Is (7x)^1/2 an integer?

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

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.

(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.

(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.

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:

GMAT assassins aren't born, they're made,
Rich

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

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,
Brent

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?

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.

GMAT assassins aren't born, they're made,
Rich

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.

Thanks Bunuel!

That was a very silly mistake from my end

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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