Apologies for raising the doubt, are you sure you have posted the correct question?

As per the given fact, x=5k+2, where k is an integer.

Now, only x/17 can result into an integer when k=3

2logj wrote:If when a certain integer x is divided by 5
the reminder is 2,then all the following could
also be an integer except
A X/17
B X/11
C X/10
D X/6
E X/3

x=5k+2; when k=3; x=17 which is divisible by 17,
when k=15, x=77, which is divisible by 11;
when k=2, x=12 which is divisible by 6;
when k=2, x=12 which is divisible by 3,
hence answer should be C, although there is no need to go for all these steps,we can straight away reason that last digit of 5k+2 will be either 2 or 7, depending upon whether k is even or odd, hence it will never be divisible by 10..!!!

i hope it helps...!!!

2logj wrote:If when a certain integer x is divided by 5
the reminder is 2,then all the following could
also be an integer except
A X/17
B X/11
C X/10
D X/6
E X/3

Make a list of values that satisfy the conditions given.
On the GMAT, this problem likely would be restricted to positive integers.

When X is divided by divisor D, the remainder is R means that the smallest possible value of X is R.
To generate the other possible values of X, just keep adding multiples of the divisor D to R.

In the problem above:
The smallest possible value of X is R=2.
Adding multiples of the divisor D=5 to R=2, we get the following list:

2,7,12,17,22,27...

Each of the numbers above is 2 more than a multiple of 5; thus, when each number is divided by 5, R=2.

Now compare the numbers in the list to the expressions in the answer choices:
12 is a multiple of 6 and of 3.
Eliminate D and E.
17 is a multiple of 17.
Eliminate A.
22 is a multiple of 11.
Eliminate B.

The correct answer is C.