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x^2 + y^2 divided by 5

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x^2 + y^2 divided by 5 Post Wed May 16, 2012 10:23 am
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  • Lap #[LAPCOUNT] ([LAPTIME])
    If x and y are integers, what is the remainder when x^2 + y^2 is divided by 5?

    1) When x-y is divided by 5, the remainder is 1
    2) When x+y is divided by, the remainder is 2

    OA C

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    Post Wed May 16, 2012 10:55 am
    massi2884 wrote:
    If x and y are integers, what is the remainder when x^2 + y^2 is divided by 5?

    1) When x-y is divided by 5, the remainder is 1
    2) When x+y is divided by, the remainder is 2

    OA C
    As is clear that from Statement 1 and 2 that by alone, we cannot infer the result.

    So Lets combine.

    To make the calculations easier, assume x - y = 1 & x + y = 5k + 2. Where k is a natural no.

    This yields x = (5k+3)/2 & y = (5k+1)/2.

    So, x^2 + y^2 = [(5k+3)/2]^2 + [(5k+1)/2]^2 = 1/4.(50 k^2 + 40 K + 10); This expression is multiple of 5, hence remainder will be 0. Answer C.

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    Post Wed May 16, 2012 11:01 am
    If x and y are integers, what is the remainder when x^2 + y^2 is divided by 5?

    Quote:
    1) When x-y is divided by 5, the remainder is 1
    If x = 6 and y = 5 then x^2 + y^2 = 36 + 25 and the remainder when x^2 + y^2 is divided by 5 is 1
    If x = 4 and y = 3 then x^2 + y^2 = 16 + 9 = 25 and the remainder when x^2 + y^2 is divided by 5 is 0
    Quote:
    2) When x+y is divided by, the remainder is 2
    If x = 7 and y = 5 then x^2 + y^2 = 49 + 25 and the remainder when x^2 + y^2 is divided by 5 is 3
    If x = 4 and y = 3 then x^2 + y^2 = 16 + 9 = 25 and the remainder when x^2 + y^2 is divided by 5 is 0
    Quote:
    From 1 + 2
    1) When x-y is divided by 5, the remainder is 1. Implies that x - y is of the form 5k + 1, where k is a non negative integer.
    2) When x+y is divided by, the remainder is 2. Implies that x + y is of the form 5l + 2, where l is a non negative integer.
    So, x + y = 5l + 2and x - y = 5k + 1
    Squaring and adding, we get
    x^2 + y^2 + 2xy + x^2 + y^2 - 2xy = 25*l^2 + 4 + 20*l + 25*k^2 + 1 + 10k
    2*(x^2 + y^2) = 25*l^2 + 20*l + 25*k^2 + 10k + 5
    2*(x^2 + y^2) = 5(5*l^2 + 4*l + 5*k^2 + 2k + 1)
    2*(x^2 + y^2) = 5* An integer
    So, we can say that x^2 + y^2 is a multiple of 5 and that if x^2 + y^2 is divided by 5, it leaves a remainder of 0

    Hence C

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    Post Wed May 16, 2012 12:09 pm
    Quote:
    1) When x-y is divided by 5, the remainder is 1
    If (x-y)/5 -> gives a reminder 1. That means that (x-y) should be some number like 6, 11, 16, 21...etc.

    6/5 -> Remainder 1
    11/5 -> Remainder 1
    16/5 -> Remainder 1

    Quote:
    2) When x+y is divided by 5, the remainder is 2
    If (x+y)/5 -> gives a reminder 2. That means that (x+y) should be some number like 7, 12, 17, 22...etc.

    7/5 -> Remainder 2
    12/5 -> Remainder 2
    17/5 -> Remainder 2

    Quote:
    From 1 + 2
    Plugging in the values

    1) if x-y = 6 (Can be 6,11,16,21...)
    x+y = 12(Can be 7,12,17,22...)
    Solving for x
    2x = 18 x=9
    if x=9, y = 3
    (x^2 + y^2)/5 = (9^2 + 3^2)/5 = (81+9)/5 = 90/5 -> Remainder =0

    2) Cross check with another set of values
    if x-y = 11 (Can be 6,11,16, 21...)
    x+y = 17 (Can be 7,12,17,22...)
    Solving for x
    2x = 28 x=14 y=3
    (14^2 + 3^2)/5 = (196 + 9)/5 = 205/5 -> Remainder =0

    Hence Together sufficient -> C[/quote]

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    Post Tue May 22, 2012 9:55 pm
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    Last edited by vikram4689 on Tue May 22, 2012 11:05 pm; edited 1 time in total

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    Post Tue May 22, 2012 11:00 pm
    massi2884 wrote:
    If x and y are integers, what is the remainder when x^2 + y^2 is divided by 5?

    1) When x-y is divided by 5, the remainder is 1
    2) When x+y is divided by, the remainder is 2

    OA C
    (1) When x - y is divided by 5, the remainder is 1.
    x - y = 5a + 1, so x - y can be 1, 6, 11, ...
    If x = 2, y = 1, x - y = 1, then x² + y² = 5. So, remainder = 0.
    If x = 3, y = 2, x - y = 1, then x² + y² = 13. So, remainder = 3.
    No definite answer; NOT sufficient.

    (2) When x + y is divided by, the remainder is 2.
    x + y = 5b + 2, so x + y can be 2, 7, 12, ...
    If x = 1, y = 1, x + y = 2, then x² + y² = 2. So, remainder = 2.
    If x = 5, y = 2, x + y = 7, then x² + y² = 29. So, remainder = 4.
    No definite answer; NOT sufficient.

    Combining (1) and (2), x - y = 5a + 1 and x + y = 5b + 2
    (x - y)² = (5a + 1)² or x² - 2xy + y² = 25a² + 10a + 1
    (x + y)² = (5b + 2)² or x² + 2xy + y² = 25b² + 20b + 4
    Adding the 2 equations, we get
    2(x² + y²) = 5(5a² + 5b² + 2a + 4b + 1), which clearly implies that 2(x² + y²) is divisible by 5 with remainder = 0 and so x² + y² is also divisible by 5 with remainder = 0; SUFFICIENT.

    The correct answer is C.

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    Post Wed May 23, 2012 7:38 am
    (C) QED

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    Post Tue Dec 15, 2015 9:16 am
    IMO this question is a walk in the park if we know the following rules:

    You can always ignore the original number for remainder calculation. If you know a/n gives remainder p and b/n gives remainder of q then-
    i) remainder of (a+b)/n = remainder of (p+q)/n
    ii) remainder of (a-b)/n = remainder of (p-q)/n
    iii) remainder of (a^2+b^2)/n = remainder of (p^2+q^2)/n
    iv) remainder of (a^2-b^2)/n = remainder of (p^2-q^2)/n
    v) remainder of (a*b)/n = remainder of (p*q)/n
    vi) remainder of (a/b)/n = remainder of (p/q)/n
    .
    .
    .
    etc

    But Im not quite sure about the accuracy of these rules.

    So, Experts:

    How do the above rules look neat or preposterous ?Wink Any corrections, additions, exceptions? Thanks

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    Post Thu Dec 17, 2015 11:09 am
    Can some expert comment on this, please? Thanks

    Nina1987 wrote:
    IMO this question is a walk in the park if we know the following rules:

    You can always ignore the original number for remainder calculation. If you know a/n gives remainder p and b/n gives remainder of q then-
    i) remainder of (a+b)/n = remainder of (p+q)/n
    ii) remainder of (a-b)/n = remainder of (p-q)/n
    iii) remainder of (a^2+b^2)/n = remainder of (p^2+q^2)/n
    iv) remainder of (a^2-b^2)/n = remainder of (p^2-q^2)/n
    v) remainder of (a*b)/n = remainder of (p*q)/n
    vi) remainder of (a/b)/n = remainder of (p/q)/n
    .
    .
    .
    etc

    But Im not quite sure about the accuracy of these rules.

    So, Experts:

    How do the above rules look neat or preposterous ?Wink Any corrections, additions, exceptions? Thanks

    Post Thu Dec 17, 2015 2:05 pm
    HI Nina1987,

    One of the great aspects of the GMAT is that you can approach most questions in a variety of different ways. To that end, 'your way' of doing things is fine as long as it helps you to hit your score goals.

    1) How have you been scoring on your CATs (including the Quant and Verbal Scaled Scores)?
    2) What is your goal score?

    3) How long have you been studying?
    4) When are you planning to take the GMAT?

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    Post Mon Dec 21, 2015 9:20 pm
    Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.


    If x and y are integers, what is the remainder when x^2 + y^2 is divided by 5?

    1) When x-y is divided by 5, the remainder is 1
    2) When x+y is divided by 5, the remainder is 2

    In the original condition, there are 2 variables(x,y), which should match with the number of equations. So you need 2 more equations. For 1) 1 equation, for 2) 1 equation, which is likely to make C the answer. In 1) & 2), it becomes x-y=1,6,11,..... x+y=2,7,12... and you get x=4, y=3 from it. So, from x^2+y^2=4^2+3^2=25, the remainder of 25 divided by 5 is 0, which is unique and sufficient. Therefore, the answer is C.


    -> For cases where we need 2 more equations, such as original conditions with “2 variables”, or “3 variables and 1 equation”, or “4 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 70% chance that C is the answer, while E has 25% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, D or E.

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    Post Sun Mar 20, 2016 5:36 pm
    I substituting number is a time consuming and lengthy process. Atleast for me Smile My way of solving this

    (X+Y) =5m+1
    X-Y = 5m+2
    We need x ^2 +Y^2

    and when we square it all the variables will contain 5m except the last constant .

    so without expannding it we will knw the value .
    i.e
    1+4 =5 remeinder of 5 is zero

    option C

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