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For integers x and y, which of the following MUST be an integer?

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For integers x and y, which of the following MUST be an integer? [#permalink]

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New post 08 Jul 2017, 05:45
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For integers x and y, which of the following MUST be an integer?


A. \(\sqrt{25x^2+30xy+36y^2}\)

B. \(\sqrt{49x^2−84xy+36y^2}\)

C. \(\sqrt{16x^2−y^2}\)

D. \(\sqrt{64x^2−64xy−64y^2}\)

E. \(\sqrt{81x^2+25xy+16y^2}\)

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Re: For integers x and y, which of the following MUST be an integer? [#permalink]

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New post 11 Jul 2017, 11:41
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Smokeybear00 wrote:
Is there an approach to this if one does not immediately recognize (x^2 - 2xy +y^2) ?


You'll have more flexibility in general if you can recognize how to factor in questions like this, but as a fallback, an alternative approach is to plug in some simple numbers for x and y. The right answer needs to always give us an integer result, so if we find that an answer does not give an integer result for some numbers we pick, that cannot be the right answer. And if you plug in x=0 and y=1, you get a negative number under the square root for C and D, so you get something undefined (or 'imaginary' in advanced math language), so those can't be right answers. If you plug in x=1 and y=1, you get √91 for A, and √122 for E, which aren't integers either, so those can't be right. That only leaves D, which does give us an integer in both cases.

If you ever need to use an approach like this, try to focus on the simplest possible sets of numbers, so you don't spent too long doing calculations. Zero (when allowed) can be a very convenient choice, and sometimes, as in this question, just plugging in 1 for everything will get you the answer (we actually could have just plugged in x=1 and y=1 and that would have been enough here - only D gives an integer - but plugging in x=0 rules out two answers so quickly I prefer to do that). At the very least, you usually will rule out three wrong answers by using very simple numbers, and then will only need to plug slightly more complicated numbers into two answer choices instead of five.
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Re: For integers x and y, which of the following MUST be an integer? [#permalink]

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New post 08 Jul 2017, 05:48
Ans : B

B can be written as = (( 7x - 6y) ^2)^1/2 = 7x - 6y = integer
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Re: For integers x and y, which of the following MUST be an integer? [#permalink]

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New post 09 Jul 2017, 07:30
its a direct application of formula...
(a+b)^2 = a^2 + 2ab + b^2

(a-b)^2 = a^2 -2ab + b^2

ans B
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Re: For integers x and y, which of the following MUST be an integer? [#permalink]

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New post 11 Jul 2017, 11:09
Is there an approach to this if one does not immediately recognize (x^2 - 2xy +y^2) ?
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For integers x and y, which of the following MUST be an integer? [#permalink]

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New post 11 Jul 2017, 13:31
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Bunuel wrote:
For integers x and y, which of the following MUST be an integer?


A. \(\sqrt{25x^2+30xy+36y^2}\)

B. \(\sqrt{49x^2−84xy+36y^2}\)

C. \(\sqrt{16x^2−y^2}\)

D. \(\sqrt{64x^2−64xy−64y^2}\)

E. \(\sqrt{81x^2+25xy+16y^2}\)


If the term inside the square root is a perfect square, then the whole term will be an integer.
We need to try and write the terms in the for of x^2 +- 2ax + a^2

A. \(\sqrt{25x^2+30xy+36y^2}\) = \(\sqrt{25x^2+5*6 xy+36y^2}\) this cannot be a perfect square

B. \(\sqrt{49x^2−84xy+36y^2}\) = \(\sqrt{49x^2−2*6*7*xy+36y^2}\) = \(\sqrt{(7x−6y)^2}\)
This is a perfect square.

Correct Option: B
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Re: For integers x and y, which of the following MUST be an integer? [#permalink]

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New post 12 Jul 2017, 05:19
The options are the expanded form of identities which are commonly tested on GMAT.

B is the answer and the identity used in B is: (a-b)^2= a^2+b^2-2ab
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Re: For integers x and y, which of the following MUST be an integer? [#permalink]

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New post 16 Jul 2017, 17:22
Bunuel wrote:
For integers x and y, which of the following MUST be an integer?


A. \(\sqrt{25x^2+30xy+36y^2}\)

B. \(\sqrt{49x^2−84xy+36y^2}\)

C. \(\sqrt{16x^2−y^2}\)

D. \(\sqrt{64x^2−64xy−64y^2}\)

E. \(\sqrt{81x^2+25xy+16y^2}\)


We need to determine which of the answer choices must be an integer.

Let’s take a look at answer choice B:

√(49x^2 - 84xy + 36y^2)

√(7x - 6y)(7x - 6y)

√(7x - 6y)^2 = |7x - 6y|

Since x and y are integers, |7x - 6y| is an integer.

Answer: B
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Re: For integers x and y, which of the following MUST be an integer? [#permalink]

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New post 08 Sep 2017, 20:35
Can we not apply the same rule to options A and D?
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Re: For integers x and y, which of the following MUST be an integer? [#permalink]

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New post 08 Sep 2017, 20:37
I mean option A and E
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For integers x and y, which of the following MUST be an integer? [#permalink]

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New post 13 Sep 2017, 15:21
Bunuel wrote:
For integers x and y, which of the following MUST be an integer?

A. \(\sqrt{25x^2+30xy+36y^2}\)

B. \(\sqrt{49x^2−84xy+36y^2}\)

C. \(\sqrt{16x^2−y^2}\)

D. \(\sqrt{64x^2−64xy−64y^2}\)

E. \(\sqrt{81x^2+25xy+16y^2}\)

santro789 wrote:
Can we not apply the same rule to options A and D E?

santro789 , I'm not sure which rule you mean. In fact, I'm slightly confused by your question. Under either rule, how do you see A and E as possible answers?

OptimusPrepJanielle wrote
Quote:
If the term inside the square root is a perfect square, then the whole term will be an integer.
We need to try and write the terms in the form of x^2 +- 2ax + a^2

ScottTargetTestPrep wrote
Quote:
√(7x - 6y)^2 = |7x - 6y|

Since x and y are integers, |7x - 6y| is an integer.

The answer to your question is no. The most basic reason: the middle term in both A and E prevents both from being perfect squares.

If A were a perfect square, looking at its terms' coefficients, it would be
\(\sqrt{(5x + 6y)^2}\)= \(\sqrt{25x^2 + 60xy + 36y^2}\)

Answer A's middle term is 30xy, not 60xy. That means it's not a perfect square.

Answer E has the same problem. Looking at its coefficients, if it were a perfect square it would be
\(\sqrt{(9x + 4y)^2}
=\sqrt{81x^2 + 72xy + 16y^2}\)

The middle term in E is 25xy, not 72xy. In neither A nor E can we get a perfect square under the square root sign. If we could, we would get an integer: the square root of a perfect square is an integer. Just think about a couple of numeric values: \(\sqrt{2^2} = 2\), and \(\sqrt{41^2} = 41\)

Without being able to factor A and E into \((a + b)^2\) or \((a - b)^2\) because their middle terms prevent them from being perfect squares, we certainly cannot use absolute value analysis to prove what they are not.

Hope that helps.
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Re: For integers x and y, which of the following MUST be an integer? [#permalink]

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New post 13 Sep 2017, 15:34
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Bunuel wrote:
For integers x and y, which of the following MUST be an integer?


A. \(\sqrt{25x^2+30xy+36y^2}\)

B. \(\sqrt{49x^2−84xy+36y^2}\)

C. \(\sqrt{16x^2−y^2}\)

D. \(\sqrt{64x^2−64xy−64y^2}\)

E. \(\sqrt{81x^2+25xy+16y^2}\)


Another approach:

The question is asking us to determine which expression MUST be an integer for ALL integer values of x and y.
So, let's TEST a pair of values.
Let's plug in x = 1 and y = 1
If an expression evaluates to be a non-integer, we can ELIMINATE that answer choice.

We get...
A)√81 = 9. This IS an integer. So, keep A
B)√1 = 1. This IS an integer. So, keep B
C)√15. This does NOT evaluate to be an integer. ELIMINATE C
D)√-64. Cannot evaluate. ELIMINATE D
E)√122. This does NOT evaluate to be an integer. ELIMINATE E

We're left with A and B
Try another set of values
Let's plug in x = 1 and y = -1
We get...
A)√31. This does NOT evaluate to be an integer. ELIMINATE A

By the process of elimination, the correct answer is B

Cheers,
Brent
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Re: For integers x and y, which of the following MUST be an integer?   [#permalink] 13 Sep 2017, 15:34
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