Is n divisible 35? 1) 4n^2/15 = Integer 2)2n/7=Integer

Question

Is n divisible 35? 1) \frac{4n^2}{15} = Integer 2) \frac{2n}{7} = Integer Source : https://www.GMATinsight.com

Archit3110 · Answer

prime factorize and solve
35 = 5*7
so value of n should have factors of 35 to test its validity of being divisible by 35
#1
 \frac{4n^2}{15} = Integer 
here least value of n has to be 15 or a multiple value of 15 which can include 7 as well so insufficient
#2
 \frac{2n}{7} = Integer 
least value of n has to be 7 or its multiple insufficient
from 1 &2 
n has to be multiple of 3,5,7 , n will be divisible by 35
Or value of n is 105/2 which gives no 
Option E

1)  \frac{4n^2}{15} = Integer 
2)  \frac{2n}{7} = Integer

wali786 · Answer

n has to be multiple of 3,5,7 , n will be divisible by 35
Or value of n is 105/2 which gives no
Option E

Can you please explain how did you get 105/2?

agv2501 · Answer

If both statements prove that n has at least one factor of 3,5 and 7 each, then how is the OA E? shouldnt it be C? Please explain

TestPrepUnlimited · Answer

To be divisible by 35, we need n to be divisible by 5 and 7 simultaneously.  Note we are not given that n is an integer.

Statement 1:

We can rewrite this as  n^2 = Integer * \frac{15}{4} , so we are allowed to take values such as  n^2 = \frac{15}{4} , which is not div by 35. On the other hand, n could contain as many factors as we want by stuffing it all in the "integer" portion, then n could still be divisible by 35, insufficient.

Statement 2:

Similar as above,  n = Integer * \frac{7}{2}  so we could have  n = \frac{7}{2}  or n =35*7 if we let  Integer = 2*35 . Insufficient.

Combined:

Combined we have  n^2 = I * \frac{15}{4}  and  n = J * \frac{7}{2}  (note I use different variables now). Square the second equation to get  n^2 = J^2 * \frac{49}{4} . Thus we can let J = 15, and I = whatever we need to fill in, and n could still be a fraction. Insufficient.

Ans: E

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Is n divisible 35? 1) 4n^2/15 = Integer 2)2n/7=Integer

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