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

Question

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}

IanStewart · Accepted Answer

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.

BrentGMATPrepNow · Answer

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)√91.This does NOT evaluate to be an integer. ELIMINATE C
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

By the process of elimination, the correct answer is B

Cheers,
Brent

Leo8 · Answer

Ans : B

B can be written as = (( 7x - 6y) ^2)^1/2 = 7x - 6y = integer

mohshu · Answer

its a direct application of formula...
(a+b)^2 = a^2 + 2ab + b^2

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

ans B

Smokeybear00 · Answer

Is there an approach to this if one does not immediately recognize (x^2 - 2xy +y^2) ?

OptimusPrepJanielle · Answer

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

Shiv2016 · Answer

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

ScottTargetTestPrep · Answer

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

santro789 · Answer

Can we not apply the same rule to options A and D?

santro789 · Answer

I mean option A and E

generis · Answer

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 
 ScottTargetTestPrep  wrote 
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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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?

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