If N is a positive integer

Question

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.

This is  Question 10 of  The e-GMAT Number Properties Marathon

Go to   Question 11 of the Marathon

EgmatQuantExpert · Accepted Answer

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

Hatakekakashi · Answer

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

Mo2men · Answer

Hi  Hatakekakashi

Pleasse review highlighted

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

Hatakekakashi · Answer

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

MathRevolution · Answer

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.

Neeyah · 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).

SchruteDwight · Answer

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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paikyang · Answer

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

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

Any prime above 2 is odd, therefore for N^2+3 = Prime, N has to be Even (only even^2 = even)

Even N means N is a multiple of 2

n^3 / 4 = 2k * 2k * 2k / 2*2

The 2's cancel out which means N^3/4 is an Integer (SUFFICIENT)

2. N is the number of odd factors of 6.

Factors of 6 = 1 * 6, 2 * 3. Odd factors = 1 & 3

2 number of odd factors. Therefore n = 2

n^3 / 4 = 2 * 2 * 2 / 2 *2

again the 2's cancel out which means N^3/4 is an Integer (SUFFICIENT)

Answer : D

robu1 · Answer

For statement 2 , we get definite value of N . No need to calculate. It’s sufficient.

bumpbot · Answer

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If N is a positive integer

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GMAT Data Sufficiency (DS) Questions

If N is a positive integer

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