If x is a positive integer, is x^(1/2) an integer

SOLUTION:

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

The square root of any positive integer is either an integer or an irrational number. So, \(\sqrt{x}=\sqrt{integer}\) cannot be a fraction, for example it cannot equal to 1/2, 3/7, 19/2, ... It MUST be an integer (0, 1, 2, 3, ...) or irrational number (for example \(\sqrt{2}\), \(\sqrt{3}\), \(\sqrt{17}\), ...). So, as given that \(x\) is a positive integer then \(\sqrt{x}\) is either an integer itself or an irrational number.


(1) \(\sqrt{4x}\) is an integer --> \(2\sqrt{x}=integer\) --> \(2\sqrt{x}\) to be an integer \(\sqrt{x}\) must be an integer or integer/2, but as \(x\) is an integer, then \(\sqrt{x}\) cannot be integer/2, hence \(\sqrt{x}\) is an integer. Sufficient.

(2)\(\sqrt{3x}\) is not an integer --> if \(x=9\), condition \(\sqrt{3x}=\sqrt{27}\) is not an integer satisfied and \(\sqrt{x}=3\) IS an integer, BUT if \(x=2\), condition \(\sqrt{3x}=\sqrt{6}\) is not an integer satisfied and \(\sqrt{x}=\sqrt{2}\) IS NOT an integer. Two different answers. Not sufficient.

Answer: A.

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

(1)\(\sqrt{4x}\) is an integer.
(2)\(\sqrt{3x}\) is not an integer.


Statement 1

\(\sqrt{4x} = k\), \(k\) is an integer
Or,\(\sqrt{x}= k/2\)

If \(k/2\) is not an integer, then \((k/2)^2\) or \(k^2/4\) is also not an integer and this implies that x is not an integer as \(x = k^2/4\).
But, it is given that \(x\) is an integer.

Therefore, \(k^2/4\) is an integer => \(k/2\) is an integer =>\(\sqrt{x}\) is an integer...........Sufficient....(B)(C)(E)


Statement 2

\(\sqrt{3x}=k\), \(k\) is not an integer
Or, \(\sqrt{x} = k/\sqrt{3}\)

Thus, \(\sqrt{x}\) = integer, when \(k\) is a multiple of \(\sqrt{3}\)
and \(\sqrt{x}\) is not an integer when \(k\) is not a multiple of \(\sqrt{3}\)

As the value of \(\sqrt{x}\) can not be uniquely determined, statement (2) is not sufficient.................(D)

Answer: (A)

Hello
Easiest way i can think of:
S1-\(\sqrt{4x}\) is an integer =\(\sqrt{4}*\sqrt{x}\) has to be integer
We know that \(\sqrt{4}\) is 2 - int.
so\(\sqrt{x}\) has to be int ( because Non Integer*integers is ALWAYS a Non integer, which is not the case here)
Sufficient (12TEN) or (ABCDE)

S2- \(\sqrt{3x}\) is NOT an integer = \(\sqrt{3}*\sqrt{x}\) is not an integer
We know that \(\sqrt{3}\) is not an integer
But Non-Integer * Integer OR Non-integer = Non- integer
So Insufficient (1E) or (AD)

thoughts?

Bunuel could you please tell me if this approach is correct/incorrect?

No. That's not correct.

(Non-integer)*(integer) could yield an integer. For example, 1/2*2 = 1.

(Non-integer)*(non-integer) could also yield an integer. For example, \(\sqrt{2}*\sqrt{2}=2\).
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[quote="Bunuel"]The Official Guide For GMAT® Quantitative Review, 2ND Edition

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

(1) \(\sqrt{4x}\) is an integer.
(2) \(\sqrt{3x}\) is not an integer.

Given x>0 & x=Integer

Find is \(\sqrt{x}\) an integer
=> let \(\sqrt{x}=k\)---------- 'k' is integer
=> OR \(x=k^2\)
Therefore Question can be re-phrased as " IS 'x' A PERFECT SQUARE "

Statement 1 \(\sqrt{4x}\) is an integer
=> let \(\sqrt{4x}=p\) ---------------- 'p' is Integer
=> \(4x=p^2\)
=> \(x=\frac{p^2}{4}\)
=> OR \(x=(\frac{p}{2})^2\) --------- equ (1)
=> Now Since (given) x>0 & x=Integer
=> Therefore 'p' HAS TO BE 2,4,6,8....OR any even number >0
=> substituting 'p' in equ (1) makes \(x=1^2 or 2^2 or 3^2\) and so on..
=> Therefore x=Perfect Square.
=> SUFFICIENT

Statement 2 \(\sqrt{3x}\) is not an integer.
=> So \(x\neq{3, 3^3, 3^5...}\)

=> BUT x=1,2,4,5,6,9,....

=> So 'x' can be Prefect Square ( when x=4,9,16....) and 'x' NOT a Perfect Square (when x=2,5,6...)
=> Therefore INSUFFICIENT

Therefore 'A'

Thanks
Dinesh

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

In other words:
Is x a perfect square?

Statement 1:
Make a list of integer options for \(\sqrt{4x}\) and simplify to isolate x:
\(\sqrt{4x}\) = 1, 2, 3, 4, 5, 6, 7, 8...
\(2\sqrt{x}\) = 1, 2, 3, 4, 5, 6, 7, 8...
\(\sqrt{x}\) = 1/2, 1, 3/2, 2, 5/2, 3, 7/2, 4...
x = 1/4, 1, 9/4, 4, 25/4, 9, 49/4, 16...
Integer options yielded in the resulting list:
x = 1, 4, 9, 16...
In every case, x is a perfect square, so the answer to the question stem is YES.
SUFFICIENT.

Statement 2:
Every value in the red list above satisfies Statement 2.
If x=1/4, the answer to the question stem is NO.
If x=1, the answer to the question stem is YES.
Since the answer is NO in the first case but YES in the second case, INSUFFICIENT.


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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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BrentGMATPrepNow

For your mention of
"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." If x is 6, then there would not be an even number of primes because root(24)=2*2*2*3
That is two threes and one 3?

@avitgutman
What are your thoughts about this?
Many thanks!

woohoo921 2*2*2*3 is three twos and one 3. You're correct that there would not be an even number of each prime; both of the powers are odd (2 to the 3rd power, and 3 to the 1st power).
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If x = 6, then 4x = 24
24 = (2)(2)(2)(3), which is three 2's and one 3.
Since we have an odd number of 2's and an odd number of 3's, √24 is a non-integer

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Hi

“ as X is an integer, then sq rootX cannot be integer/2” why is this?

Because no x/2, where the fraction is reduced to its lowest terms and x is a nonzero integer, yields an integer when squared.
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