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Math Expert V
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For how many ordered pairs (x , y) that are solutions of the equation  [#permalink]

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Question Stats: 27% (01:43) correct 73% (02:16) wrong based on 262 sessions

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$$(x - y)^2 + 2y^2 = 18$$

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 This question was provided by Math Revolution for the Game of Timers Competition _________________
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GMAT 1: 690 Q50 V34 For how many ordered pairs (x , y) that are solutions of the equation  [#permalink]

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7
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?

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

Let us take y^2=0 => (x-y)^2 = 18 NOT FEASIBLE since it is not a perfect square. (No integer solution)
Let us take y^2= 1 => 2y^2 = 2 & (x-y)^2=16 (1). FEASIBLE since it is a perfect square (integer solutions)
Let us take y^2= 4 => 2y^2=8 => (x-y)^2 = 10 NOT FEASIBLE since it is NOT a perfect square (No integer solutions)
Let us take y^2=9 => 2y^2 =18 => (x-y)^2 =0 (2) FEASIBLE since it is a perfect square (integer solutions)

Take equation (1)
2y^2 = 2=> y^2 = 1 => y=1 or y=-1
(x-y)^2 = 16 => x-y = 4 or x-y = -4
If y=1 and x-y=4 => x =5, y=1 (5,1) (i)
If y=1 and x-y=-4 => x=-3, y=1 (-3,1). (ii)
If y=-1 and x-y=4 => x=3 y=-1 (3,-1). (iii)
If y=-1 and x-y=-4 => x= -5 y=-1 (-5,-1) (iv)
Equation (1) provides 4 ordered pairs

Take equation (2)
(x-y)^2=0 => x=y
2y^2=18 => y=3 or y=-3
If y=3 => x=3 (3,3)
If y=-3 => x=-3 (-3,-3)
Equation (2) provides 2 ordered pairs

Total six (6) ordered pairs are solution

IMO C

Originally posted by Kinshook on 16 Jul 2019, 08:29.
Last edited by Kinshook on 16 Jul 2019, 10:12, edited 1 time in total.
##### General Discussion
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Re: For how many ordered pairs (x , y) that are solutions of the equation  [#permalink]

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(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

Observing the equation carefully.
We can have either 2y^2 = 18 when x = y which means x=y=3
Only one perfect square of integer when multiplied by 2. subtracted from 18 will leave another perfect square.
When y =1 we will have 2y^2 = 2 hence it will leave (x-y)^2 = 16.
For no other value of y we will have perfect square for (x-y)^2.

hence correct answer is 2 pairs.
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Re: For how many ordered pairs (x , y) that are solutions of the equation  [#permalink]

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With the possible cases : x=3,y=3

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

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

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(x−y)^2 + 2(y)^2 = 18

Now both the terms must be +ve on the LHS.
y can't be >3 (OR <-3) because 2(4)^2 = 32 which is > RHS

Let y be 3, then

(3-3) ^2 + 2*9 =18 (3,3) is a soln.

Let y be 2, then

(x−y)^2 + 8 = 18, but (x−y)^2 can't be 10 for any integer pair (x,2)

Let y be 1, then

16 + 2 = 18, possible for (5,1)

If y=0, x^2 =18 won't give an integer.

If y=-1 then x can be 3, So (3,-1)

Again y can't be -2 But,

For y= -3, x can be -3 so that (x−y)^2 + 2(y)^2 = 18 (-3,-3)
4 pairs of soln. in total.

Originally posted by LeoN88 on 16 Jul 2019, 08:24.
Last edited by LeoN88 on 16 Jul 2019, 08:32, edited 2 times in total.
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Re: For how many ordered pairs (x , y) that are solutions of the equation  [#permalink]

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1
1
C

Clearly y can't be greater than 3. In order for x to be integer, y can only take +/- 1 and +/- 3

For y =1, we get x = 5 and -3 - two solutions
For y = -1, we get x = 3 and -5 - two solutions
For y = 3, we get x = 3
For y = -3, we get x = -3

Total 6 solutions.
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For how many ordered pairs (x , y) that are solutions of the equation  [#permalink]

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1
Square + even no = 18 - therefore the square should also be even or 0.

The only possibilities are
0 + 18
4 + 14
16 + 2

First scenario, 0 + 18
(x-y)^2 = 0 and 2y^2=18
x=y and y=+/-3
(x,y) = (3,3) or (-3,-3)

Second scenario, 4 + 14
2y^2 =14, y^2 =7 which is not possible since y is an integer

Third scenario, 16 + 2
(x-y)^2=16 and 2y^2=2
x-y=4 and y=+/-1
x=5,y=1 or X=3,y=-1

(3,3)
(3,-3)
(5,1)
(3,-1)
4 pairs

Option B

Posted from my mobile device

Originally posted by prashanths on 16 Jul 2019, 08:36.
Last edited by prashanths on 16 Jul 2019, 22:38, edited 2 times in total.
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Re: For how many ordered pairs (x , y) that are solutions of the equation  [#permalink]

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Solving the equation we find it's an equation of circle and the only integers that fit the value of x and y are 3,-3 so totally there would be 4 combinations of x and y. (3,3),(-3,-3),(3,-3),(-3,3).

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

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3
(x−y)^2+2y^2=18 - Includes both +ve and -ve Integers

(x−y)^2 can have Perfect squares values of 0,1,4,9,16 only.
If (x−y)^2 is odd then 2*y^2 must be odd - which is not possible. So (x−y)^2 can take only 0,4 or 16.

If (x−y)^2 is 4 then 2*y^^2=14; y^2=7 not possible.
Hence (x-y)^2 can be 16 which gives y=1 or -1. If Y=1 then X=5 or -3 and Y=-1 then X= 3 or -5.
When (x-y)^2 is 0, 2y^=18, y=3,-3. Then x=3,-3
Hence 4 ordered pairs are possible (5,1),(-3,1),(3,-1) , (-5,-1), (3,3) and (-3,-3).

IMO C

Originally posted by Arvind42 on 16 Jul 2019, 08:38.
Last edited by Arvind42 on 16 Jul 2019, 22:32, edited 1 time in total.
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Re: For how many ordered pairs (x , y) that are solutions of the equation  [#permalink]

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$$(x−y)^2+2y^2=18$$

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

Solution -

$$(x−y)^2 and y^2$$ are always positive.

So , $$(x−y)^2+2y^2 =18$$ is possible in only one way. x-y = 0 => x=y

$$x =y =3$$
or
$$x =y =-3$$

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Re: For how many ordered pairs (x , y) that are solutions of the equation  [#permalink]

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(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

I think that the answer is 6 C.
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For how many ordered pairs (x , y) that are solutions of the equation  [#permalink]

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3
(x-y)^2 and 2y^2 are always non negative

So let's see what are the cases we can get a perfect square + 2*perfect square = 18 (We take perfect squares since we need x and y to be integers)

Perfect squares: 0, 1, 4, 9, 16 ------- (1)
2*Perfect squares: 0, 2, 8, 18 ------- (2)

So only cases when an element from list 1 and an element from list 2 add up to 18 is (0+18) and (16+2)

(0+18) - Possible when (x,y) = (3,3) and (-3,-3) - 2 cases
(16+2) - Possible when (x,y) = (5,1), (-5,-1), (-3,1) and (3,-1) - 4 cases

Therefore we have 6 cases in total

Originally posted by firas92 on 16 Jul 2019, 08:41.
Last edited by firas92 on 16 Jul 2019, 09:18, edited 1 time in total.
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Re: For how many ordered pairs (x , y) that are solutions of the equation  [#permalink]

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(x−y)^2+2y^2=18
(x−y)^2=18-2y^2

The left hand side of the equation is a perfect square, so the the RHS has to be a perfect square as well.
Given that X & Y both are integers.

so,if we put Y=1 or -1, then RHS=16 which is a perfect square, so solving the LHS & replacing the value of RHS , (x−y)^2 = 16; we can have 2 different values of X;

If we put Y=+-2, then, RHS is not a perfect square; ignore
If we put Y=+-3,then RHS is equal to 0; which is a perfect square; so we have 2 different sets of value for X & Y again.

hence IMO correct answer is B
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Re: For how many ordered pairs (x , y) that are solutions of the equation  [#permalink]

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IMO Answer is 2 pairs, A

(x−y)^2+2y^2=18

we know that (x-y)^2 >0 and y^2>0
if x and y are integers, possible values of (x-y)^2 --> 0,1,4,9,16
possible values of 2y^2 --> 0,2,8,18

the sum of above two is 18.
looking at numbers we can easily omit, 1,4,9 from X values as there is no 17,14 or 9 in required y values.

only possible x numbers are 0,16
which pairs are 18,2 for 2y^2

so 2 pairs or A
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Re: For how many ordered pairs (x , y) that are solutions of the equation  [#permalink]

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Let's list out all the set of squares of integers that are less than 18. - 1, 4, 9, 16

For the equation $$(x-y)^2 + 2y^2 = 18$$, lets say, $$z= x-y$$

Possible values of $$z^2$$ and $$y^2$$ are $$(16,1)$$ & $$(0,9)$$

So, the possible values of x and y are $$(0,3); (0,-3); (3,-1); (-5,-1); (-3,1)$$ and $$(5,1)$$

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Re: For how many ordered pairs (x , y) that are solutions of the equation  [#permalink]

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The solutions to the equation are

(5,1)

(-3,1)

(3,3)

Now reversing the signs

(-5,-1)

(-3,-3)

(3,-1)

C it is for.me

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Re: For how many ordered pairs (x , y) that are solutions of the equation  [#permalink]

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Since x and y are both integers x-y will also be an integer.

Since 2y^2 shall always be positive. (x-y)^2 shall be less than 18.

(x-y)^2 can be 1,4,9,16
see that it cant be 1 or 9 as even-odd will give an odd number and 2y^2 is even

for 4, we have- 2y^2=14 which gives a non integer result for y. hence discarded

for 16, 2y^2=18-16=2
y^2=1. y can be +1 or -1

when y=+1, x is 5 as (x-y)^2=16
when y=-1. x is 3

so there are 2 ordered pairs

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Re: For how many ordered pairs (x , y) that are solutions of the equation  [#permalink]

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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

(x-y)ˆ2=18-2yˆ2… (x-y)ˆ2=perfect square;
18-2yˆ2…(2*3ˆ2)-2yˆ2…2(3ˆ2-yˆ2)…2(9-yˆ2)={1,4,16,25,36…}
only possible integer solution for y is when 2(9-yˆ2)=16…9-yˆ2=8…-yˆ2=-1…y=1;
(x-y)ˆ2=16… (x-1)ˆ2=16…(x-3)(x+1)=0;
possible pairs are (3,1) and (-1,1) = 2 pairs;

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Re: For how many ordered pairs (x , y) that are solutions of the equation  [#permalink]

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I will use plugin method since there will be less option that will satisfy the above equation.

For y=1
(x-y)^2 = 16
ordered pair can be (5,1) (-3,1)

for y = -1
(x-y)^2 = 16
ordered pair can be (3,-1) (-5,-1)

y=2 no ordered pair (x-y)^2 not equal to 8

for y=3
ordered pair can be (3,3) (-3,-3)

Total 6 pairs.
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Re: For how many ordered pairs (x , y) that are solutions of the equation  [#permalink]

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From the statement we have that maximum and minimum values for Y are:

2y^2=18 so, Y =3 or - 3, so x=0 for both (we have 2 sets here)

Then, with this restrictions of y, we have 2 more combinations that comply with the formula: (5,1) and (-3, -1). Re: For how many ordered pairs (x , y) that are solutions of the equation   [#permalink] 16 Jul 2019, 08:58

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