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# If x is a positive integer, is (x^2 + y^2)^1/2 an integer?

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If x is a positive integer, is (x^2 + y^2)^1/2 an integer?  [#permalink]

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24 Mar 2015, 04:30
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If x is a positive integer, is $$\sqrt{x^2 + y^2}$$ an integer?

(1) $$\sqrt{x + y}$$ is an integer

(2) 8x^2 - y^2 = 0

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If x is a positive integer, is (x^2 + y^2)^1/2 an integer?  [#permalink]

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24 Mar 2015, 05:23
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Bunuel wrote:
If x is a positive integer, is $$\sqrt{x^2 + y^2}$$ an integer?

(1) $$\sqrt{x + y}$$ is an integer

(2) 8x^2 - y^2 = 0

$$\sqrt{x^2 + y^2}$$ will be an integer only when Y is an integer , why ? because square root of a fractional number cannot be an integer , and X and Y are Pythagorean pair (eg: 3,4 or 6,8 etc.) .

(1) $$\sqrt{x + y}$$ is an integer
this tells us that Y is an integer . NOT SUFFICIENT.

(2) 8x^2 - y^2 = 0
if we substitute 8X^2= Y^2 in $$\sqrt{x^2 + y^2}$$ = 3X (we cannot have -3X as X is a positive integer) .
SUFFICIENT.

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Re: If x is a positive integer, is (x^2 + y^2)^1/2 an integer?  [#permalink]

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24 Mar 2015, 22:03
Hi All,

DS questions are often built around distinct patterns (even if you don't immediately recognize that a pattern is there). Often, the way to prove that there's a pattern is to TEST VALUES and pay careful attention to the results.

Here, we're told that X is a POSITIVE INTEGER. We're asked if \sqrt{(X^2 + Y^2} is an integer. This is a YES/NO question.

Fact 1: \sqrt{(X+Y)} is an integer.

IF....
X = 1
Y = 0
Then \sqrt{1} IS an integer and the answer to the question is YES.

IF....
X = 1
Y = 3
Then \sqrt{10} is NOT an integer and the answer to the question is NO.
Fact 1 is INSUFFICIENT

Fact 2: 8(X^2) - Y^2 = 0

We can manipulate this equation into...

8(X^2) = Y^2

At first glance, you might not know if there's a pattern in this information, so let's TEST a few values and see if a pattern emerges....

IF...
X = 1
Y = +-\sqrt{8}
Then \sqrt{9} IS an integer and the answer to the question is YES

IF....
X = 2
Y = +-\sqrt{32}
Then \sqrt{36} IS an integer and the answer to the question is YES

IF....
X = 3
Y = +-\sqrt{72}
Then \sqrt{81} IS an integer and the answer to the question is YES

Looking at these first 3 examples, it appears that the resulting calculation will ALWAYS be a perfect square, so the answer to the question is ALWAYS YES.
Fact 2 is SUFFICIENT

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Special Offer: Save $75 + GMAT Club Tests Free Official GMAT Exam Packs + 70 Pt. Improvement Guarantee www.empowergmat.com/ ***********************Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!*********************** Manager Joined: 26 Dec 2012 Posts: 146 Location: United States Concentration: Technology, Social Entrepreneurship WE: Information Technology (Computer Software) Re: If x is a positive integer, is (x^2 + y^2)^1/2 an integer? [#permalink] ### Show Tags 24 Mar 2015, 23:08 X is +ve integer, root(X^2+Y^2) =int or not? 1. root(X+y) = integer; X+Y =integer^2; still we are not sure Y^2+X^2 will be perfect square or not. 2. 8X^2-Y^2=0; therefore Y^2=8X^2; therefore X^2+Y^2=9X^2 which is perfect square, yes sufficient Hence answer is B Thanks, Manager Joined: 06 Jan 2014 Posts: 64 Re: If x is a positive integer, is (x^2 + y^2)^1/2 an integer? [#permalink] ### Show Tags 25 Mar 2015, 00:42 (1) \sqrt{x + y} is an integer Not Sufficient (2) 8x^2 - y^2 = 0 y^2= 8x^2 \sqrt{x^2 + y^2} = (9x^2)^1/2 Sufficient Ans B _________________ ______________________________ Liked the Post !!!! KUDOs Plzzzzz Math Expert Joined: 02 Sep 2009 Posts: 49208 Re: If x is a positive integer, is (x^2 + y^2)^1/2 an integer? [#permalink] ### Show Tags 30 Mar 2015, 04:36 1 1 Bunuel wrote: If x is a positive integer, is $$\sqrt{x^2 + y^2}$$ an integer? (1) $$\sqrt{x + y}$$ is an integer (2) 8x^2 - y^2 = 0 MAGOOSH OFFICIAL SOLUTION: Attachment: rootofsquaresum_text.PNG [ 27.53 KiB | Viewed 2093 times ] _________________ Board of Directors Joined: 17 Jul 2014 Posts: 2697 Location: United States (IL) Concentration: Finance, Economics GMAT 1: 650 Q49 V30 GPA: 3.92 WE: General Management (Transportation) Re: If x is a positive integer, is (x^2 + y^2)^1/2 an integer? [#permalink] ### Show Tags 17 Mar 2016, 18:57 Bunuel wrote: If x is a positive integer, is $$\sqrt{x^2 + y^2}$$ an integer? (1) $$\sqrt{x + y}$$ is an integer (2) 8x^2 - y^2 = 0 the question basically asks: is x^2 + y^2 a perfect square? i don't know why, but first thing that popped into my mind = Pythagorean triplets..since there it is tested the sum of the 2 squares = a perfect square... 1. sqrt(x+y)= integer. suppose x=1, and y=3. x+y=4, and sqrt(4)=2. so the condition is satisfied. but 1^2 + 3^2 = 10, and sqrt(10) is not an integer. since the numbers can change greatly, it is impossible to deduct anything about x^2 + y^2. 2. 8x^2 - y^2 = 0 y^2 = 8x^2 now: x^2 +8x^2 = 9x^2. this under square root = 3x. so 2 alone is sufficient. Current Student Joined: 12 Aug 2015 Posts: 2651 Schools: Boston U '20 (M) GRE 1: Q169 V154 Re: If x is a positive integer, is (x^2 + y^2)^1/2 an integer? [#permalink] ### Show Tags 10 Aug 2016, 07:58 Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 6205 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: If x is a positive integer, is (x^2 + y^2)^1/2 an integer? [#permalink] ### Show Tags 10 Aug 2016, 20:51 If we modify the original condition and the question, we can square both sides of sqrt(x^2+y^2)=int?. Then we get x^2+y^2=int^2?. Hence, we know x^2 and y^2 are squares of certain integers. Hence, if we look at the condition 2), from y^2=8x^2, we get x^2+y^2=x^2+8x^2=9x^2=(3x)^2. The answer is always yes and the condition is sufficient. Hence, the correct answer is B. Attachments variable approach's answer probability.jpg [ 219.74 KiB | Viewed 1348 times ] _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$99 for 3 month Online Course"
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Re: If x is a positive integer, is (x^2 + y^2)^1/2 an integer?  [#permalink]

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11 Aug 2016, 02:10
Top Contributor
Bunuel wrote:
If x is a positive integer, is $$\sqrt{x^2 + y^2}$$ an integer?

(1) $$\sqrt{x + y}$$ is an integer

(2) 8x^2 - y^2 = 0

(1) $$\sqrt{x + y}$$ is an integer;
when x=4,y=-3 then $$\sqrt{x^2 + y^2}$$ is an integer.
When x=4,y=5 then $$\sqrt{x^2 + y^2}$$ is not an integer,Not Sufficient

(2) $$8x^2 - y^2$$=0 ,or $$y^2=8x^2$$; so $$\sqrt{x^2 + y^2}$$=$$\sqrt{x^2 + 8x^2}$$=$$\sqrt{9x^2}$$=3x,Since x is a positive integer ,Thus 3x is an integer.Sufficient

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Re: If x is a positive integer, is (x^2 + y^2)^1/2 an integer?  [#permalink]

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02 Jan 2018, 01:29
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