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If x is a positive integer, is (x^2 + y^2)^1/2 an integer?
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24 Mar 2015, 03:30
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If x is a positive integer, is (x^2 + y^2)^1/2 an integer?
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24 Mar 2015, 04:23
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. answer B
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Re: If x is a positive integer, is (x^2 + y^2)^1/2 an integer?
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24 Mar 2015, 21:03
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 Final Answer: GMAT assassins aren't born, they're made, Rich
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Re: If x is a positive integer, is (x^2 + y^2)^1/2 an integer?
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24 Mar 2015, 22:08
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^2Y^2=0; therefore Y^2=8X^2; therefore X^2+Y^2=9X^2 which is perfect square, yes sufficient
Hence answer is B
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Re: If x is a positive integer, is (x^2 + y^2)^1/2 an integer?
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24 Mar 2015, 23:42
(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
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Re: If x is a positive integer, is (x^2 + y^2)^1/2 an integer?
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30 Mar 2015, 03:36



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Re: If x is a positive integer, is (x^2 + y^2)^1/2 an integer?
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17 Mar 2016, 17:57
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.



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Re: If x is a positive integer, is (x^2 + y^2)^1/2 an integer?
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Re: If x is a positive integer, is (x^2 + y^2)^1/2 an integer?
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10 Aug 2016, 19:51
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.
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Re: If x is a positive integer, is (x^2 + y^2)^1/2 an integer?
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11 Aug 2016, 01:10
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. SufficientCorrect Answer B
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