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

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

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16 Jul 2019, 07:00
1
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95% (hard)

Question Stats:

27% (01:45) correct 73% (02:17) wrong based on 275 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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For how many ordered pairs (x , y) that are solutions of the equation  [#permalink]

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Updated on: 16 Jul 2019, 09:12
8
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
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Originally posted by Kinshook on 16 Jul 2019, 07:29.
Last edited by Kinshook on 16 Jul 2019, 09:12, 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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16 Jul 2019, 07:33
2
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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Updated on: 16 Jul 2019, 21:32
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 Lampard42 on 16 Jul 2019, 07:38.
Last edited by Lampard42 on 16 Jul 2019, 21:32, edited 1 time in total.
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For how many ordered pairs (x , y) that are solutions of the equation  [#permalink]

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Updated on: 16 Jul 2019, 08:18
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, 07:41.
Last edited by firas92 on 16 Jul 2019, 08:18, edited 1 time in total.
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For how many ordered pairs (x , y) that are solutions of the equation  [#permalink]

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Updated on: 16 Jul 2019, 12:02
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, 09:33.
Last edited by nick1816 on 16 Jul 2019, 12:02, 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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16 Jul 2019, 10:42
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.

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

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16 Jul 2019, 10:46
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

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

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16 Jul 2019, 20:15
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, 20:15