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Remainder

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outty Just gettin' started!
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Remainder Post Mon May 19, 2014 2:40 pm
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  • Lap #[LAPCOUNT] ([LAPTIME])
    The integers x and y are both positive, the remainder when x is divided by 12 is 7, and the remainder when y is divided by 12 is 3. Each of the following is a possible value of 2x+ y EXCEPT

    a 125
    b 101
    c 77
    d 63
    e 53

    d

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    Post Mon May 19, 2014 2:57 pm
    outty wrote:
    The integers x and y are both positive, the remainder when x is divided by 12 is 7, and the remainder when y is divided by 12 is 3. Each of the following is a possible value of 2x+ y EXCEPT

    a 125
    b 101
    c 77
    d 63
    e 53

    d
    When x is divided by 12, the remainder is 7.
    Thus, x is a MULTIPLE OF 12 plus 7:
    x = 12a + 7.

    When y is divided by 12, the remainder is 3.
    Thus, y is a MULTIPLE OF 12 plus 3:
    y = 12b + 3.

    Thus:
    2x + y =

    = 2(12a + 7) + 12b + 3

    = 24a + 14 + 12b + 3

    = 24a + 12b + 12 + 5

    = 12(2a + b + 1) + 5.

    Implication:
    2x + y is a MULTIPLE OF 12 plus 5.

    A: 125 = 12*10 + 5.
    B: 101 = 12*8 + 5.
    C: 77 = 12*6 + 5.
    D: 63 = 12*5 + 3.
    E: 53 = 12*4 + 5.

    Only D is not a multiple of 12 plus 5.

    The correct answer is D.

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    Post Mon May 19, 2014 3:51 pm
    outty wrote:
    The integers x and y are both positive, the remainder when x is divided by 12 is 7, and the remainder when y is divided by 12 is 3. Each of the following is a possible value of 2x+ y EXCEPT

    a 125
    b 101
    c 77
    d 63
    e 53
    Another approach is to test values.

    IMPORTANT: 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.

    Okay, onto the question....

    The remainder when x is divided by 12 is 7
    Possible values of x = 7, 19, 31, 43, 55, ...
    So, possible values of 2x = 14, 38, 62, 86, 110, ...

    The remainder when y is divided by 12 is 3
    Possible values of y = 3, 15, 28, 40, 52, ...

    Each of the following is a possible value of 2x + y EXCEPT

    A) 125 This equals 110 + 15. ELIMINATE A
    B) 101 This equals 86 + 15. ELIMINATE B
    C) 77 This equals 62 + 15. ELIMINATE C
    D) 63
    E) 53 This equals 38 + 15. ELIMINATE E

    Answer: D

    Cheers,
    Brent

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    Post Mon May 19, 2014 5:31 pm
    Hi outty,

    I'm a big fan of TESTing VALUES on this question. Brent's approach showcases this tactic nicely. There is one aspect I would add to it: since this is an EXCEPT question, once you find the exception, you can stop working.

    I would start with answer E because it's smallest (so it would have the least number of possible sums that could equal it).

    53 = 38 + 15. It's possible, so it's NOT what we're looking for.

    63 though…using the possible values of 2X…

    2X = 14; Y would have to be 49.
    2X = 38; Y would have to be 25.
    2X = 62; Y would have to be 1.

    Y cannot be any of these options, so 63 is the option that is NOT possible…

    Final Answer: D

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    Post Thu Dec 24, 2015 12:33 pm
    Hi Brent,

    I think that there is a little mistake in your calculations here:

    The remainder when y is divided by 12 is 3
    Possible values of y = 3, 15, 28, 40, 52, ...

    possible values of y should be 3, 15, 27,39,51,63. etc.....

    You added 13 instead of 12 to 15 making you reach 28 instead of 27 and hence whatever followed was based on that, hence the following numbers need to be revised.

    Cheers



    Brent@GMATPrepNow wrote:
    outty wrote:
    The integers x and y are both positive, the remainder when x is divided by 12 is 7, and the remainder when y is divided by 12 is 3. Each of the following is a possible value of 2x+ y EXCEPT

    a 125
    b 101
    c 77
    d 63
    e 53
    Another approach is to test values.

    IMPORTANT: 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.

    Okay, onto the question....

    The remainder when x is divided by 12 is 7
    Possible values of x = 7, 19, 31, 43, 55, ...
    So, possible values of 2x = 14, 38, 62, 86, 110, ...

    The remainder when y is divided by 12 is 3
    Possible values of y = 3, 15, 28, 40, 52, ...

    Each of the following is a possible value of 2x + y EXCEPT

    A) 125 This equals 110 + 15. ELIMINATE A
    B) 101 This equals 86 + 15. ELIMINATE B
    C) 77 This equals 62 + 15. ELIMINATE C
    D) 63
    E) 53 This equals 38 + 15. ELIMINATE E

    Answer: D

    Cheers,
    Brent

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