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from i, if n= 2, [(N^x)-2N] is not divisible by 3. if n= 3, [(N^x)-2N] is divisible by 3.
from ii, if x= 2, [(N^x)-2N] is divisible by 3. if x= 3, [(N^x)-2N] is not divisible by 3.

the answer is C now that i read the thread and discovered that the posted question was wrong

A) X=3 => 5(3^3)-2N
45-2N.
Now, 45 is divisible by 3 but 2N?? huh ... N can be 1 so it makes indivisible
N=2 makes indivisible .. but N=3 makes it divisible(39) so I can't say from A

B) N=5 => 5(5^x)-10
5(5^x-2)
let take x = 1 => 5(5^1 - 2) => 13 not divisible by 3
x = 2 => 5(5^2 - 2) => 38 not divisible
x = 3 => not divisible
x= 4 => divisible!!

I can't really say from B as well !!

plugin both values of x and N i get 615 => Divisible!! so I need both conditions!!

(1) x=3 ---> 5( N^3) - 2N = 3N^3 + 2N^3 - 2N
We have: 3N^3 is divisible by 3, now consider
2N^3-2N = 2 N( N-1) ( N+1) . The product of three consecutive number must be divisible by 3 ---> 2N^3- 2N is divisible by 3

---> 5 N^3 - 2N is divisible by 3 --> answer to the question : YES!
--->suff

(2) not suff as proved by others.

---> A it is.

BUT, it's the case in which N is integer. How's if N is not an integer?!!

laxieqv, you said that the product of 3 consecutive integers is always divisible by 3.
what if N=0 ?then we have (N-1)N(N+1)=0
if N=1, we have: (N-1)N(N+1)=0
If N=-1, which is also an integer, (N-1)N(N+1)=0

I know that the sum of 3 consecutive integers will always be divisible by 3, but it's not so obvious that the product of 3 consecutive integers will be too, unless we exclude the 0 among our consecutive integers.

(1) x=3 ---> 5( N^3) - 2N = 3N^3 + 2N^3 - 2N We have: 3N^3 is divisible by 3, now consider 2N^3-2N = 2 N( N-1) ( N+1) . The product of three consecutive number must be divisible by 3 ---> 2N^3- 2N is divisible by 3

---> 5 N^3 - 2N is divisible by 3 --> answer to the question : YES! --->suff

(2) not suff as proved by others.

---> A it is.

BUT, it's the case in which N is integer. How's if N is not an integer?!!

Anyways, A is the OA.

That OA seems to assume N and x would take on integer values and not fractional. But nothing in the stem suggests that this should be the case....