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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, 13:02
(xy)^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 \((xy)^2\)=0 and \(y^2\)=9 \((xy)^2=0\) or xy=0.....(1) \(y^2=9\) y= 3 or 3 We have 2 distinct integral values of y and 1 integral value of xy. Hence number of solutions possible= 2*1=2 Case 2  When \((xy)^2=16\) and \(y^2=1\) \((xy)^2=16\) (xy)=4 or 4 and \(y^2=1\) y=1 or 1 We have 2 distinct integral values of xy and 2 distinct integral values of y Number of solutions possible= 2*2=4Total 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 (xy) 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.



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For how many ordered pairs (x , y) that are solutions of the equation
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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 (\((xy)^2\)) and another is a multiple of 2 (2* \(y^2\)) Only possible pair is 16 + 2 = 18 Now either \((xy)^2 =16\) and \(2y^2 = 2\) or \((xy)^2\)= 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, \((xy)^2 =16\) and \(2y^2 = 2\) \(2y^2 = 2\) therefore y = +1,1 and for \((xy)^2 =16\) (xy) = +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



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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, 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 (XY)^2 will always be integer and that too of an even number,
since (xy)^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
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, 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)



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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, 11:42
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.
Answer: C



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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, 11:46
(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?
(xy)^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 (x1)^2 = 16 > x1 = (+/)4 > x = 5 or 3. Hence two pairs (5,1) and (3,1) If y = 2, then solving we get (x2)^2 = 10...not a perfect square. If y = 3, then solving we get (x3)^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
Answer Choice: C



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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, 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.
\((xy)^2\)+2\(y^2\)=18 Substittuing , x=5 and y=1
\((51)^2\)+2*\(1^2\)=\(4^2\)+2=18
x=3 and y=1 \((31)^2\)+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



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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, 12:15
\((x−y)^2+2y^2=18\) For how many ordered pairs (x , y) that are solutions of the equation above are x and y both integers? 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 \((xy)^2\) = 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 SQUAREAlso, 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 square1+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 square16+2 = 18..... This is possible because 2/2 is perfect squarePossible combinations of A and B are: A = 0 and B = 18.... I A = 16 and B = 2..... II IA=0 \((xy)^2\) = 0 xy = 0 x=yB=18 \(2y^2=18\) y=+ or  3If y = 3, x = 3....(1) If y = 3, x=3......(2)IIA=16 \((xy)^2=16\) xy=+ or 4 x=y+4 or x = y4B=2 \(2y^2=2\) y=+ or  1If 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.



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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, 12:48
\((x−y)^2\)+2\(y^2\) = 18 (1)
\((x−y)^2\) = 182\(y^2\)
since \((x−y)^2\) ≥ 0 so 182\(y^2\)≥ 0 9\(y^2\)≥ 0 (y+3)(y3)≤ 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 \((x1)^2\) = 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 so 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, 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 (xy)^2 +2y^2 = 18 (xy)^2 = 2*9  2y^2 (xy)^2 = 2*(9y^2)
From the given equation we see that (9y^2) after multiplying by 2 must equals something squared (= (xy)^2). We need y and x to be integers, so 2*(9y^2) may equals 0, 1, 4, 9, 16, 25, 36 etc. From 2*(9y^2) = 0 and (xy)^2 = 2*(9y^2) we have y = 3, x = 3 OR y = 3, x = 3 From 2*(9y^2) = 1 we can't have integer y From 2*(9y^2) = 2 we can't have integer y From 2*(9y^2) = 4 we can't have integer y From 2*(9y^2) = 9 we can't have integer y From 2*(9y^2) = 16 and (xy)^2 = 2*(9y^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*(9y^2) = 25), because 2*(9y^2) can't be more than 18.
So the answer is 4 pairs  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, 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 xy = 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.
Answer: D
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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, 13:39
(xy)^2+2y^2=18 We are to find integer values of x and y that satisfy the above equation. (xy)>4 will produce LHS>RHS So our combination integer values of x and y which can satisfy this equation must be such that (xy)<=4.
When y=1, (x+1)^2+2=18 x^2+2x15=0 (x3)(x+5)=0 Hence x=5 or x=3 So we have (x,y) pairs of (5,1) and (3,1)
When y=1, (x1)^2+2=18 x^22x15=0 (x+5)(x3)=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.
The 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, 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
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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, 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



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For how many ordered pairs (x , y) that are solutions of the equation
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16 Jul 2019, 17:34
solving for the given eqn ; we get x^2+3y^22xy=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
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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, 18:51
\((x−y)^2 + 2y^2 = 18\) 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 noninteger 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 \((x3)^2 =0\) 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 \((x1)^2 =16\) and \(x=5\) or \(3\) > \((5,1)\) and \((3,1)\) are both solution.
Answer is (C)
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.



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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, 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



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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, 20:02
(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 or3 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.



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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, 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 , (51)^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



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For how many ordered pairs (x , y) that are solutions of the equation
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16 Jul 2019, 21:15
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.
Answer C




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