If n is an integer, is n odd? (1) n/3 is divisible by 3 (2)

Question

If  n  is an integer, is  n  odd?
(1)  \frac{n}{3}  is divisible by  3  
(2)  2n  has twice as many factors as  n

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s1lntz · Answer

If n/3=3, n is odd and n/3 is divisible by 3. If n/3=6, n is even and n/3 is divisible by 3. So statement 1 is insufficient.

The only way I can think for statement 2 to be true is prime numbers other than 2.
if n=5 (factors 1, 5); 2n=10 (factors 1, 2, 5, 10)
if n=7 (factors 1, 7); 2n=14 (factors 1, 2, 7, 14)
So, statement 2 alone is sufficient. answer is B.

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. Visit https://www.mathrevolution.com/gmat/lesson for details. Since we have 1 variable ( n ) and 0 equations, D is most likely to be the answer. So, we should consider each condition on its own first. Condition 1) We have an expression \frac{n}{3} = 3k for some integer k . Then we have n = 3^2k . If k = 1 , then n = 3^2 = 9 , n is an odd number and the answer is 'yes'. If k = 2 , then n = 3^2 \cdot 2 = 18 , n is an even number and the answer is 'no'. Since condition 1) does not yield a unique solution, it is not sufficient. Condition 2) Assume n = p^a \cdot q^b \cdot r^c where p, q and r are different prime numbers. Then the number of factors of n is (a+1)(b+1)(c+1) . If p, q and r are odd prime numbers, then 2n = 2^1 \cdot p^a \cdot q^b \cdot r^c and the number of factors of 2n is (1+1)(a+1)(b+1)(c+1) = 2(a+1)(b+1)(c+1) which is twice as many factors as n . Assume p = 2 which means one of prime factors of n is 2 . Then 2n = 2p^a \cdot q^b \cdot r^c = 2^{a+1} \cdot q^b \cdot r^c and the number of factors of 2n is (a+2)(b+1)(c+1) . We have (a+2)(b+1)(c+1) = 2(a+1)(b+1)(c+1) or a+2 = 2a + 2 . Then we have a = 0 and n is an odd number. Since condition 2) yields a unique solution, it is sufficient. Therefore, D is the answer. If the original condition includes “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations” etc., one more equation is required to answer the question. If each of conditions 1) and 2) provide an additional equation, there is a 59% chance that D is the answer, a 38% chance that A or B is the answer, and a 3% chance that the answer is C or E. Thus, answer D (conditions 1) and 2), when applied separately, are sufficient to answer the question) is most likely, but there may be cases where the answer is A,B,C or E.

chetan2u · Answer

(1)  \frac{n}{3}  is divisible by  3  
 \frac{n}{3}=3x , where x is an integer. 
 n=9x 
n could be even or odd depending on x. 
Insufficient

(2)  2n  has twice as many factors as  n 
Logically: Each factor of n including 1 will get multiplied by 2 when there is no 2 present in n as factor. But if n is even, multiplying with 2 will not double the factors as 1*2 is 2, but 2 as a factor would already be there. Thus, n is odd. 
Number properties: 
 Say n is odd  and equal to  a^3b^2 , where a and b are odd prime numbers.  You could take anything else but the answer will remain the same.  
So, number of factors of n are  (3+1)(2+1)=12 
2n= 2^1*a^3b^2 , so number of factors =  (1+1)(3+1)(2+1)=24 , double of 12. 
 Say n is even  and equal to  2^3b^2 , where b is odd prime number.  You could take anything else but the answer will remain the same.  
So, number of factors of n are  (3+1)(2+1)=12 
2n= 2^1*2^3b^2=2^4b^2 , so number of factors =  (4+1)(2+1)=15 , which is not double of 12. 
Thus, n is odd.

B

If n is an integer, is n odd? (1) n/3 is divisible by 3 (2)

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GMAT Data Sufficiency (DS) Questions

If n is an integer, is n odd? (1) n/3 is divisible by 3 (2)

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