swerve wrote:When the positive integer x is divided by 6, the remainder is 4. Each of the following could also be an integer EXCEPT

A. x/2
B. x/3
C. x/7
D. x/11
E. x/17

----ASIDE-------------------
When it comes to remainders, we have a nice rule that says:
If N divided by D leaves remainder R, then the possible values of N are R, R+D, R+2D, R+3D,. . . etc.
For example, if k divided by 5 leaves a remainder of 1, then the possible values of k are: 1, 1+5, 1+(2)(5), 1+(3)(5), 1+(4)(5), . . . etc.
----ONTO THE QUESTION!----------

When the positive integer x is divided by 6, the remainder is 4.
So, the possible values of x are: 4, 10, 16, 22, 28, 34, . . . etc

Now let's test some of these possible values of x.

If x = 4, then x/2 = 4/2 = 2, which IS an integer. ELIMINATE A
If x = 10, none of the remaining answers are integers. Keep going.
If x = 16, none of the remaining answers are integers. Keep going.
If x = 22, then x/11 IS an integer. ELIMINATE D
If x = 28, then x/7 IS an integer. ELIMINATE C
If x = 34, then x/17 IS an integer. ELIMINATE E

By the process of elimination, the answer is B

Cheers,
Brent

swerve wrote:When the positive integer x is divided by 6, the remainder is 4. Each of the following could also be an integer EXCEPT

A. x/2
B. x/3
C. x/7
D. x/11
E. x/17

APPROACH#2

When the positive integer x is divided by 6, the remainder is 4
In other words, x is 4 greater than some multiple of 6
We can write: x = 6k + 4, where k is an integer

If x = 6k + 4, we can also write: x = 6k + 3 + 1
Factor to get: x = 3(2k + 1) + 1
Since 2k+1 must be an integer, we know that 3(2k + 1) is a multiple of 3
This means 3(2k + 1) + 1 is 1 greater than some multiple of 3
This means x is NOT a multiple of 3
This means x is NOT divisible by 3

So, x/3 cannot be an integer

Answer: B

Cheers,
Brent

Good catch - thanks Fabio!
I've edited my response accordingly.

Cheers,
Brent

swerve wrote:When the positive integer x is divided by 6, the remainder is 4. Each of the following could also be an integer EXCEPT

A. x/2
B. x/3
C. x/7
D. x/11
E. x/17

The OA is B

Source: Manhattan Prep

Since, when x is divided by 6, the remainder is 4, x can be values such as:

4, 10, 16, 22, 28, 34

So we see that x/2, x/7, x/11, and x/17 can all be integers (let x be 4, 28, 22, and 34, respectively). Thus x/3 cannot be an integer.

Alternate Solution:

We can write x = 6s + 4 for some integer s. Notice that x = 6s + 4 = 6s + 3 + 1 = 3(2s + 1) + 1. Since x is one more than an integer multiple of 3 and since consecutive integers never share factors, x cannot be divisible by 3.

Answer: B