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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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(1) \(\sqrt{x + y}\) is an integer
(2) 8x^2 - y^2 = 0
Pure algebraic questions like this one are no longer part of the GMAT Data Sufficiency syllabus.
For relevant Data Sufficiency questions, please check the Data Sufficiency forum:
Data Sufficiency Forum
Data Sufficiency Forum
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Bunuel
\(\sqrt{x^2 + y^2}\) 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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Expert reply
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
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
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,
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,
(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
(2) 8x^2 - y^2 = 0
y^2= 8x^2
\sqrt{x^2 + y^2} = (9x^2)^1/2
Sufficient
Ans B
Bunuel
MAGOOSH OFFICIAL SOLUTION:
Attachment:
rootofsquaresum_text.PNG [ 27.53 KiB | Viewed 6723 times ]
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Bunuel
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.
GMAT 1: 760 Q51 V42
Posts: 10,057
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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Bunuel
(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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