If x and y are integers and if (x-y)^2 + 2*y^2 ...

Question

If  x  and  y  are integers and if  (x-y)^2 + 2*y^2 = 27 , which of the following could be the value of  x ?

A) -2
B) -1
C) 3
D) 4
E) 5

PD: Is there a way to solve this without having to try every value of  x  and solving the equation for  y ?

Bunuel · Accepted Answer

Since 2(y^2) is even, (x - y)^2 must be odd for the sum to total the odd number 27.

Odd perfect squares less than 27 are 1, 9, and 25.

• If (x - y)^2 = 1, then 2(y^2) must be 26, giving y^2 = 13, which is not possible because in this case, y will not be an integer.

• If (x - y)^2 = 9, then 2(y^2) must be 18, giving y^2 = 9. This implies y = 3 or y = -3.
 ○ If y = 3, then from (x - y)^2 = 9, we get (x - 3)^2 = 9, which gives x = 0 or x = 6. 
○ If y = -3, then from (x - y)^2 = 9, we get (x + 3)^2 = 9, which gives x = 0 or x = -6.

• If (x - y)^2 = 25, then 2(y^2) must be 2, giving y^2 = 1. This implies y = 1 or y = -1. 
 ○ If y = 1, then from (x - y)^2 = 25, we get (x - 1)^2 = 25, which gives x = -4 or x = 6. 
○ If y = -1, then from (x - y)^2 = 25, we get (x + 1)^2 = 25, which gives x = -6 or  x = 4.

Answer: D.

egmat · Answer

Watch this solution to see how you can solve a complex question such as this one by drawing inferences and by visualizing the equations on the number line instead of actually solving algebraically for all the cases.

https://www.youtube.com/watch?v=IV_bvo_uKJk

gmatophobia · Answer

A few observations that we can make -

(x-y)^2  and  2y^2  will always be a non-negative integers.
 
 The value of  2y^2  will be same for  \pm y

Let's assume some values of  y  and obtain the corresponding value for  (x-y)^2  -

1)  y = \pm1

(x-y)^2 + 2 = 27

(x - y)^2 = 25

(x - y) = \pm 5

As  x - y  is an integer, we can have  y = \pm1 . Let's look at a few other values.

2)  y = \pm2

(x-y)^2 + 8 = 27

(x-y)^2 = 27 - 8 = 19

Eliminate this possibility, as  x - y  will not be an integer.

3)  y = \pm3

(x-y)^2 + 18 = 27

(x-y)^2 = 9

As  x - y  is an integer, we can have  y = \pm3

(x-y) = \pm3

We further observe that  y  cannot be greater than  3  or less than  -3 , as that would result in  (x - y)^2  = negative. This is not possible.

Let's take the first case-

y = \pm1

(x - y)^2 = 25

(x - y) = \pm 5

Case 1(A) : y = 1, x = 5

x - 1 = 5

x = 6 → We don't have this in the option. Hence, eliminate

Case 1(B) : y = -1, x = 5

x + 1 = 5

x = 4 → Matches option D.

Hence, we need not try other possibilities.

Option D

GMATGuruNY · Answer

(x-y)^2 + 2y^2 = 27

Since x and y are integers,  (x-y)^2  and  y^2  are both perfect squares.

Since the left side must sum to 27,  y^2  must be equal to a perfect square less than 27, yielding the following options for  y^2 :
0, 1, 4, 9, 16, 25

To yield a sum of 27, one possible option for  y^2  is 1, with the result that  (x-y)^2=25  and that  x-y=±5 .
If  y=±1 , only D among the answer choices will yield a viable option for  x :
 x=4  and  y=-1 , with result that  x-y = 4-(-1) = 5

D

MartyMurray · Answer

If  x  and  y  are both integers and if  (x-y)^2 + 2(y^2) = 27 , which of the following could be the value of x?

a. -2
b. -1
c. 3 
d. 4 
e. 5

This question could seem impossible to answer since we have just one equation and two variables. However, we have two additional constraints.

The first is that  x  and  y  must both be integers.

The second is that the possible values of  x  are limited to the set of answer choices.

So, since this is a GMAT Quant question, we'll know that we'll be able to answer it by using those constraints.

Given that  x  and  y  are integers and that  (x-y)^2 + 2(y^2) = 27 , what we have are one times one perfect square + two times a second perfect square adding up to  27 . So, one way we can answer this question is to simply consider all the perfect squares that are less than  27 , identify which ones add up to  27  and find a value of  x  that works.

It's not a super elegant way of answering the question, but it demonstrates a key GMAT Quant characteristic, which is that GMAT Quant questions are designed to test our skill in simply finding a way to get to an answer. So, if you don't quickly see an elegant mathematical way to answer a question, it's likely that you'll be fine just using a brute force method to get it done.

There are six perfect squares less than  27 :  0 ,  1 ,  4 ,  9 ,  16 ,  25 .

Scanning them, we quickly see that only  0 ,  1 ,  4 , and  9  can be doubled since  2  x  16  and  2  x  25  are greater than  27 .

Then, considering possible ways of getting to  27 , we see that there are only two:  25 + 2(1) = 27  and  9 + 2(9) = 27 .

In the first case,  y^2 = 1 . So,  y  must be  1  or  -1 . Then, for  (x - y)^2  to equal  25 ,  x - y  must be  5  or  -5 .

So, if  y = 1 , then  x  must be  6    or  -4 . Since neither  6  nor  -4  are among the answer choices, this path to  27  won't work.

If  y = -1 , then for  x - y  to equal  5  or  -5 , x can be  4  or  -6 .

We see that  4  is among the answer choices.

So, the correct answer is (D).

Bunuel · Answer

__________________
y^2 could also be 9.

GMATGuruNY · Answer

Given the answer choices, 9 is not a viable option for  y^2  because  y=±3  would require that  x=0  or  x=6 , neither of which is among the five options.
I've edited my response to clarify that 1 is one possible option for  y^2 .
Since this option yields a value for  x  that is among the five answer choices, no need to look any further.

Bunuel · Answer

Yes, y^2 = 9 does not yield an x value that is among the answer choices, nor was I implying that it does. I was simply pointing out that 1 is not "the only possible option for y^2".

GMATGuruNY · Answer

Agreed. As the Wordle bot states when the bot and I come to the same conclusion, "We are as one."

AnkurGMAT20 · Answer

Bunuel , if we pick a value of X from the given option, will that be a long procedure to find out the solution? Also, if we pick a value from the options, is there a way to do a faster calculation?

Bunuel · Answer

I wouldn't approach this question by plugging in options. You'll end up with ugly quadratics for y, trying to figure out if there's an integer solution.

Oppenheimer1945 · Answer

3y^2-2xy+x^2-27=0 1) Reframe the eq as quadaratic in y 2) Find Discrimiant as function of x D(x)=4x^2-4*3(x^2-27)= 4(81-2x^2) D(x)>=0 and of form N^2 D(4)=4(81-32)=14^2 D(x)>=0 follows that -9/rt(2)<=x<=9/rt(2) => -5<=x<=5 D(x) is integer at x=0,4,-4,6,-6 Tick x=4

ashdank94 · Answer

This is one of the hardest problems I have encountered. Is it normal to get a question like this as your first question on the GMAT Focus? It took me way too long to solve this

HarshavardhanR · Answer

https://gmatclub.com/forum/download/file.php?id=82403 GMAT-Club-Forum-9fqjpkw8.png

korteraz · Answer

Solved using Quadratic formula( maybe longer but fewer guesses and makes more sense to me)

y1,2 = (-b (+ or -) (b^2 - 4 * a * c)^(1/2)) / 2a
so I want to find when (b^2 - 4 * a * c) ^ 1/2 is an integer because: an integer (+ or -) non integer is not an integer

x^2 -2xy +3y^2 -27 = 0

when x= -2
3y^2 + 4y - 23 = 0
using the Quadratic formula (b^2 - 4*a*c)^(1/2) => (16 + 4*3*23)^(1/2) = 292^(1/2) is not an integer

when x= -1
(4+26*4*3)^(1/2) = (316)^(1/2) not an integer

when x= 3
(36+ 3*4*18)^(1/2) = 252^1/2 not an integer

when x= 4
(64+12*11)^1/2 = 196^1/2 is an integer

when x=5
(100+ 12*2)^1/2 = 124^1/2 not an integer

Regor60 · Answer

This shouldn't be too difficult because the first obvious test yields the answer.

With 27 on the RHS and a square and multiple of two on LHS, seems logical to assume Y=1 and subtract 2 from RHS yielding a nice 25 and solving for (X-Y)^2 on left gives X=6.

Well, that not being among answer choices, recognize Y= -1 also works and solving again yields X = 4.

Posted from my mobile device

KarishmaB · Answer

We are given a second degree equation in 2 variables and asked the possible value of one of the variables. We know that there are infinite values in such a case. But here we are given that both variables must be integers and that reduces our number of acceptable values.

x will vary as y will take different values. I have a 2y^2 on the left and another square. On the right I have 27 which is a very small number. This means my y^2 will be less than 14 and hence the only options for y I have are 0, 1, 2, 3 and negatives. Since y is squared in 2y^2, I will ignore the negatives to begin with.

Put y = 0, we get  (x-y)^2 = 27 . Eliminate

Put y = 1, we get  (x - y)^2 = 25

So (x-y) can be 5 or -5. 
y can be 1 or -1. Possible values are 6, -4, 4, -6

Answer (D)

Had we not got an answer, we would have continued with other 2 values y = 2 and y = 3.

Kinshook · Answer

pablovaldesvega

If  x  and  y  are integers and if  (x-y)^2 + 2*y^2 = 27 , which of the following could be the value of  x ?

(x-y)^2 + 2*y^2 = 27

y = 1; (x-y)^2 = 25; x -y = 5 ; x = 6; y - x = 5; x = -4
y = 3; (x-y)^2 = 9 ; x - y = 3; x = 6; y - x = 3; x = 0
y = -1 ; (x-y)^2 = 25; x -y = 5 ;  x = 4 ; y - x = 5; x = -5

IMO D

bumpbot · Answer

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