If when a certain integer x is divided by 5, the remainder

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If when a certain integer x is divided by 5, the remainder is 2, then each of the following could also be an integer EXCEPT:

A. x/17
B. x/10
C. x/6
D. x/3
E. x/11

The OA is B

Source: Princeton Review

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by GMATGuruNY » Tue Jan 22, 2019 2:24 pm
swerve wrote:If when a certain integer x is divided by 5, the remainder is 2, then each of the following could also be an integer EXCEPT:

A. x/17
B. x/10
C. x/6
D. x/3
E. x/11
Since dividing x by 5 yields a remainder, x is not a multiple of 5 and thus not a multiple of 10.
Since x is not a multiple of 10, x/10 cannot yield an integer value.

The correct answer is B.
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by Brent@GMATPrepNow » Tue Jan 22, 2019 2:32 pm
swerve wrote:If when a certain integer x is divided by 5, the remainder is 2, then each of the following could also be an integer EXCEPT:

A. x/17
B. x/10
C. x/6
D. x/3
E. x/11

The OA is B

Source: Princeton Review
Here's another (longer) approach:

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.

When a certain integer x is divided by 5, the remainder is 2
So, the possible values of x are as follows: 2, 7, 12, 17, 22, 27, 32, . . . etc
Let's use these values to ELIMINATE answer choices

If x = 12, then x/3 is an integer. ELIMINATE D
If x = 12, then x/6 is an integer. ELIMINATE C
If x = 17, then x/17 is an integer. ELIMINATE A
If x = 22, then x/11 is an integer. ELIMINATE E

By the process of elimination, the correct answer is B

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Brent
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by fskilnik@GMATH » Wed Jan 23, 2019 3:30 am
swerve wrote:If when a certain integer x is divided by 5, the remainder is 2, then each of the following could also be an integer EXCEPT:

A. x/17
B. x/10
C. x/6
D. x/3
E. x/11
Source: Princeton Review
$$x = 5M + 2,\,\,\,M\,\,\operatorname{int} \,\,\,\,\,\,\left( * \right)$$
$$?\,\,\,:\,\,\,{\text{not}}\,\,\,{\text{integer}}$$

$$\left( {\text{A}} \right)\,\,M = 3\,\,{\text{refutes}}$$
$$\left( {\text{B}} \right)\,\,\frac{x}{{10}} = \operatorname{int} \,\,\,\mathop \Rightarrow \limits^{ \cdot \,\,2} \,\,\,\,\frac{x}{5} = \operatorname{int} \,\,\,\, \Rightarrow \,\,\,x\,\,{\text{divisible}}\,\,{\text{by}}\,\,5\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,\,{\text{impossible}}$$


We follow the notations and rationale taught in the GMATH method.

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by Scott@TargetTestPrep » Sun Jan 27, 2019 6:38 pm
swerve wrote:If when a certain integer x is divided by 5, the remainder is 2, then each of the following could also be an integer EXCEPT:

A. x/17
B. x/10
C. x/6
D. x/3
E. x/11

The OA is B

Source: Princeton Review
Since, when x is divisible by 5, the remainder is 2, x can be values such as:

2, 7, 12, 17, 22

So we see that x/17, x/6, x/3, and x/11 can all be integers. Thus x/10 cannot be an integer.

Answer: B

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