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For how many ordered pairs (x , y) that are solutions of the equation
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16 Jul 2019, 08:00
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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
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For how many ordered pairs (x , y) that are solutions of the equation
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Updated on: 16 Jul 2019, 10:12
(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 => (xy)^2 = 18 NOT FEASIBLE since it is not a perfect square. (No integer solution) Let us take y^2= 1 => 2y^2 = 2 & (xy)^2=16 (1). FEASIBLE since it is a perfect square (integer solutions) Let us take y^2= 4 => 2y^2=8 => (xy)^2 = 10 NOT FEASIBLE since it is NOT a perfect square (No integer solutions) Let us take y^2=9 => 2y^2 =18 => (xy)^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 (xy)^2 = 16 => xy = 4 or xy = 4 If y=1 and xy=4 => x =5, y=1 (5,1) (i) If y=1 and xy=4 => x=3, y=1 (3,1). (ii) If y=1 and xy=4 => x=3 y=1 (3,1). (iii) If y=1 and xy=4 => x= 5 y=1 (5,1) (iv) Equation (1) provides 4 ordered pairs
Take equation (2) (xy)^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.




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Re: For how many ordered pairs (x , y) that are solutions of the equation
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16 Jul 2019, 08:14
(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 (xy)^2 = 16. For no other value of y we will have perfect square for (xy)^2.
hence correct answer is 2 pairs. Answer = A



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Re: For how many ordered pairs (x , y) that are solutions of the equation
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16 Jul 2019, 08:16
Answer is Option B
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
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Updated on: 16 Jul 2019, 08:32
(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
(33) ^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.
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Re: For how many ordered pairs (x , y) that are solutions of the equation
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16 Jul 2019, 08:33
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
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Updated on: 16 Jul 2019, 22:38
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 (xy)^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 (xy)^2=16 and 2y^2=2 xy=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
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Originally posted by prashanths on 16 Jul 2019, 08:36.
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Re: For how many ordered pairs (x , y) that are solutions of the equation
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16 Jul 2019, 08:38
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
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Updated on: 16 Jul 2019, 22:32
(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 (xy)^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 (xy)^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
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16 Jul 2019, 08:38
\((x−y)^2+2y^2=18\)
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. xy = 0 => x=y
\(x =y =3\) or \(x =y =3\)
Answer  A



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Re: For how many ordered pairs (x , y) that are solutions of the equation
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16 Jul 2019, 08:39
(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
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Updated on: 16 Jul 2019, 09:18
(xy)^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
Answer is (C)
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
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16 Jul 2019, 08:44
(x−y)^2+2y^2=18 (x−y)^2=182y^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
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16 Jul 2019, 08:44
IMO Answer is 2 pairs, A
(x−y)^2+2y^2=18
we know that (xy)^2 >0 and y^2>0 if x and y are integers, possible values of (xy)^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
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16 Jul 2019, 08:45
Let's list out all the set of squares of integers that are less than 18.  1, 4, 9, 16
For the equation \((xy)^2 + 2y^2 = 18\), lets say, \(z= xy\)
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 \((5,1)\)
So, the answer is 6



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Re: For how many ordered pairs (x , y) that are solutions of the equation
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16 Jul 2019, 08:47
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
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16 Jul 2019, 08:50
Since x and y are both integers xy will also be an integer.
Since 2y^2 shall always be positive. (xy)^2 shall be less than 18.
(xy)^2 can be 1,4,9,16 see that it cant be 1 or 9 as evenodd 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=1816=2 y^2=1. y can be +1 or 1
when y=+1, x is 5 as (xy)^2=16 when y=1. x is 3
so there are 2 ordered pairs
Answer A



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Re: For how many ordered pairs (x , y) that are solutions of the equation
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16 Jul 2019, 08:53
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 (xy)ˆ2=182yˆ2… (xy)ˆ2=perfect square; 182yˆ2…(2*3ˆ2)2yˆ2…2(3ˆ2yˆ2)…2(9yˆ2)={1,4,16,25,36…} only possible integer solution for y is when 2(9yˆ2)=16…9yˆ2=8…yˆ2=1…y=1; (xy)ˆ2=16… (x1)ˆ2=16…(x3)(x+1)=0; possible pairs are (3,1) and (1,1) = 2 pairs; Answer (A).



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Re: For how many ordered pairs (x , y) that are solutions of the equation
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16 Jul 2019, 08:56
I will use plugin method since there will be less option that will satisfy the above equation.
For y=1 (xy)^2 = 16 ordered pair can be (5,1) (3,1)
for y = 1 (xy)^2 = 16 ordered pair can be (3,1) (5,1)
y=2 no ordered pair (xy)^2 not equal to 8
for y=3 ordered pair can be (3,3) (3,3)
Total 6 pairs. Hence, C is the answer.



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Re: For how many ordered pairs (x , y) that are solutions of the equation
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16 Jul 2019, 08:58
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).
So (B) is the answer.




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