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
When the positive integer x is divided by 6, the remainder
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----ASIDE-------------------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
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
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APPROACH#2swerve 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
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
Last edited by Brent@GMATPrepNow on Mon Nov 26, 2018 11:30 am, edited 1 time in total.
- fskilnik@GMATH
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Brent´s second solution is the one I value most.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
Source: Manhattan Prep
Just a small correction: x = 6k + 3 + 1 = 3 (2k+1) + 1.
(Since Q=2k+1 must be an integer, x=3Q+1 is not divisible by 3 and it is done.)
Regards,
Fabio.
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Good catch - thanks Fabio!
I've edited my response accordingly.
Cheers,
Brent
I've edited my response accordingly.
Cheers,
Brent
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- Scott@TargetTestPrep
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Since, when x is divided by 6, the remainder is 4, x can be values such as: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
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
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