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805+ Level|   Number Properties|   Roots|      
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Is (7x)^1/2 an integer? Updated on: 05 Jun 2013, 03:49
Originally posted by alex1233 on 13 Jan 2013, 08:22.
Last edited by Bunuel on 05 Jun 2013, 03:49, edited 3 times in total.
Renamed the topic and edited the question.
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Is \(\sqrt{7x}\) an integer?

(1) \(\sqrt{\frac{x}{7}}\) is an integer

(2) \(\sqrt{28x}\) is an integer

Show Spoiler
Can someone please explain why point 2 is not correct? What I did is that 28 is factored into => 2*2*7 so therefore I concluded x must have at least one 7. so if x has one seven then \sqrt{7x} must be an integer. Am I doing something wrong here?

Thanks!
Alex
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Is (7x)^1/2 an integer? 13 Jan 2013, 08:58
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Is \(\sqrt{7x}\) an integer?

Notice that we are not told that x is an integer.

(1) \(\sqrt{\frac{x}{7}}\) is an integer. Given that \(\sqrt{\frac{x}{7}}=integer\) --> square it: \(\frac{x}{7}=integer^2\) --> \(x=7*integer^2\). So, \(\sqrt{7x}=\sqrt{7*(7*integer^2)}=7*integer=integer\). Sufficient.

(2) \(\sqrt{28x}\) is an integer. If \(x=\frac{1}{28}\), then \(\sqrt{7x}=\frac{1}{2}\neq{integer}\) BUT if \(x=0\), then \(\sqrt{7x}=0=integer\). Not sufficient.

Answer: A.
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Is (7x)^1/2 an integer? 05 Jun 2013, 23:53
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alex1233
Is \(\sqrt{7x}\) an integer?

(1) \(\sqrt{\frac{x}{7}}\) is an integer

(2) \(\sqrt{28x}\) is an integer
Show more

From F.S 1 , we know that \(\sqrt{\frac{x}{7}}\) = Integer(I)--> \(\sqrt{\frac{x*7}{7*7}}\) = \(\sqrt{7x}*\frac{1}{7}\) = I--> \(\sqrt{7x}\) = 7*I. Thus, an integer. Sufficient.

From F.S 2, we know that \(\sqrt{28x}\) = Integer(I) --> \(\sqrt{7*4x}\) = 2*\(\sqrt{7x}\) = I.
Thus, \(\sqrt{7x}\) =\(\frac{I}{2}\). Depending on I being odd/even, the given expression will be a non-integer/integer. Insufficient.

A.
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Is (7x)^1/2 an integer? 13 Jan 2013, 09:16
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Is \(\sqrt{7x}\) an integer?

Notice that we are not told that x is an integer.

(1) \(\sqrt{\frac{x}{7}}\) is an integer. Given that \(\sqrt{\frac{x}{7}}=integer\) --> square it: \(\frac{x}{7}=integer^2\) --> \(x=7*integer^2\). So, \(\sqrt{7x}=\sqrt{7*(7*integer^2)}=7*integer=integer\). Sufficient.

(2) \(\sqrt{28x}\) is an integer. If \(x=\frac{1}{28}\), then \(\sqrt{7x}=\frac{1}{2}\neq{integer}\) BUT if \(x=0\), then \(\sqrt{7x}=0=integer\). Not sufficient.

Answer: A.
Show more


Thanks sorry for asking again... But could you please explain if my method/ logic was completely wrong? ie if we break out 28 int primes we see that it has 2,2,7 so x must have at least one 7? therefore we can assume that \sqrt{7x} should be integer?

Thanks again!
Alex
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Is (7x)^1/2 an integer? 13 Jan 2013, 09:29
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Bunuel
Is \(\sqrt{7x}\) an integer?

Notice that we are not told that x is an integer.

(1) \(\sqrt{\frac{x}{7}}\) is an integer. Given that \(\sqrt{\frac{x}{7}}=integer\) --> square it: \(\frac{x}{7}=integer^2\) --> \(x=7*integer^2\). So, \(\sqrt{7x}=\sqrt{7*(7*integer^2)}=7*integer=integer\). Sufficient.

(2) \(\sqrt{28x}\) is an integer. If \(x=\frac{1}{28}\), then \(\sqrt{7x}=\frac{1}{2}\neq{integer}\) BUT if \(x=0\), then \(\sqrt{7x}=0=integer\). Not sufficient.

Answer: A.


Thanks sorry for asking again... But could you please explain if my method/ logic was completely wrong? ie if we break out 28 int primes we see that it has 2,2,7 so x must have at least one 7? therefore we can assume that \sqrt{7x} should be integer?

Thanks again!
Alex
Show more

If x is NOT an integer you cannot make its prime factorization.

If we were told that x IS an integer, then \(\sqrt{28x}=2\sqrt{7x}=integer\). Thus x must be of the form \(7^{odd}*integer^2\), so in this case \(\sqrt{7x}=\sqrt{7*(7^{odd}*integer^2)}=\sqrt{7^{even}*integer^2}=integer\).

Hope it's clear.
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Is (7x)^1/2 an integer? 26 May 2013, 21:17
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alex1233
Is \(\sqrt{7x}\) an integer?

(1) \(\sqrt{\frac{x}{7}}\) is an integer

(2) \(\sqrt{28x}\) is an integer



Can someone please explain why point 2 is not correct? What I did is that 28 is factored into => 2*2*7 so therefore I concluded x must have at least one 7. so if x has one seven then \sqrt{7x} must be an integer. Am I doing something wrong here?

Thanks!
Alex
Show more


Quote:
\(\sqrt{7x}\) = integer ? or 7x= int^2 ?
Show more


(1) \(\sqrt{\frac{x}{7}}\) is an integer ---- lets say b
x/7 = b^2 --> where b is an integer
x= 7 *b^2
7x = (7*b) ^2
hence 7x=int^2--- sufficient


(2) \(\sqrt{28x}\) is an integer ---- lets say c

28x= c^2
7x=c^2/4 = (c/2)^2
If C is odd , then ans is No and if C is even then yes . ( If you knew that x was an integer , then this would have sufficed)
Insufficient.
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Is (7x)^1/2 an integer? 03 Feb 2014, 06:29
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(1) \(\sqrt{\frac{x}{7}}\) is integer.

Let this integer be p. then

\(\sqrt{\frac{x}{7}} = p\)

\(\frac{x}{7} = p^2\)

\(x=7*p^2\)

\(7x=7^2*p^2\)

\(\sqrt{7x}=7*p\) an integer. (1) is sufficient

(2) \(\sqrt{28x}\) is integer.

Let this integer be q.

\(\sqrt{28x} = q\)

\(\sqrt{4*7x} = q\)

\(2*\sqrt{7x} = q\)

\(\sqrt{7x} = \frac{q}{2}\)

It is given that q is integer, but \(\frac{q}{2}\) may or may not be an integer. not sufficient.

A is the soln.
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Is (7x)^1/2 an integer? 16 Apr 2015, 11:54
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Hi All,

This question can be solved by TESTing VALUES, but you have to really be thorough with your TESTs (and think about what X COULD be given the information in the Facts).

We're asked if Root(7X) is an integer. This is a YES/NO question.

Fact 1: Root(X/7) is an integer.

This Fact tells us that X has to be a specific type of multiple of 7....

IF....
X = 7
the Root(7/7) = 1 and is an integer
Root(49) = 7 and the answer to the question is YES

IF....
X = 28
the Root(28/7) = 2 and is an integer
Root(196) = 14 and the answer to the question is YES

This hints at the pattern that X will be a "perfect square times a multiple of 7', which means that the answer to the question is ALWAYS YES. You can also use prime-factorization to prove that this is the case.
Fact 1 is SUFFICIENT

Fact 2: Root(28X) is an integer.

This Fact tells us that X has to be a specific type of multiple of 7....

IF....
X = 7
the Root(196) = 14 and is an integer
Root(49) = 7 and the answer to the question is YES

IF....
X = 1/28
the Root(1) = 1 and is an integer
Root(7/28) = 1/2 and the answer to the question is NO
Fact 2 is INSUFFICIENT

Final Answer:
Show Spoiler
A

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Is (7x)^1/2 an integer? 15 May 2016, 09:45
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sahilhanda
Is \(\sqrt{(7x)}\) an integer?

(1) \(\sqrt{(x/7)}\) is an integer.

(2) \(\sqrt{(28x)}\)is an integer.

Statement (1) BY ITSELF is sufficient to answer the question, but statement (2) by itself is not.
Statement (2) BY ITSELF is sufficient to answer the question, but statement (1) by itself is not.
Statements (1) and (2) TAKEN TOGETHER are sufficient to answer the question, even though NEITHER statement BY ITSELF is sufficient.
EITHER statement BY ITSELF is sufficient to answer the question.
Statements (1) and (2) TAKEN TOGETHER are NOT sufficient to answer the question, requiring more data pertaining to the problem.
Show more

Hi
lets see the what info Q has..
Is \(\sqrt{(7x)}\) an integer?..
possible -
a) if x ia an integer and of type y^2*7..
b) if x is a fraction of type y^2/7..

lets see the statements
(1) \(\sqrt{(x/7)}\) is an integer.
this tells us that x = y^2*7, where y is an integer, SAME as case (a) above
Suff


(2) \(\sqrt{(28x)}\)is an integer.
this tells us that \(\sqrt{(4*7*x)}\)is an integer.
so x = y^2/28..
if y is multiple of 2, ans is YES
if y is not a multiple of 2, ans is No
Insuff

A
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Is (7x)^1/2 an integer? 14 Oct 2018, 16:34
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Is \(\sqrt{7x}\) an integer?

(1) \(\sqrt{\frac{x}{7}}\) is an integer

(2) \(\sqrt{28x}\) is an integer
Show more

Target question: Is \(\sqrt{7x}\) an integer?

Statement 1: \(\sqrt{\frac{x}{7}}\) is an integer
Let's say that √(x/7) = k, where k is an integer
Square both sides to get: x/7 = k²
Multiply both side by 7 to get: x = 7k²

This means 7x = 7(7k²) = 49k²
So, √(7x) = √(49k²) = 7k
Since k is an integer, we know that 7k is an integer.
So, the answer to the target question is YES, √(7x) IS an integer
Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Statement 2: \(\sqrt{28x}\) is an integer
There are several values of x that satisfy statement 2. Here are two:
Case a: x = 28, which means, √(7x) = √(196) = 14. So, in this case , the answer to the target question is YES, √(7x) IS an integer
Case b: x = 9/28, which means, √(7x) = √(63/28) = some non-integer. So, in this case , the answer to the target question is NO, √(7x) is NOT an integer
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Answer: A

Cheers,
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Is (7x)^1/2 an integer? 28 Jan 2019, 08:29
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For statement 1, x=7 and x=343 both satisfy the condition, so how can the answer be A?

For statement 2, x= 1/28 and x=7 satisfy the condition,

But when both combined, x=1/28 is not valid. and x=7 is the unique answer.

So Answer is C, how is the answer A, can anyone guide how my method is incorrect?
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Is (7x)^1/2 an integer? 28 Jan 2019, 12:38
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anoopdev
For statement 1, x=7 and x=343 both satisfy the condition, so how can the answer be A?

For statement 2, x= 1/28 and x=7 satisfy the condition,

But when both combined, x=1/28 is not valid. and x=7 is the unique answer.

So Answer is C, how is the answer A, can anyone guide how my method is incorrect?
Show more

Hi anoopdev,

When dealing with DS questions, you have to make sure that you are answering the question that is ASKED (and define whether the answer is consistent - meaning that it stays the same OR that it is inconsistent - meaning that it changes).

Based on the information in Fact 1, you listed two possible values for X. Both of those values lead to the SAME answer to the question (re: YES, you would end up with an integer). That is a consistent result. All of the other potential values for X that 'fit' the information in Fact 1 will lead to the SAME answer to the given question, so Fact 1 is SUFFICIENT.

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Is (7x)^1/2 an integer? 20 Dec 2019, 10:58
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Bunuel
Is \(\sqrt{7x}\) an integer?

Notice that we are not told that x is an integer.

(1) \(\sqrt{\frac{x}{7}}\) is an integer. Given that \(\sqrt{\frac{x}{7}}=integer\) --> square it: \(\frac{x}{7}=integer^2\) --> \(x=7*integer^2\). So, \(\sqrt{7x}=\sqrt{7*(7*integer^2)}=7*integer=integer\). Sufficient.

(2) \(\sqrt{28x}\) is an integer. If \(x=\frac{1}{28}\), then \(\sqrt{7x}=\frac{1}{2}\neq{integer}\) BUT if \(x=0\), then \(\sqrt{7x}=0=integer\). Not sufficient.

Answer: A.
Show more

Bunuel,

Can you tell me the mistake I am making?

In A,
i. Let \(x=7\) then \(\sqrt{\frac{x}{7}}\) will be an integer and \(\sqrt{7x}\) \(=\) \(\sqrt{7*7}\) \(=\) \(\sqrt{7^2}\) will be an integer
ii. Let \(x=21\) then \(\sqrt{\frac{x}{7}}\) will be an integer but \(\sqrt{7x}\) \(=\) \(\sqrt{7*7*3}\) will not be an integer

Doesn't this make A insufficient?
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Is (7x)^1/2 an integer? 20 Dec 2019, 11:05
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Bunuel
Is \(\sqrt{7x}\) an integer?

Notice that we are not told that x is an integer.

(1) \(\sqrt{\frac{x}{7}}\) is an integer. Given that \(\sqrt{\frac{x}{7}}=integer\) --> square it: \(\frac{x}{7}=integer^2\) --> \(x=7*integer^2\). So, \(\sqrt{7x}=\sqrt{7*(7*integer^2)}=7*integer=integer\). Sufficient.

(2) \(\sqrt{28x}\) is an integer. If \(x=\frac{1}{28}\), then \(\sqrt{7x}=\frac{1}{2}\neq{integer}\) BUT if \(x=0\), then \(\sqrt{7x}=0=integer\). Not sufficient.

Answer: A.

Bunuel,

Can you tell me the mistake I am making?

In A,
i. Let \(x=7\) then \(\sqrt{\frac{x}{7}}\) will be an integer and \(\sqrt{7x}\) \(=\) \(\sqrt{7*7}\) \(=\) \(\sqrt{7^2}\) will be an integer
ii. Let \(x=21\) then \(\sqrt{\frac{x}{7}}\) will be an integer but \(\sqrt{7x}\) \(=\) \(\sqrt{7*7*3}\) will not be an integer

Doesn't this make A insufficient?
Show more

If x = 21, then \(\sqrt{\frac{x}{7}}=\sqrt{\frac{21}{7}}=\sqrt{3}\), which is not an integer.
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Is (7x)^1/2 an integer? 20 Dec 2019, 11:12
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Bunuel
Is \(\sqrt{7x}\) an integer?

Notice that we are not told that x is an integer.

(1) \(\sqrt{\frac{x}{7}}\) is an integer. Given that \(\sqrt{\frac{x}{7}}=integer\) --> square it: \(\frac{x}{7}=integer^2\) --> \(x=7*integer^2\). So, \(\sqrt{7x}=\sqrt{7*(7*integer^2)}=7*integer=integer\). Sufficient.

(2) \(\sqrt{28x}\) is an integer. If \(x=\frac{1}{28}\), then \(\sqrt{7x}=\frac{1}{2}\neq{integer}\) BUT if \(x=0\), then \(\sqrt{7x}=0=integer\). Not sufficient.

Answer: A.

Bunuel,

Can you tell me the mistake I am making?

In A,
i. Let \(x=7\) then \(\sqrt{\frac{x}{7}}\) will be an integer and \(\sqrt{7x}\) \(=\) \(\sqrt{7*7}\) \(=\) \(\sqrt{7^2}\) will be an integer
ii. Let \(x=21\) then \(\sqrt{\frac{x}{7}}\) will be an integer but \(\sqrt{7x}\) \(=\) \(\sqrt{7*7*3}\) will not be an integer

Doesn't this make A insufficient?

If x = 21, then \(\sqrt{\frac{x}{7}}=\sqrt{\frac{21}{7}}=\sqrt{3}\), which is not an integer.
Show more

Thanks Bunuel!

That was a very silly mistake from my end :|
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Is (7x)^1/2 an integer? 30 May 2020, 01:34
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Is \(\sqrt{7x}\) an integer?

(1) \(\sqrt{\frac{x}{7}}\) is an integer

(2) \(\sqrt{28x}\) is an integer

Show Spoiler
Can someone please explain why point 2 is not correct? What I did is that 28 is factored into => 2*2*7 so therefore I concluded x must have at least one 7. so if x has one seven then \sqrt{7x} must be an integer. Am I doing something wrong here?

Thanks!
Alex
Show more

Sharing my approach to this problem,

Statement-1 \(\sqrt{\frac{x}{7}}\) = Int
Now let's try to modify \(\sqrt{7x}\) in order to get \(\sqrt{\frac{x}{7}}\)
\(\sqrt{7x}\) = \(\sqrt{7x * \frac{7}{7}}\)
\(\sqrt{\frac{x}{7}}\) * \(\sqrt{49}\)
We are give that \(\sqrt{\frac{x}{7}}\) is an integer and \(\sqrt{49}\) is also an integer hence \(\sqrt{7x}\) is an integer (Sufficient)

Statement-2 \(\sqrt{28x}\) = Int
Now let's try to modify \(\sqrt{7x}\) in order to get \(\sqrt{28x}\)
\(\sqrt{7x}\) = \(\sqrt{7x * \frac{4}{4}}\)
\(\sqrt{28x}\) * \(\sqrt{\frac{1}{4}}\)
We are give that \(\sqrt{28x}\) is an integer (let's call it \(k\)) and \(\sqrt{\frac{1}{4}}\) yields \(\frac{1}{2}\)

Now, for \(k\) * \(\frac{1}{2}\)

If \(k\) \(=\) \(even\) then the expression is an integer and if \(k\) \(=\) \(odd\) the expression will not be an integer (Not Sufficient)

Ans. A
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Is (7x)^1/2 an integer? 06 Apr 2025, 04:50
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Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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