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If x is a positive integer, is sqrt(x) an integer?

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Rospino Just gettin' started!
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If x is a positive integer, is sqrt(x) an integer? Post Mon Mar 31, 2014 4:47 pm
Elapsed Time: 00:00
  • Lap #[LAPCOUNT] ([LAPTIME])
    If x is a positive integer, is sqrt(x) an integer?
    (1) sqrt(4x) is an integer
    (2) sqrt(3x) is not an integer

    OA: A

    I used the plug in number approach. However, this problem took me about 3.5 mins - 1.5 mins too long. Any quicker ways?
    (1) sqrt(4x) is an integer
    a. if x = 1, then sqrt(4x) = 2, then sqrt(x) = 1, which is an integer.
    b. if x = 4, then sqrt(4x) = 4, then sqrt(x) = 2, which is an integer.
    c. if x = 9, then sqrt(4x) = 6, then sqrt(x) = 3, which is an integer.

    Sufficient. Eliminate BCE; ADRemains.

    (2) sqrt(3x) is not an integer
    a. if x = 2, then sqrt(3x) = sqrt(6), then sqrt(x) = sqrt(2), which is not an integer.
    b. if x = 4, then sqrt(3x) = sqrt(12), then sqrt(x) = sqrt(4), which is an integer.

    Insufficient. Eliminate D

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    Post Mon Mar 31, 2014 4:53 pm
    Rospino wrote:
    If x is a positive integer, is √x an integer?
    (1) √(4x) is an integer
    (2) √(3x) is not an integer
    Target question: Is √x an integer?

    Given: x is a positive integer

    Statement 1: √(4x) is an integer
    IMPORTANT CONCEPT: If K is an integer, then √K will be an integer if the prime factorization of K has an even number of each prime.
    Some examples:
    √144 = 12 (integer), and 144 = (2)(2)(2)(2)(3)(3) [four 2's and two 3's]
    √1600 = 40 (integer), and 1600 = (2)(2)(2)(2)(2)(2)(5)(5) [six 2's and two 5's]
    √441 = 21 (integer), and 441 = (3)(3)(7)(7)[two 3's and two 7's]
    √12 = some non-integer, and 12 = (2)(2)(3)[two 2's and one 3's]

    So, if √(4x) is an integer, then the prime factorization of 4x has an even number of each prime.
    Since 4x = (2)(2)(x) we can see that the prime factorization of x must have an even number of each prime.
    If the prime factorization of x has an even number of each prime, then √x must be an integer.
    Since we can answer the target question with certainty, statement 1 is SUFFICIENT

    Statement 2: √(3x) is not an integer.
    There are several values of x that meet this condition. Here are two:
    Case a: x = 4. This means that √(3x) = √12, which is not an integer. In this case, √x is an integer.
    Case b: x = 5. This means that √(3x) = √15, which is not an integer. In this case, √x is not an integer.
    Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

    Answer = A

    Cheers,
    Brent

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    Rospino Just gettin' started!
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    Post Mon Mar 31, 2014 5:04 pm
    Brent,

    Thank you very much for the explanation. It was very clear. Also, the concept you mentioned will definitely save me time on the exam.

    Quote:
    IMPORTANT CONCEPT: If K is an integer, then √K will be an integer if the prime factorization of K has an even number of each prime.
    Regards,
    Ricardo

    Post Mon Mar 31, 2014 5:05 pm
    You're welcome.
    It's a frequently-tested concept, so good to know.

    Cheers,
    Brent

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    Amrabdelnaby Really wants to Beat The GMAT! Default Avatar
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    Post Sun Nov 22, 2015 2:00 pm
    Brent,

    Question though!

    Now √4x is equal to 2√x, so if √x yields to 1.5 or 2.5 or 3.5 etc... √4x or 2√x will be always an integer.

    so how come this statement is sufficient?

    why did we assume that √x must be an integer if √4x is an integer although √4x =2√x

    Brent@GMATPrepNow wrote:
    Rospino wrote:
    If x is a positive integer, is √x an integer?
    (1) √(4x) is an integer
    (2) √(3x) is not an integer
    Target question: Is √x an integer?

    Given: x is a positive integer

    Statement 1: √(4x) is an integer
    IMPORTANT CONCEPT: If K is an integer, then √K will be an integer if the prime factorization of K has an even number of each prime.
    Some examples:
    √144 = 12 (integer), and 144 = (2)(2)(2)(2)(3)(3) [four 2's and two 3's]
    √1600 = 40 (integer), and 1600 = (2)(2)(2)(2)(2)(2)(5)(5) [six 2's and two 5's]
    √441 = 21 (integer), and 441 = (3)(3)(7)(7)[two 3's and two 7's]
    √12 = some non-integer, and 12 = (2)(2)(3)[two 2's and one 3's]

    So, if √(4x) is an integer, then the prime factorization of 4x has an even number of each prime.
    Since 4x = (2)(2)(x) we can see that the prime factorization of x must have an even number of each prime.
    If the prime factorization of x has an even number of each prime, then √x must be an integer.
    Since we can answer the target question with certainty, statement 1 is SUFFICIENT

    Statement 2: √(3x) is not an integer.
    There are several values of x that meet this condition. Here are two:
    Case a: x = 4. This means that √(3x) = √12, which is not an integer. In this case, √x is an integer.
    Case b: x = 5. This means that √(3x) = √15, which is not an integer. In this case, √x is not an integer.
    Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

    Answer = A

    Cheers,
    Brent

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    Post Sun Nov 22, 2015 2:31 pm
    Now √4x is equal to 2√x, so if √x yields to 1.5 or 2.5 or 3.5 etc... √4x or 2√x will be always an integer.

    so how come this statement is sufficient?

    why did we assume that √x must be an integer if √4x is an integer although √4x =2√x

    Brent@GMATPrepNow wrote:
    Rospino wrote:
    If x is a positive integer, is √x an integer?
    (1) √(4x) is an integer
    (2) √(3x) is not an integer
    Hi Amrabdelnaby,

    I am not Brent Smile but let's adress this statement: "Now √4x is equal to 2√x, so if √x yields to 1.5 or 2.5 or 3.5 etc... √4x or 2√x will be always an integer."

    If √x = 1.5
    x = (1.5)^2 = 2.25

    The prompt tells us that x is a positive number: therefore x could never be 2.25 or any non-integer

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    Post Sun Nov 22, 2015 2:38 pm
    Oh my!

    I missed that part of the question!

    what a shame Smile

    I totally agree with you CEO.. if x itself is an +ve integer and √4x is also an integer then √x must be an integer because x must contain an even number of prime factors in this case, since we eliminated the possibility of x being a fraction.

    I see what you are saying now.

    Thanks man Smile

    theCEO wrote:
    Now √4x is equal to 2√x, so if √x yields to 1.5 or 2.5 or 3.5 etc... √4x or 2√x will be always an integer.

    so how come this statement is sufficient?

    why did we assume that √x must be an integer if √4x is an integer although √4x =2√x

    Brent@GMATPrepNow wrote:
    Rospino wrote:
    If x is a positive integer, is √x an integer?
    (1) √(4x) is an integer
    (2) √(3x) is not an integer
    Hi Amrabdelnaby,

    I am not Brent Smile but let's adress this statement: "Now √4x is equal to 2√x, so if √x yields to 1.5 or 2.5 or 3.5 etc... √4x or 2√x will be always an integer."

    If √x = 1.5
    x = (1.5)^2 = 2.25

    The prompt tells us that x is a positive number: therefore x could never be 2.25 or any non-integer

    Post Sun Nov 22, 2015 2:38 pm
    Amrabdelnaby wrote:
    Brent,

    Question though!

    Now √4x is equal to 2√x, so if √x yields to 1.5 or 2.5 or 3.5 etc... √4x or 2√x will be always an integer.

    so how come this statement is sufficient?

    why did we assume that √x must be an integer if √4x is an integer although √4x =2√x

    √x CAN NEVER equal something.5
    If √x WERE to equal something.5, then x = (something.5)² = somenumber.25
    Here we have a contradiction, because the given information that says x is an INTEGER, and an integer cannot end in .25

    Cheers,
    Brent

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    Post Fri Nov 27, 2015 1:35 am
    An algebraic way of seeing this:

    Suppose that √x = m + .5, where x and m are integers.

    Squaring both sides, we have

    x = m² + m + .25

    But this says x = (integer)² + (integer) + .25, a contradiction. Hence x and m cannot both be integers.

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