Bunuel wrote:
If n is an integer greater than 1, is 3^n-2^n divisible by 35?
(1) n is divisible by 15.
(2) n is divisible by 18.
Explanation to follow.
This particular question of defining whether the expression would be an integer value or not can be solved by using the cyclicity rule:
first factorize 35: 7*5 , and we know the unit cyclicity of 3 & 2 is 4; viz 3^1:3, 3^2:9: 3^3: 7: 3^4:1, 3^5:3 and for 2 ; 2^1:2 , 2^2:4, 2^3: 8, 2^4: 6,2^5:2..
from stmt 1: we can say that n can be 15,30,45,60 ..
therefore 3^n-2^n to be divisible by 7*5 it has to have a unit digit either 5 or 0, which is a yes and no case , so it is insufficient
for stmt 1: n would be 18,36,54..
and for all values we get unit digits as divisible by 5, hence yes and sufficient ..
IMO B is correct...