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by mohit11 » Sat Sep 25, 2010 1:01 am
If x is a positive integer, is sqrt X an integer

(1) sqrt 36x is an integer

(2) sqrt 3x +4 is an integer

[spoiler]OA : A[/spoiler]

[spoiler]
Ah, i figured out the question trap while typing this thread. Initially i thought in 1. 6 Sqrt X can be an integer if X is 1.5 so A is not suff. However, the trap is that X is a positive integer so Sqrt X cannot be 1.5 because then X will not be a positive integer.

Another way to look at it is that sqrt 2 is 1.4 and sqrt 3 is 1.7 and 1.5 lies between 1.4 and 1.7 therefore x must lie between 2 and 3 and thus it will not be a positive integer. Same will be these with 2.5 , 3.5 , 4.5 etc.[/spoiler]
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by saurabhmahajan » Sat Sep 25, 2010 1:31 am
Yeap answer is A

i plugged in with numbers
sqrt 36x :
i tried with 4 16 36 and found that stmt 1 is true

sqrt 3x+4:
i plugged in 4 16 36 and also 20
stmt 2 not sufficient.
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by chendawg » Thu Jan 20, 2011 10:39 am
ahh good question.
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by Brian@VeritasPrep » Thu Jan 20, 2011 6:00 pm
Great question - thanks for posting!

One other way to look at statement 1 is this:

If sqrt (36x) is an integer, then you can break apart the multiplication inside the square root to say that:

sqrt 36x = integer

sqrt 36 * sqrt x = integer

6 * sqrt x = integer

So the square root of x must either be an integer or a fraction like 2/3. But since we know that x is an integer, it can't be a fraction, so the square root of x must be an integer.


The takeaway here - when square roots involve multiplication under the radical sign, it can be extremely helpful to factor out perfect squares to chip away at what's left.
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by Zerks87 » Mon Jan 31, 2011 9:06 am
I want to take as much algebra out of this as possible.

Stmt (1) sqrt 36x is an integer.

The prime factors of 36 are 3 * 2 * 3 *2 (since 6*6=36). This means that both prime factors have a pair that make it a perfect square. Therefore, in order for 36x to be a perfect square, x must also be a perfect square, therefore sqrt x is an integer.

SUFFICIENT

Stmt (2) sqrt (3x + 4) is an integer

For this we must think about what number when added to 4 is a perfect square and then see what you need to multiply by 3 in order to get that number.

16 is the first perfect square and 3(4) + 4 = 16
64 is also a perfect square 3(20) + 4 = 64

Since x can equal both 4 and 20, it can be both a perfect square or not a perfect square, INSUFF.

Choose A
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by tgou008 » Tue May 03, 2011 9:02 am
Brian@VeritasPrep wrote:Great question - thanks for posting!

One other way to look at statement 1 is this:

If sqrt (36x) is an integer, then you can break apart the multiplication inside the square root to say that:

sqrt 36x = integer

sqrt 36 * sqrt x = integer

6 * sqrt x = integer

So the square root of x must either be an integer or a fraction like 2/3. But since we know that x is an integer, it can't be a fraction, so the square root of x must be an integer.


The takeaway here - when square roots involve multiplication under the radical sign, it can be extremely helpful to factor out perfect squares to chip away at what's left.
What difficulty level do you think this question is?
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by saketk » Wed Nov 02, 2011 7:29 pm
May be a 650 level question.
The key to solve this question is to remember that x is an integer
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by prashant misra » Fri Nov 18, 2011 6:08 am
the answer to this question is option A i did this question by picking numbers.
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by Sharma_Gaurav » Thu Mar 01, 2012 10:49 pm
Straight a . Simple question .
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by krusta80 » Fri Mar 02, 2012 5:04 am
Interesting side question I posed to myself while looking at part 2 of this question is (and I warn anyone trying to stay on topic to AVOID the rest of this post):

How many square integers, x, exist such that 3x+4 is also a perfect square?
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by ronnie1985 » Thu Mar 29, 2012 8:04 am
(A) QED
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by shubhamkumar » Wed Apr 04, 2012 9:27 am
krusta80 wrote:Interesting side question I posed to myself while looking at part 2 of this question is (and I warn anyone trying to stay on topic to AVOID the rest of this post):

How many square integers, x, exist such that 3x+4 is also a perfect square?
Very True,That made me spend a minute more on this because Statement 2 is actually contradicting statement 1.as per statement 1, x is a square.considering stmt 2, I was not able to get any value from it which makes x a square.
Experts Please correct me if I am wrong.
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by Bill@VeritasPrep » Wed Apr 04, 2012 9:37 am
krusta80 wrote:Interesting side question I posed to myself while looking at part 2 of this question is (and I warn anyone trying to stay on topic to AVOID the rest of this post):

How many square integers, x, exist such that 3x+4 is also a perfect square?
x=4 (3x + 4 = 16 = 4^2)

x=64 (3x + 4 = 196 = 14^2)

I'm sure there are others, but those are two I found.
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by Lifetron » Sat Aug 25, 2012 7:29 pm
I chose A ! But had a lil doubt ! Great question !
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by mp2469 » Wed Aug 29, 2012 12:05 pm
Brian@VeritasPrep wrote:Great question - thanks for posting!

One other way to look at statement 1 is this:

If sqrt (36x) is an integer, then you can break apart the multiplication inside the square root to say that:

sqrt 36x = integer

sqrt 36 * sqrt x = integer

6 * sqrt x = integer

So the square root of x must either be an integer or a fraction like 2/3. But since we know that x is an integer, it can't be a fraction, so the square root of x must be an integer.


The takeaway here - when square roots involve multiplication under the radical sign, it can be extremely helpful to factor out perfect squares to chip away at what's left.
Why can't X=1 for statement 1?
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