If x and y are integers and 2x + 9y = 57, then the least possible valu

Question

If x and y are integers and 2x + 9y = 57, then the least possible value of x^2 + 2xy + y^2 is:

A. 1
B. 9
C. 16
D. 121
E. 196

AnthonyRitz · Accepted Answer

Here's the Official Solution:

origen87 · Answer

Question asks for the least possible value, so lets go from A-E
1.(x+y)=+/- 1
Solve this with the equation 2x+9y=57, we can quickly find that the solutions wont be integers
2. (x+y)=+/- 3
With (x+y=3, solutions wont be integers
now substitute (x+y=-3) in 2x+9y=57; y=9,x=-12

Ans:B

quantumliner · Answer

Given: 2x + 9y = 57, x^2 + 2xy + y^2

x^2 + 2xy + y^2 can be written as (x+y)^2

2x + 9y = 57 can be written as 2(x+y) + 7y = 57

(x+y) + 7y/2 = 57/2

(x+y) = 57/2 - 7y/2 = (57-7y)/2

(x+y)^2 =   ^2

For (x+y)^2 to be the least, the difference to (57-7y) has to be least.

Case 1 : Closest multiple of 7 to 57 is 56 that means y = 8.

Since the condition given is both x and y are integers, replacing y = 8 in the equation 2x + 9y = 57, we get 2x = 57 - 72 = -15. So x is not an integer. Hence Y=8 cannot be a valid value.

Case 2: Next Closest multiple of 7 to 57 is 49 that means y = 7

Replacing y = 7 in the equation 2x + 9y = 57, we get 2x = 57 - 63 = -14. So x is an integer (-7). Hence Y=7 is a valid value.

Putting y=7 in the equation   ^2 =   ^2 =   ^2

=   ^2   ^2 = 16

Answer is C. 16

vmelgargalan · Answer

is there any easier way to solve this?

AnthonyRitz · Answer

Yes. There's a better way. :wink:

Stay tuned!

To clarify, not trying to be a dick, but Bunuel posted with the answer secret so people could take an unbiased crack at it, and I don't want to spoil that. If you want to be spoiled, I'm willing to give out the official solution by PM. Or else it will definitely show up here after the correct answer is revealed.

vmelgargalan · Answer

Hehe, I might take you up on that offer. I am basically assuming that x^2 + 2xy + y^2 is at its smallest when y is as large as it can be and x as small as it can be. So by trial and error we get a value of y=5 (9x5=45) and x=6 (6x2=12), if we plug this into the equation it is (6)^2 + 2 (5)(6) + (5)^2 which gives 121.

Trying to see from a different angle mhmmm.... because even if we have negative numbers we do not get any of the available options

sarugiri · Answer

Since it says x and y are integers , y has to be maximum and x has to be minimum in 2x+9y=57 Taking y=9 and x=-12 and applying the values to x^2+2xy+y^2, =(-12)^2+2(-12)(9)+9^2 =144-216+81 =9 Hi Bunuel, I worked on the above steps after seeing the OA. But, actually I ended choosing C. How do we confirm in these type of questions that the answers we choose are the minimum values.?Since the wrong answers we arrive is also there in the options.

Prayersmith2017 · Answer

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enantiodromia · Answer

I didn't understand from the second step onwards 
Can you please simplify it and tell me?

Posted from my mobile device

spence11 · Answer

Three things to note:

1)  2x + 9y = 57 , where  x  and  y  are integers is a linear Diophantine equation. Linear Diophantine equations have special properties
2)  x^2 + 2xy + y^2 = (x + y)^2 
3) The minimum value of a perfect square is  0 , and in order to approach this limit the absolute value of the expression under the square should be minimized

We are looking for the smallest value of  |x + y|

With linear Diophantine equations, once one combination of  x  and  y  are found, other values can be found by concurrently adding the  x  coefficient to the  y  solution and subtracting the  y  coefficient from the  x  solution OR concurrently adding the  y  coefficient to the  x  solution and subtracting the  x  coefficient from the  y  solution. With more difficult linear Diophantine, one can use the extended Euclidean algorithm to find an initial solution, but with simple equations like this, just look and try a few numbers. Here's one:

x=5,y=6,9(5)+2(6)=57

Now move through other possible values while trying to minimize  |x+y|

x=5+2=7,y=6-9=-3,|x+y|=4  
  x=7+2=9,y=-3-9=-12,|x+y|=3

Testing solutions in either direction only produces  |x+y|>3 , so the smallest possible value of  (x + y)^2=3^2=9

Answer  B

InfiniteLoop80 · Answer

I worked this backwards.

We know 2x + 9y = 57, and we need to get the minimum value of (x+y)^2 that satisfies this.
ie we need to get minimum value of (x+y) that works to satisfy the equation
Rewriting the equation, 7y + 2(x+y) = 57 
 y = (57 - 2(x+y)) / 7

Plugging in the from the lowest value of x+y that would give me an integer for y, 
 x+y = +1 
y = (57 - 2) / 7 ===> Not divisible and hence not integer
  x+y =-1 
y = 59/7 ==> Not divisible and hence not an integer
 
x+y = +3 
y = 57-6/7 = 49/7 = 7. 
Clearly (x+y)^2 = 9 is the right answer

Hope this helps some people.

broall · Answer

Solution.

We will find the roots for equation $2x+9y=57$ with $x,y$ are integer.

First, $9y$ and $57$ are divisible by $3$, so $2x$ is divisible by $3$, or $x$ is divisible by $3$.

Set $x=3x_1 \implies 6x_1+9y=57 \implies 2x_1+3y=19$

Note that $2x_1$ is even, $19$ is odd so $3y$ is odd or $y$ is odd.

Set $y=2y_1+1 \implies 2x_1+3(2y_1+1)=19 \implies 2x_1+6y_1=16 \implies x_1+3y_1=8$

Note that $\frac{8}{3}=2$ and $\frac{3y_1}{3}=0$ so $\frac{x_1}{3}=2$.

Set $x_1=3t+2 \implies (3t+2)+3y_1=8 \implies 3t+3y_1=6 \implies t+y_1 =2 \implies y_1 = 2-t$

Hence we have $x=3x_1=3(3t+2)=9t+6$ and $y=2y_1+1=2(2-t)+1=4-2t+1=5-2t$

The roots of this equation is $(x,y)=(9t+6,5-2t)$ with $t$ is an integer.

Now we have $(x+y)^2=(9t+6+5-2t)^2=(7t+11)^2$

Note that $|7t+11| \geq 0$ but we can't get the value $0$ since $t$ is integer.

Note that $-11/7=-1.57$. 
If $t=-1 \implies |7t+11|=5$
If $t=-2 \implies |7t+11|=3$

We choose $t=-2 \implies (7t+11)^2=9$, this is the answer.

broall · Answer

Solution 2. We will try all answer choice:

A. $1$
This choice gives us $(x+y)^2=1 \implies x+y=1$ or $x+y=-1$

With $x+y=1$ we have
$\Big\{ \begin {array}{lr}
2x + 9y = 57 \
x+y=1
\end{array} \implies \Big\{ \begin {array}{lr}
2x + 9(1-x) = 57 \
x+y=1
\end{array} \implies \Big\{ \begin {array}{lr}
-7x = 48 \
x+y=1
\end{array} \implies \Big\{ \begin {array}{lr}
-7x = 48 \
x+y=1
\end{array} \implies \Big\{ \begin {array}{lr}
x = \frac{-48}{7} \
x+y=1
\end{array}$

Eliminate this case because $x$ is not integer.

With $x+y=-1$ we have
$\Bigg\{ \begin {array}{lr}
2x + 9y = 57 \
x+y=-1
\end{array} \implies \Big\{ \begin {array}{lr}
2x + 9(-1-x) = 57 \
x+y=1
\end{array} \implies \Big\{ \begin {array}{lr}
-7x = 66 \
x+y=1
\end{array} \implies \Big\{ \begin {array}{lr}
-7x = 66 \
x+y=1
\end{array} \implies \Big\{ \begin {array}{lr}
x = \frac{-66}{7} \
x+y=1
\end{array}$

Eliminate this case because $x$ is not integer.

B. $9$
This choice gives us $(x+y)^2=9 \implies x+y=3$ or $x+y=-3$

With $x+y=3$ we have
$\Big\{ \begin {array}{lr}
2x + 9y = 57 \
x+y=3
\end{array} \implies \Big\{ \begin {array}{lr}
2x + 9(3-x) = 57 \
x+y=3
\end{array} \implies \Big\{ \begin {array}{lr}
-7x = 30 \
x+y=3
\end{array} \implies \Big\{ \begin {array}{lr}
7x = -30 \
x+y=3
\end{array} \implies \Big\{ \begin {array}{lr}
x = \frac{-30}{7} \
x+y=3
\end{array}$

Eliminate this case because $x$ is not integer.

With $x+y=-3$ we have
$\Bigg\{ \begin {array}{lr}
2x + 9y = 57 \
x+y=-3
\end{array} \implies \Big\{ \begin {array}{lr}
2x + 9(-3-x) = 57 \
x+y=-3
\end{array} \implies \Big\{ \begin {array}{lr}
-7x = 84 \
x+y=-3
\end{array} \implies \Big\{ \begin {array}{lr}
x = -12 \
y=9
\end{array}$

This one satisfies the given conditions. Hence this choice is the correct answer.

gbatistoti · Answer

spence11

I think I understood something wrong in the red part. You say we can add the x coefficient to the y solution and subtract the y coefficient from the x solution but when you solve it you add the x coefficient to the x solution and subtract the y coefficient from the y solution, right ?

Neville44 · Answer

(x+y)^2 will have the least value when at least one of them is negative and the difference between x and y is minimum.
So check values of 9 such that 9y is greater than 57 (also check only for odd multiples for obvious reason).
So start with 9*7=63 so x= -4. So (x+y= 7-4=3). Try for another value i.e 9*9 again x+y will be 3. If you go for 9*11 then the difference between x and y keeps increasing so (x+y)^2 would also increase. Hence 3^2 = 9 is the answer

hkkat · Answer

Do you have more similar questions? 
TIA

TarunKumar1234 · Answer

Given, 2x + 9y = 57 and x and y are integers, they might be positive or negative. we are not sure yet. We have to get min (x+y)^2 value.

x= 24 and y= 1 (Trying with min y)
x= (24-9)= 15 and y= 1+2= 3 (Exchange A and B from equation Ax +By =C)
x= (15-9)= 6 and y= 3+2= 5
x= (6-9)= -3 and y= 5+2= 7
x= (-3-9)=-12 and y= 7+2= 9
x= (-12-9)=-21 and y= 9+2= 11

Min (x+y)^2 = (-12+9)^2= (-3)^2= 9, So, I think B

sm1510 · Answer

Given:  2x + 9y = 57 ,  x^2 + 2xy + y^2

x^2 + 2xy + y^2  can be written as  (x+y)^2

2x + 9y = 57  can be written as 2(x+y) + 7y = 57

(x+y) + \frac{7y}{2} = \frac{57}{2}

(x+y) = \frac{57}{2} - \frac{7y}{2} = \frac{(57-7y)}{2}

(x+y)^2  =    ^2

For  (x+y)^2  to be the least, the difference to (57-7y) has to be least.

Case 1 : Closest multiple of 7 to 57 is 56 that means y = 8.

Since the condition given is both x and y are integers, replacing  y = 8  in the equation  2x + 9y = 57 , we get  2x = 57 - 72 = -15 . So x is not an integer. Hence  Y=8  cannot be a valid value.

Case 2: Next Closest multiple of 7 to 57 is 49 that means  y = 7

Replacing  y = 7  in the equation  2x + 9y = 57 , we get  2x = 57 - 63 = -6 . So x is an integer (-3). Hence  Y=7  is a valid value.

Putting  y=7  in the equation    ^2 =   ^2

=    ^2   =    ^2 = 16

Case 3: Question asks for the least value of  (x+y)^2   hence try the other closest multiple of 7 too :  y = 9

Replacing  y = 9  in the equation  2x + 9y = 57 , we get  2x = 57 - 81 = -24 . So x is also an integer (-12). Hence  Y=9  is also a valid value.

Putting  y=9  in the equation    ^2 =   ^2

=    ^2  =     ^2 = 9

16 (Case 2) > 9 (case 3)

Answer is B. 9

jantwarog · Answer

IMO the simplest solution to this question is to plug in the values for X and Y.

1 step: Work out the fact that x^2 + 2xy + y^2 is actually (x+y)^2

2 step: If the question says that both values are integers it means that y has to be odd, try it:

- If y=1 then x=24
- If y=3 then x=15
- If y=5 then x=6, this is the moment when you can start counting => (5+6)^2 = 121, hold it and keep plugging.
- If y=7 then x=-3 => (-3+7)^2=16
- If y=9 then x=-12 => (-12+9)^2=9
- If y=11 then x=-21 => (-21+11)^2=100 since now this value will start to grow, hence the lowest possible result is 9.

Answer B

If x and y are integers and 2x + 9y = 57, then the least possible valu

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If x and y are integers and 2x + 9y = 57, then the least possible valu

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