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M32-08

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M32-08  [#permalink]

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New post 17 Jul 2017, 04:41
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A
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C
D
E

Difficulty:

  95% (hard)

Question Stats:

33% (01:06) correct 67% (01:36) wrong based on 24 sessions

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Re M32-08  [#permalink]

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New post 17 Jul 2017, 04:41
Official Solution:


If \(x\) and \(y\) are integers and \(-x \leq y \leq x\), does \(\sqrt{x^2 - y^2} = x + y\)?



This is a hard question. You should pay attention to every detail and read the solution very carefully



First of all, \(-x \leq y \leq x\) ensures two things:

1. \(x^2-y^2\geq 0\), so the square root of this number will be defined.

2. \(x+y\geq 0\), so the square root won't be equal to negative number.



Next, \(-x \leq x\) implies that \(x \geq 0\).



And finally, before moving to the statements, let's rephrase the question:

Does \(\sqrt{x^2 - y^2} = x + y\)?

Square both sides: does \(x^2 - y^2 = x^2+2xy + y^2\)?

Does \(xy+y^2=0\)? Notice here that we cannot reduce this by \(y\), because we'll loose a possible root: \(y=0\).

Does \(y(x+y)=0\)?

Does \(y=0\) or \(x=-y\)?



(1) \(xy\) is NOT a square of an integer.



If \(y = 0\) were true, then \(xy\) would be 0, which is a square of an integer.

However, if \(x = -y\) and \(x ≠ 0\) were true, for example if \(x = -y = 1\), then \(xy\) would be \(-x^2\), so negative, which cannot be a square of an integer. So, \(x = -y\) IS possible.

Of course, \(x\) can be not equal to \(-y\), and the statemented still could hold true. For example, \(x = 2\) and \(y =1\).



Thus, we have have two different answers to the question: Not sufficient.



(2) Point \((x, y)\) is NOT on x-axis.



If \(y = 0\) were true, then point \((x, y)\) would be ON the x-axis. So, we know that \(y ≠ 0\).

However, if \(x = -y\) and \(x ≠ 0\) were true, for example if \(x = -y = 1\), then then point \((x, y)\) would be \((x, -x)\), so (positive, negative), which would mean that it's below x-axis. So, \(x = -y\) IS possible.

Of course, \(x\) can be not equal to \(-y\), and the statemented still could hold true. For example, \(x = 2\) and \(y =1\).



(1)+(2) We could have the same cases: if \(x = -y\) and \(x ≠ 0\), for example if \(x = -y = 1\), then the answer is YES but if say \(x = 2\) and \(y =1\), then the answer is NO. Not sufficient.


Answer: E
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Re: M32-08  [#permalink]

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New post 11 Sep 2017, 07:25
Can this question be solved by plugging in values?
the approach explained by bunuel is tough to comprehend does someone else has an easy method or trick for this question?
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Re: M32-08  [#permalink]

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New post 11 Sep 2017, 07:38
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M32-08  [#permalink]

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New post 27 Oct 2018, 06:56
1
By plugging in numbers -

Hey Bunuel amazing question.
I solved it a bit differently. Please correct me if i am wrong.

Reducing the equation.
\(\sqrt{x^2 - y^2}\) \(= x + y\)
\(x^2 - y^2 = x^2 + y^2 + 2xy\)
\(-2y^2 = 2xy\)
\(y^2 = -xy\) ?


Statement 1
- xy is NOT a square of an integer.
It means x is not equal to y.
If x = 3 and y = -3. Then equation condition satisfied.
But, if x = 3 and y = -2 then equation condition not satisfied.
So INSUFFICIENT

Statement 2 -
- Point (x,y) is NOT on x-axis.
From this we can infer that y is not equal to zero.
Still the above to examples are possible.
So INSUFFICIENT.

Both statements together -

We can still see that the above examples can happen and hence it is not Sufficient.

Answer - E
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M32-08   [#permalink] 27 Oct 2018, 06:56
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