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 [spoiler]BCE; AD[/spoiler]Remains.
(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
If x is a positive integer, is sqrt(x) an integer?
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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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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.
Ricardo
Thank you very much for the explanation. It was very clear. Also, the concept you mentioned will definitely save me time on the exam.
Regards,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.
Ricardo
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You're welcome.
It's a frequently-tested concept, so good to know.
Cheers,
Brent
It's a frequently-tested concept, so good to know.
Cheers,
Brent
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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
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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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
Hi Amrabdelnaby,
I am not Brent 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
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 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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Oh my!
I missed that part of the question!
what a shame
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
I missed that part of the question!
what a shame
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
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 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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√x CAN NEVER equal something.5Amrabdelnaby 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
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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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.
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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Solution:
Question Stem Analysis:
We need to determine whether √x is an integer, given that x is a positive integer. Notice that √x is an integer if x is a perfect square.
Statement One Alone:
Since √(4x) = 2√x is an integer, √x is either itself an integer or it is 1/2 of some integer. If √x is the former case, we are done. If √x is the latter case, we can let √x = k/2 where k is some positive integer. Squaring both sides, we have:
x = k^2/4
Since x is a positive integer, we see that k has to be even. If k is even, then k/2, or √x, will be an integer. Statement one alone is sufficient.
Statement Two Alone:
Statement two is not sufficient. For example, if x = 4, then √x = 2 is an integer (notice that √(3x) = √12 is not an integer). However, if x = 5, then √x = √5 is not integer (notice that √(3x) = √15 is also not an integer). Statement two alone is not sufficient.
Answer: A
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