 = 3x2 i.e factors = 1,2,3,6

1 is odd and 3 is odd

n=2

2^3/4= 2(integer)

sufficient

(B) imo


Hi Hatakekakashi

Pleasse review highlighted

If n=1 ....then 1+3 = 4 s it is NOT prime number as statement states.
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Re: If N is a positive integer  [#permalink]

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New post 28 Feb 2018, 11:24
Mo2men wrote:
Hatakekakashi wrote:
EgmatQuantExpert wrote:
Question:

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 \(6\).

    A) Statement (1) ALONE is sufficient, but statement (2) ALONE is not sufficient.
    B) Statement (2) ALONE is sufficient, but statement (1) ALONE is not sufficient.
    C) Both statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
    D) EACH statement ALONE is sufficient.
    E) Statement (1) and (2) TOGETHER are NOT sufficient.



we need to know if n is a multiple of 2

1) n^2 + 3 is prime

if n =1
n=2

the statement is true for both

but n^3/4 is not a integer if n=1 and an integer when n=2

2) n= number of odd factors of 6

6 = 3x2 i.e factors = 1,2,3,6

1 is odd and 3 is odd

n=2

2^3/4= 2(integer)

sufficient

(B) imo


Hi Hatakekakashi

Pleasse review highlighted

If n=1 ....then 1+3 = 4 s it is NOT prime number as statement states.


my bad..was at work :)
thanks for pointing out :D

i guess (D) then

n^2+ 3 is prime n can't be odd as odd+odd =even and not prime as value >2

n=2 value 7
n=4 value 19
n=6 39 (not prime)
n=8 67 prime
n=10 103 (prime)
we can see that n^2 + 3 is prime

n^3/4 will most likely be an integer
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Re: If N is a positive integer  [#permalink]

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New post 28 Feb 2018, 12:35

Solution:



We need to find, whether \(N^3/4\) is an integer or not?
Now, \(N^3/4\) is an integer only when \(N\) is even number.
Thus, we only need to find, whether \(N\) is an even number or not?
Statement 1:
"\(N^2+3\) is a prime number.”
Since \(N\) is a positive integer, therefore, \(N^2+3\) is always greater than \(3\).
We know, all Prime numbers greater than \(3\) are odd.
Thus,
    \(N^2+3\) = \(odd\)
    \(N^2\)+ \(odd=odd\)
    \(N^2\)= \(even\)
Therefore, \(N^2+3\) is a prime number when \(N\) is an even number.
Thus, \(N^3/4\) is an integer.
Therefore, Statement 1 alone is sufficient to answer the question.

Statement 2:
“\(N\) is the number of odd factors of \(6\).”
      \(6= 2*3\)
To calculate odd factors of a number, we do not consider the even prime number. Therefore,
    Odd factors of \(6\)= power of \(3+1\)
    Odd factors of\(6\)= \(1+1\)
    Odd factors of \(6=2\)
Thus, \(N=2\).
Therefore, \(N^3/4\) is an integer.
Therefore, Statement 2 alone is sufficient to answer the question.

Therefore, we can find the answer by EACH of the statement alone.
Answer: Option D
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Re: If N is a positive integer  [#permalink]

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New post 03 Mar 2018, 13:40
EgmatQuantExpert wrote:
Question:

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 \(6\).

    A) Statement (1) ALONE is sufficient, but statement (2) ALONE is not sufficient.
    B) Statement (2) ALONE is sufficient, but statement (1) ALONE is not sufficient.
    C) Both statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
    D) EACH statement ALONE is sufficient.
    E) Statement (1) and (2) TOGETHER are NOT sufficient.


Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

The first step of the VA (Variable Approach) method is to modify the original condition and the question, and then recheck the question.

We can modify the original condition and question as follows.

The question that \(N^3/4\) is an integer is equivalent to \(N\) is an even integer.

Condition 1)
In order for \(N^2 + 3\) to an prime number, \(N^2\) and \(N\) must be an even integer.
The condition 1) is sufficient.

Condition 2)
\(N = 1\) or \(N = 3\).
For both cases, \(N^3/4\) is \(1/4\) or \(27/4\).
They are not integers.
Since ‘no’ is also a unique answer by CMT (Common Mistake Type) 1, the condition 2) is sufficient.

Therefore, D is the answer.
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Re: If N is a positive integer  [#permalink]

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New post 03 Mar 2019, 10:58
MathRevolution wrote:
EgmatQuantExpert wrote:
Question:

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 \(6\).

    A) Statement (1) ALONE is sufficient, but statement (2) ALONE is not sufficient.
    B) Statement (2) ALONE is sufficient, but statement (1) ALONE is not sufficient.
    C) Both statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
    D) EACH statement ALONE is sufficient.
    E) Statement (1) and (2) TOGETHER are NOT sufficient.


Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

The first step of the VA (Variable Approach) method is to modify the original condition and the question, and then recheck the question.

We can modify the original condition and question as follows.

The question that \(N^3/4\) is an integer is equivalent to \(N\) is an even integer.

Condition 1)
In order for \(N^2 + 3\) to an prime number, \(N^2\) and \(N\) must be an even integer.
The condition 1) is sufficient.

Condition 2)
\(N = 1\) or \(N = 3\).
For both cases, \(N^3/4\) is \(1/4\) or \(27/4\).
They are not integers.
Since ‘no’ is also a unique answer by CMT (Common Mistake Type) 1, the condition 2) is sufficient.

Therefore, D is the answer.


For condition 2, why did you state that N=1 or N=3? Condition 2 says N is the number of odd factors of 6. The number of odd factors of 6 is 2 (3 and 1).
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Re: If N is a positive integer  [#permalink]

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New post 05 Apr 2019, 01:23
Alternative approach to statement I:

any prime >3 can be expressed as 6n+1, hence, N^2+3=6n+1=6n-2, now, 6n-2= 2(3n-1), hence N^2 will always be divisible by 2^2

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Re: If N is a positive integer   [#permalink] 05 Apr 2019, 01:23
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