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If x is a positive integer, is (x^2 + y^2)^1/2 an integer? 24 Mar 2015, 04:30
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B
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55% (hard)

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63% (01:55) correct 37% (02:21) wrong based on 447 sessions

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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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B
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If x is a positive integer, is (x^2 + y^2)^1/2 an integer? 24 Mar 2015, 05:23
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Bunuel
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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\(\sqrt{x^2 + y^2}\) 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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If x is a positive integer, is (x^2 + y^2)^1/2 an integer? 24 Mar 2015, 22:03
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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:
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B

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If x is a positive integer, is (x^2 + y^2)^1/2 an integer? 24 Mar 2015, 23:08
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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

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If x is a positive integer, is (x^2 + y^2)^1/2 an integer? 25 Mar 2015, 00:42
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(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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If x is a positive integer, is (x^2 + y^2)^1/2 an integer? 30 Mar 2015, 04:36
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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
Show more

MAGOOSH OFFICIAL SOLUTION:
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If x is a positive integer, is (x^2 + y^2)^1/2 an integer? 17 Mar 2016, 18:57
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Bunuel
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
Show more

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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If x is a positive integer, is (x^2 + y^2)^1/2 an integer? 10 Aug 2016, 20:51
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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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If x is a positive integer, is (x^2 + y^2)^1/2 an integer? 11 Aug 2016, 02:10
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Bunuel
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
Show more


(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

Correct Answer B
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If x is a positive integer, is (x^2 + y^2)^1/2 an integer? 07 Aug 2024, 03:30
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