GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 22 Sep 2019, 01:15

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# For how many ordered pairs (x , y) that are solutions of the equation

Author Message
TAGS:

### Hide Tags

Director
Joined: 19 Oct 2018
Posts: 880
Location: India
For how many ordered pairs (x , y) that are solutions of the equation  [#permalink]

### Show Tags

Updated on: 16 Jul 2019, 13:02
1
(x-y)^2+2y^2=18

As $$18=0^2+2*3^2$$
and $$18= 4^2+2*1^2$$

Hence there are 2 cases possible

Case 1- when $$(x-y)^2$$=0 and $$y^2$$=9
$$(x-y)^2=0$$
or x-y=0.....(1)

$$y^2=9$$
y= 3 or -3

We have 2 distinct integral values of y and 1 integral value of x-y.
Hence number of solutions possible= 2*1=2

Case 2 - When $$(x-y)^2=16$$ and $$y^2=1$$
$$(x-y)^2=16$$
(x-y)=4 or -4

and $$y^2=1$$
y=1 or -1

We have 2 distinct integral values of x-y and 2 distinct integral values of y
Number of solutions possible= 2*2=4

Total number of solutions possible= 2+4=6

IMO C

[There are couple of reasons for why we don't need to find the solutions of (x, y)
1. If (x-y) and y are integers, x will always be an integer.
2. Values of y are different in both cases, hence there is no overlap possible.
3 it will save some precious time. ]

Originally posted by nick1816 on 16 Jul 2019, 10:33.
Last edited by nick1816 on 16 Jul 2019, 13:02, edited 1 time in total.
Manager
Joined: 21 Jan 2019
Posts: 100
For how many ordered pairs (x , y) that are solutions of the equation  [#permalink]

### Show Tags

16 Jul 2019, 10:40
Quote:
(x−y)2+2y2=18(x−y)2+2y2=18

For how many ordered pairs (x , y) that are solutions of the equation above are x and y both integers?

A. 2
B. 4
C. 6
D. 8
E. 10

18 can be a sum of 2 terms wherein one is a square ($$(x-y)^2$$) and another is a multiple of 2 (2* $$y^2$$)

Only possible pair is 16 + 2 = 18

Now either $$(x-y)^2 =16$$ and $$2y^2 = 2$$ or $$(x-y)^2$$= 2 and $$2y^2 = 16$$

As for $$2 y^2= 16$$, y wont be an integer so we rule out scenario 2.
Hence for first scenario, $$(x-y)^2 =16$$ and $$2y^2 = 2$$
$$2y^2 = 2$$
therefore y = +1,-1
and for $$(x-y)^2 =16$$
(x-y) = +4, - 4
Therefore each will have 2 pairs of values (One with y=1 and another with y=-1). So total 4 pairs.
Option B
Intern
Joined: 08 Apr 2011
Posts: 12
Re: For how many ordered pairs (x , y) that are solutions of the equation  [#permalink]

### Show Tags

16 Jul 2019, 10:51
(x−y)^2+2y^2=18

For how many ordered pairs (x , y) that are solutions of the equation above are x and y both integers?

A. 2
B. 4
C. 6
D. 8
E. 10

If we look at the equation it tells us that (X-Y)^2 will always be integer and that too of an even number,

since (x-y)^2 = 18 - 2y^2->>> even
the even square less than 18 are only 16 and 4 so effectively 2 pairs can help us to reach 18 on RHS,

y = 1, x = 5
y=-1 x= 3

Manager
Joined: 17 Jan 2017
Posts: 86
Re: For how many ordered pairs (x , y) that are solutions of the equation  [#permalink]

### Show Tags

16 Jul 2019, 11:07
(x−y)^2+2y^2=18

For how many ordered pairs (x , y) that are solutions of the equation above are x and y both integers?

A. 2
B. 4
C. 6
D. 8
E. 10

(x−y)^2+2y^2=18 or, (x−y)^2 =18 - 2y^2
it shows that y can not be more than 3 as (x−y)^2 will be positive. so, there are 6 values of y are possible (-3, -2, -1, 0, 1, 2, 3).
So, plugging in the value of y, the possible integer value of x and y are: (-3,-3 ), (-5, -1), (5, 1), (3, 3)

So, the correct answer choice is (B)
Manager
Joined: 18 Jun 2013
Posts: 136
Location: India
Concentration: Technology, General Management
GMAT 1: 690 Q50 V35
GPA: 3.2
WE: Information Technology (Consulting)
Re: For how many ordered pairs (x , y) that are solutions of the equation  [#permalink]

### Show Tags

16 Jul 2019, 11:42
1
Given,

(x − y)^2 + 2*y^2 = 18

This can be seen as 18 being the sum of 2 squares.

Case 1:

=> The first perfect square less than 18 is 16. Let us see if we can have integer values of x and y such that one of the squars is 16

if y = 1, we can try to get (x - y)^2 to be 16.

=> x = 5, y = 1 satisfies the equation. Now let us play with the signs of x and y and see which combinations satisfy the equation. By doing so we get the following 3 combinations of x and y.

1. (5 , 1)
2. (-5 , 1)
3. (-5 , -1)

Case 2:

=> The second perfect square less than 18 is 9. Let us see if we can have integer values of x and y such that one of the squars is 9

if y = 3, we can try to get (x - y)^2 to be 9.

=> x = 3, y = 3 satisfies the equation. Now let us play with the signs of x and y and see which combinations satisfy the equation. By doing so we get the following 3 combinations of x and y.

4. (3 , 3)
5. (-3 , -3)
6. (-3 , 3)

Case 3:

=> The third perfect square less than 18 is 4. Let us see if we can have integer values of x and y such that one of the squars is 4

By putting multiple values of y we see that we do not have an integer pair for x and y such that the equation satisfies where one of the squares in the LHS is the perfect square 4.

Thus we conclude that we have 6 unique integer pairs of x and y which satisfy the given equation.

Manager
Joined: 10 Aug 2016
Posts: 68
Location: India
Re: For how many ordered pairs (x , y) that are solutions of the equation  [#permalink]

### Show Tags

16 Jul 2019, 11:46
1
(x−y)^2+2y^2=18

For how many ordered pairs (x , y) that are solutions of the equation above are x and y both integers?

(x-y)^2 + 2y^2 = 18
In the above equation both square terms will be positive and less than 18 respectively.

By trail and error,
If y = 0, then solving we get x^2 = 18.. not a perfect square.
If y = 1, then solving we get (x-1)^2 = 16 --> x-1 = (+/-)4 --> x = 5 or -3. Hence two pairs (5,1) and (-3,1)
If y = 2, then solving we get (x-2)^2 = 10...not a perfect square.
If y = 3, then solving we get (x-3)^2 = 0 --> x = 3. One pair (3,3)
If y = 4, then 2y^2 becomes > 18 hence y cannot take any value > 3

Similarly y can take negative numbers -1, -2, -3... anything -4 and below we will have second term > 18 hence not possible.

If y = -1, then x+1 = (+/-)4 --> x = 3, -5. Hence two pairs (3, -1) and (-5, -1)
If y = -2, then (x+2)^2 = 10. Not a perfect square.
If y = -3, then (x+3)^2 = 0 --> x = -3. One pair (-3, -3)

Hence total pairs (x, y) are (5,1); (-3,1); (3,3); (3, -1); (-5, -1); (-3, -3)
Total number of pairs = 6.

A. 2
B. 4
C. 6
D. 8
E. 10

Manager
Joined: 22 Oct 2018
Posts: 73
Re: For how many ordered pairs (x , y) that are solutions of the equation  [#permalink]

### Show Tags

16 Jul 2019, 12:09
x−y
)
2
+2
y
2
=18
(x−y)2+2y2=18

For how many ordered pairs (x , y) that are solutions of the equation above are x and y both integers?

A. 2
B. 4
C. 6
D. 8
E. 10

Solution:

This problem can be approached by substituting the values of x and y to solve for 18.

$$(x-y)^2$$+2$$y^2$$=18
Substittuing ,
x=5 and y=1

$$(5-1)^2$$+2*$$1^2$$=$$4^2$$+2=18

x=-3 and y=1
$$(-3-1)^2$$+2*$$1^2$$=(-4)^2+2=18

Although there are other values of x and y which satisfy this equation but the value of ordered pair is not integer(for eg when x=0 then y=+$$\sqrt{(6)^}$$ and -$$\sqrt{(6)^}$$.But this value of y is not an integer..Hence we have two ordered pairs of (x,y) i.e (5,1) and (-3,-1) which are the solution of the equation.

Hence A IMO
Manager
Joined: 24 Jun 2019
Posts: 108
Re: For how many ordered pairs (x , y) that are solutions of the equation  [#permalink]

### Show Tags

16 Jul 2019, 12:15
1
$$(x−y)^2+2y^2=18$$

For how many ordered pairs (x , y) that are solutions of the equation above are x and y both integers?

Another question that requires trial and error!! Whenever I see trial and error questions, I feel there must be more elegant solutions that I am missing. So I will look forward to the elegant solutions tomorrow!

Here is how I solved it:

We know that x and y are both integers...

Let $$(x-y)^2$$ = A .... Here, A is an integer because x and y are integers. Also A is non negative because A is a square of integer.
Let $$2y^2$$ = B.... Here, B is integer because y is integer. Also, B is non negative because B is 2 times square of integer.

A + B = 18

We need to find combinations of Non negative integers A and B such that A+B = 18

A and B will be less than or equal to 18 because both are non negative and A has to be a PERFECT SQUARE
Also, B/2 has to be a perfect square

Below are all possibilities with A as perfect square
0+18 = 18 .... This is possible because 18/2 is perfect square
1+17 = 18..... NOT POSSIBLE.... 17/2 is not perfect square
4+14 = 18..... NOT POSSIBLE.... 14/2 is not perfect square
9+9 = 18..... NOT POSSIBLE.... 9/2 is not perfect square

16+2 = 18..... This is possible because 2/2 is perfect square

Possible combinations of A and B are:
A = 0 and B = 18.... I
A = 16 and B = 2..... II

I
A=0
$$(x-y)^2$$ = 0
x-y = 0
x=y

B=18
$$2y^2=18$$
y=+ or - 3

If y = 3, x = 3....(1)
If y = -3, x=-3......(2)

II
A=16
$$(x-y)^2=16$$
x-y=+ or -4
x=y+4 or x = y-4

B=2
$$2y^2=2$$
y=+ or - 1

If y=1,
x = 5 .... (3)
or
x = -3.... (4)

If y = -1,
x = 3... (5)
or
x = -5.... (6)

Answer: C - 6 possible pairs of (x,y) are solutions to above equations.
Senior Manager
Joined: 31 May 2018
Posts: 306
Location: United States
Concentration: Finance, Marketing
Re: For how many ordered pairs (x , y) that are solutions of the equation  [#permalink]

### Show Tags

16 Jul 2019, 12:48
1
$$(x−y)^2$$+2$$y^2$$ = 18 ---(1)

$$(x−y)^2$$ = 18-2$$y^2$$

since $$(x−y)^2$$ ≥ 0
so 18-2$$y^2$$≥ 0
9-$$y^2$$≥ 0
(y+3)(y-3)≤ 0

values of y---- -3≤ y ≤3---(-3,-2,-1,0,1,2,3)

when y=-3
solving the equation -(1) we get x=-3
we get (-3,-3)

when y=-2 we will not get the integer value of x (given both x and y are integers)
so y=-2 is not possible

when y=-1
solving equation -(1) we get
$$(x+1)^2$$ = 16
from here we get x=3 and x=-5 (both values of x satify equation -(1) for y=-1)
so we get (3,-1) and (-5,-1)

when y=0 we will not get the integer value of x so y=0 is not possible

when y=1
solving equation -(1) we get
$$(x-1)^2$$ = 16
from here we get x=5 and x=-3 (both values of x satify equation -(1) for y= 1)
so we get (5,1) and (-3,1)

y=2 is not possible since we will not get the integer value of x

when y=3 we get x=3 solving equation --(1)
so (3,3)

we get the total ordered pair (x,y) = (-3,-3), (3,-1), (-5,-1), (5,1), (-3,1), (3,3)
total 6 ordered pair
Manager
Joined: 06 Feb 2019
Posts: 108
Re: For how many ordered pairs (x , y) that are solutions of the equation  [#permalink]

### Show Tags

16 Jul 2019, 12:59
For how many ordered pairs (x , y) that are solutions of the equation above are x and y both integers?

First of all, we need to simplify the equation
(x-y)^2 +2y^2 = 18
(x-y)^2 = 2*9 - 2y^2
(x-y)^2 = 2*(9-y^2)

From the given equation we see that (9-y^2) after multiplying by 2 must equals something squared (= (x-y)^2).
We need y and x to be integers, so 2*(9-y^2) may equals 0, 1, 4, 9, 16, 25, 36 etc.
From 2*(9-y^2) = 0 and (x-y)^2 = 2*(9-y^2) we have y = -3, x = -3 OR y = 3, x = 3
From 2*(9-y^2) = 1 we can't have integer y
From 2*(9-y^2) = 2 we can't have integer y
From 2*(9-y^2) = 4 we can't have integer y
From 2*(9-y^2) = 9 we can't have integer y
From 2*(9-y^2) = 16 and (x-y)^2 = 2*(9-y^2) we have y = 1 and x = 5 OR y = -1 and x = 3

we will not consider every squared integer number more than 16 (for example, 2*(9-y^2) = 25), because 2*(9-y^2) can't be more than 18.

So the answer is 4 pairs - B
Manager
Joined: 29 May 2019
Posts: 105
Re: For how many ordered pairs (x , y) that are solutions of the equation  [#permalink]

### Show Tags

16 Jul 2019, 13:15
(x−y)^2 + 2y^2=18

For how many ordered pairs (x , y) that are solutions of the equation above are x and y both integers?

We need sum of both (x−y)^2 and 2y^2 to be 18.
As both are squares 18 is sum of two positive integers.

If we consider y to be 1.
then 2y^2 = 2
therefore, (x−y)^2 must be 16
x-y = 4
y is 1 so x will be 5
( 5, 1 )

We can take y as
1,
$$\sqrt{2}$$
$$\sqrt{3}$$
$$\sqrt{4}$$
$$\sqrt{5}$$
$$\sqrt{6}$$
$$\sqrt{7}$$
$$\sqrt{8}$$

y can not be $$\sqrt{9}$$ as y = 3 then x will also be 3.

Its not an ordered pair.

So total values will be 8.

_________________
Pick yourself up, dust yourself off, and start again.

Success is the sum of all small efforts.

MAKE IT HAPPEN
Senior Manager
Joined: 18 May 2019
Posts: 269
Re: For how many ordered pairs (x , y) that are solutions of the equation  [#permalink]

### Show Tags

16 Jul 2019, 13:39
(x-y)^2+2y^2=18
We are to find integer values of x and y that satisfy the above equation.
(x-y)>4 will produce LHS>RHS
So our combination integer values of x and y which can satisfy this equation must be such that (x-y)<=4.

When y=-1, (x+1)^2+2=18
x^2+2x-15=0
(x-3)(x+5)=0
Hence x=-5 or x=3
So we have (x,y) pairs of (-5,-1) and (3,-1)

When y=1, (x-1)^2+2=18
x^2-2x-15=0
(x+5)(x-3)=0
Hence x=-5 or x=3
We have (x,y) integer pairs of (-5,1) and (3,1).

Every other integral combinations of (x,y) does not result in a LHS which equals 18 (RHS).

Therefore there are 4 ordered pairs (x,y) that are solutions of the equation.

Posted from my mobile device
Intern
Joined: 02 Jun 2013
Posts: 43
Re: For how many ordered pairs (x , y) that are solutions of the equation  [#permalink]

### Show Tags

16 Jul 2019, 15:44
Here (x−y)^2+2^2=18

So we can try this with number picking method too. Both elements are squared , so they have to be positive.

Now lets say y =0 then after putting this value in above equation we dont get the integer value of x.

Y= 1 , then x= 5,-3 so , here got two pair ( 5,1) (-3,1)
y= 2, then x cant be an integer
y= 3, then then x can be 3 so here the pair is (3,3) now we can have (-3,-3) too

So there are 4 pairs . Ans - B
_________________
Nothing comes easy ! Neither do i want !!
Manager
Joined: 30 Aug 2018
Posts: 89
Location: India
Concentration: Finance, Accounting
GPA: 3.36
WE: Consulting (Computer Software)
Re: For how many ordered pairs (x , y) that are solutions of the equation  [#permalink]

### Show Tags

16 Jul 2019, 16:13
Start populating with y=0 ---> x integer
y=1 ---> x not integer
y=2---->x not integer
y=3------> x integer.
therefore 1,3 satisfies also -1,-3 will satisfy
total 4 solutions
GMAT Club Legend
Joined: 18 Aug 2017
Posts: 4783
Location: India
Concentration: Sustainability, Marketing
GPA: 4
WE: Marketing (Energy and Utilities)
For how many ordered pairs (x , y) that are solutions of the equation  [#permalink]

### Show Tags

16 Jul 2019, 17:34
solving for the given eqn ; we get
x^2+3y^2-2xy=18
test with y=1 we get x=5,-3 then at x=5 y=3,7
total pairs ; IMO B ; 4

(x−y)2+2y2=18(x−y)2+2y2=18

For how many ordered pairs (x , y) that are solutions of the equation above are x and y both integers?

A. 2
B. 4
C. 6
D. 8
E. 10
_________________
If you liked my solution then please give Kudos. Kudos encourage active discussions.
Senior Manager
Joined: 30 Sep 2017
Posts: 343
Concentration: Technology, Entrepreneurship
GMAT 1: 720 Q49 V40
GPA: 3.8
WE: Engineering (Real Estate)
For how many ordered pairs (x , y) that are solutions of the equation  [#permalink]

### Show Tags

Updated on: 16 Jul 2019, 18:51
1
$$(x−y)^2 + 2y^2 = 18$$ can be rewritten as $$(x−y)^2 = 2*(3 - y)*(3+y)$$

$$(x−y)^2 \geq{0}$$, then $$2*(3 - y)*(3+y) \geq{0}$$, which limits us to $$-3 \leq{y} \leq{3}$$. Given x and y have to be both integers, y can be of only 7 possible values: $$-3, -2, -1, 0, 1, 2, 3$$

$$y=0,$$ $$y=-2$$ and $$y=2$$ all lead to non-integer x. Eliminate them.

There are six integer pairs $$(x , y)$$ that are solutions of the equation as shown below:
$$y=-3,$$ then $$(x+3)^2 =0$$ and $$x=-3$$ --> $$(-3,-3)$$ is a solution.
$$y=3,$$ then $$(x-3)^2 =0$$ and $$x=3$$ --> $$(3,3)$$ is a solution.
$$y=-1,$$ then $$(x+1)^2 =16$$ and $$x=-5$$ or $$3$$ --> $$(-5,-1)$$ and $$(3,-1)$$ are both solution.
$$y=1,$$ then $$(x-1)^2 =16$$ and $$x=5$$ or $$-3$$ --> $$(5,1)$$ and $$(-3,1)$$ are both solution.

Originally posted by chondro48 on 16 Jul 2019, 18:45.
Last edited by chondro48 on 16 Jul 2019, 18:51, edited 2 times in total.
Intern
Joined: 22 Jun 2019
Posts: 41
Concentration: Operations, Organizational Behavior
Re: For how many ordered pairs (x , y) that are solutions of the equation  [#permalink]

### Show Tags

16 Jul 2019, 18:46
(x - y)^2 + 2 y^2 = 18
x^2 - 2 x y + 3 y^2 - 18 = 0

plug in numbers
x = -5, y = -1
x = -3, y = -3
x = -3, y = 1
x = 3, y = -1
x = 3, y = 3
Manager
Joined: 28 Feb 2014
Posts: 146
Location: India
GPA: 3.97
WE: Engineering (Education)
Re: For how many ordered pairs (x , y) that are solutions of the equation  [#permalink]

### Show Tags

16 Jul 2019, 20:02
1
(x−y)^2+2y^2=18

We can start putting values of y:
y=0; (x)^2=18, does not give x as interger
y=1; (x−1)^2+2=18 --> (x−1)^2=16 -> (x−1)=+/-4 -> x=5 or-3
y=-1; (x+1)=+/-4 -> x=3 or -5
y=3; (x−3)=0 -> x=3
y=-3; (x+3)=0 -> x=-3

Ordered pairs of (x,y)
5,1
-3,1
3,-1
-5,1
3,3
-3,3

C is correct.
Senior Manager
Joined: 18 Jan 2018
Posts: 308
Location: India
Concentration: General Management, Healthcare
Schools: Booth '22, ISB '21, IIMB
GPA: 3.87
WE: Design (Manufacturing)
Re: For how many ordered pairs (x , y) that are solutions of the equation  [#permalink]

### Show Tags

16 Jul 2019, 20:12
Given equation is (x−y)^2+2y^2=18 , and x & y are integers

Instead of trying to solve this equation , we can do trail and error method

lets say (x−y)^2+2y^2=18

18 - could be written as 16+2 , 2*9,6*3 , 18*1

But Only 16+2 and 2*9 can will us solutions where x and y are integers

==> case 1 : where x = y = 3 , (3−3)^2+2*3^2=18
Case 2 : where x = y = -3 , (-3+3)^2+2(-3)^2=18
Case 3 : where x = 5 , y=1 , (5-1)^2+2*1^2=18
Case 4 : where x = 3 , y= -1 , (3+1)^2+2(-1)^2=18

Total 4 cases ( 3,3) ,(-3,-3) , (5,1) ,(4,-1)--Option B
Manager
Joined: 08 Jan 2018
Posts: 129
For how many ordered pairs (x , y) that are solutions of the equation  [#permalink]

### Show Tags

16 Jul 2019, 21:15
1
We need to find ordered pairs such that x and y both are integers.
Let us look at the equation once and see what we can determine:

$$(x – y)^2$$ will always be positive -> (a)
$$y^2$$ will always be positive -> (b)
Thus, 18 will be the result of the sum of squares of two positive numbers. -> (c)

Looking at the equation and from (b) and (c) we find that y can take a range of value such that $$-3 \leq{y} \leq{+3}$$
Thus, we can find all ordered pairs in the given range of y

y = -3
$$(x + 3)^2$$ – 18 = 0
x = 3
So (3, -3) is a ordered pair. -> [1]

y = -2
$$(x + 2)^2$$+ 8 =18
$$(x + 2)^2$$ = 10 -> 10 is not a square of any integer.

y = -1
$$(x + 1)^2$$ – 2 = 18
$$(x + 1)^2$$ = 20 -> 20 is not a square of any integer.

y = 0
$$x^2$$ = 18
x = -9 OR x = 9
So (-9, 0) is a ordered pair. -> [2]
So (9, 0) is a ordered pair. -> [3]

y = 1
$$(x – 1)^2$$ + 2 = 18
$$(x – 1)^2$$ = 16
x = 5 OR x = -3
So (-3, 1) is a ordered pair. -> [4]
So (5, 1) is a ordered pair. -> [5]

y = 2
$$(x – 2)^2$$ = 10 -> 10 is not a square of any integer.

y = 3
$$(x -3)^2$$ + 18 = 18
x = 3
So (3, 3) is a ordered pair. -> [6]

From [1], [2], [3], [4], [5] and [6] we have 6 ordered pairs.

For how many ordered pairs (x , y) that are solutions of the equation   [#permalink] 16 Jul 2019, 21:15

Go to page   Previous    1   2   3   4    Next  [ 75 posts ]

Display posts from previous: Sort by