Jul 16 03:00 PM PDT  04:00 PM PDT Join a free live webinar and find out which skills will get you to the top, and what you can do to develop them. Save your spot today! Tuesday, July 16th at 3 pm PST Jul 16 08:00 PM EDT  09:00 PM EDT Strategies and techniques for approaching featured GMAT topics. Tuesday, July 16th at 8 pm EDT Jul 15 10:00 PM PDT  11:00 PM PDT Get the complete Official GMAT Exam Pack collection worth $100 with the 3 Month Pack ($299) Jul 15 10:00 PM PDT  11:00 PM PDT Take 20% off the plan of your choice, now through midnight on Monday, July 15th Jul 19 08:00 AM PDT  09:00 AM PDT The Competition Continues  Game of Timers is a teambased competition based on solving GMAT questions to win epic prizes! Starting July 1st, compete to win prep materials while studying for GMAT! Registration is Open! Ends July 26th Jul 21 07:00 AM PDT  09:00 AM PDT Attend this webinar to learn a structured approach to solve 700+ Number Properties question in less than 2 minutes
Author 
Message 
TAGS:

Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 56239

If x and y are integers and 2x + 9y = 57, then the least possible valu
[#permalink]
Show Tags
21 Mar 2017, 05:01
Question Stats:
24% (02:46) correct 76% (02:38) wrong based on 265 sessions
HideShow timer Statistics
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
Official Answer and Stats are available only to registered users. Register/ Login.
_________________




Veritas Prep GMAT Instructor
Affiliations: Veritas Prep
Joined: 21 Dec 2014
Posts: 45
Location: United States (DC)
GPA: 3.11
WE: Education (Education)

Re: If x and y are integers and 2x + 9y = 57, then the least possible valu
[#permalink]
Show Tags
29 Mar 2017, 01:06
Here's the Official Solution: Quote: Let's begin by simply finding at least one pair of values \(x\) and \(y\) that works in the given equation.
Since \(2x\) is always even, but \(57\) is odd, we must have \(9y =\) odd, so \(y\) is odd. The first odd multiple of \(9\) below \(57\) is \(9 * 5 = 45\), which would leave \(2x = 57  45 = 12\) and \(x = 6\). Thus \((6, 5)\) is one pair of values \((x, y)\) that solves the given equation, and \(x^2 + 2xy + y^2 = (x+y)^2\) could be \(121\).
However, now note that the least common multiple of \(2\) and \(9\) is \(2 * 9 = 18\), and we can reduce \(x\) and increase \(y\) (or vice versa) in a \(9:2\) ratio. That is, increasing \(x\) by \(9\) and decreasing \(y\) by \(2\), or vice versa, would not change the total value of \(2x + 9y\).
In turn, such a swap would increase or decrease the total of \(x + y\) by \(7\) each time. So it would also be possible, for instance, to obtain \(x+y = 4\) by switching \((x, y)\) from \((6, 5)\) to \((6  9, 5 + 2) = (3, 7)\), which gives \(x^2 + 2xy + y^2 = 16\).
However, this is still not the minimum for \(x + y\), since one more swap gives \((x, y) = (39, 7+2) = (12, 9)\). Then \(x + y = 3\), and \(x^2 + 2xy + y^2 = 9\). This value actually is the minimum, since continued swaps will only ever increase or decrease the value of \(x + y\) by \(7\) each time and thus cannot move \((x+y)^2\) closer to zero. B is correct.
Here's another, slightly more algebraic approach:
Let's have the unknown sum of \(x + y\) be \(C\). We can write the system
\(2x + 9y = 57\) \(x + y = C\)
Now, multiplying the second equation by \(2\) and subtracting will eliminate \(x\):
\(2x + 9y = 57\) \(2x + 2y = 2C\) \(7y = 57  2C\)
Since the left side is a multiple of \(7\), the right side must also be a multiple of \(7\). The closest multiple of \(7\) to \(57\) is \(56\), but this would require \(C = 1/2\), which is impossible since \(x\) and \(y\) are both integers. The next closest multiple of \(7\) to \(57\) is \(63\), which occurs when \(C = 3\). This, in turn, gives \(9\) as the least possible value of \(x^2 + 2xy + y^2\). Once again, B is correct.




Manager
Status: GMAT...one last time for good!!
Joined: 10 Jul 2012
Posts: 56
Location: India
Concentration: General Management
GPA: 3.5

Re: If x and y are integers and 2x + 9y = 57, then the least possible valu
[#permalink]
Show Tags
28 Mar 2017, 08:20
Bunuel wrote: 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 Question asks for the least possible value, so lets go from AE 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
_________________
Kudos for a correct solution




Senior Manager
Joined: 24 Apr 2016
Posts: 327

Re: If x and y are integers and 2x + 9y = 57, then the least possible valu
[#permalink]
Show Tags
21 Mar 2017, 11:32
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 = (577y)/2
(x+y)^2 = [(577y)/2] ^2
For (x+y)^2 to be the least, the difference to (577y) 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 [(577*7)/2] ^2 = [(5749)/2] ^2 = [(5749)/2] ^2
= [(8)/2] ^2 [4] ^2 = 16
Answer is C. 16



Manager
Joined: 16 Jan 2017
Posts: 60

Re: If x and y are integers and 2x + 9y = 57, then the least possible valu
[#permalink]
Show Tags
23 Mar 2017, 09:00
is there any easier way to solve this?



Veritas Prep GMAT Instructor
Affiliations: Veritas Prep
Joined: 21 Dec 2014
Posts: 45
Location: United States (DC)
GPA: 3.11
WE: Education (Education)

If x and y are integers and 2x + 9y = 57, then the least possible valu
[#permalink]
Show Tags
23 Mar 2017, 09:08
vmelgargalan wrote: is there any easier way to solve this? Yes. There's a better way. 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.



Manager
Joined: 16 Jan 2017
Posts: 60

If x and y are integers and 2x + 9y = 57, then the least possible valu
[#permalink]
Show Tags
23 Mar 2017, 09:48
AnthonyRitz wrote: vmelgargalan wrote: is there any easier way to solve this? Yes. There's a better way. 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. 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



Intern
Joined: 22 Oct 2016
Posts: 21

Re: If x and y are integers and 2x + 9y = 57, then the least possible valu
[#permalink]
Show Tags
28 Mar 2017, 07:31
Since it says x and y are integers[doesnt talk about their signs], 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 =144216+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.
_________________



Intern
Joined: 28 Mar 2017
Posts: 1

If x and y are integers and 2x + 9y = 57, then the least possible valu
[#permalink]
Show Tags
30 Mar 2017, 02:14
\[\begin{align} & \text{2x+9y=57} \\ & \text{9y=57(mod2)} \\ & \text{y=1(mod2)}\to \text{y=2t+1} \\ & \text{substitute in the first eqn } \\ & \text{x=9t+24(x+y}{{\text{)}}^{\text{2}}}\text{=(7t+25}{{\text{)}}^{\text{2}}}\text{f(t)=(7t+25}{{\text{)}}^{\text{2}}} \\ & \Rightarrow \text{{f}'}(\text{t})\text{=14}(\text{7t+25}) \\ & \text{{f}'}(\text{t})\text{=0}\Rightarrow \text{t=25/7=3}\text{.5} \\ & \text{but t is integer so we take t=3 and t=4 } \\ & \text{then we find f(3) ; f(4),the least of them will be the least for all integers } \\ & \text{f(3)=16 } \\ & \text{f(4)=9} \\ & \text{And so 9 is the least value} \\ \end{align}\]



Intern
Joined: 17 May 2016
Posts: 18

Re: If x and y are integers and 2x + 9y = 57, then the least possible valu
[#permalink]
Show Tags
04 Apr 2017, 23:42
I didn't understand from the second step onwards Can you please simplify it and tell me? Prayersmith2017 wrote: \[\begin{align} & \text{2x+9y=57} \\ & \text{9y=57(mod2)} \\ & \text{y=1(mod2)}\to \text{y=2t+1} \\ & \text{substitute in the first eqn } \\ & \text{x=9t+24(x+y}{{\text{)}}^{\text{2}}}\text{=(7t+25}{{\text{)}}^{\text{2}}}\text{f(t)=(7t+25}{{\text{)}}^{\text{2}}} \\ & \Rightarrow \text{{f}'}(\text{t})\text{=14}(\text{7t+25}) \\ & \text{{f}'}(\text{t})\text{=0}\Rightarrow \text{t=25/7=3}\text{.5} \\ & \text{but t is integer so we take t=3 and t=4 } \\ & \text{then we find f(3) ; f(4),the least of them will be the least for all integers } \\ & \text{f(3)=16 } \\ & \text{f(4)=9} \\ & \text{And so 9 is the least value} \\ \end{align}\] Posted from my mobile device Posted from my mobile device



Intern
Joined: 06 Apr 2017
Posts: 29
Location: United States (OR)
Concentration: Finance, Leadership
GMAT 1: 730 Q48 V44 GMAT 2: 730 Q49 V40
GPA: 3.98
WE: Corporate Finance (Health Care)

If x and y are integers and 2x + 9y = 57, then the least possible valu
[#permalink]
Show Tags
10 May 2017, 14:47
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 \(x+y\)  \(x=5+2=7,y=69=3,x+y=4\)
 \(x=7+2=9,y=39=12,x+y=3\)
Testing solutions in either direction only produces \(x+y>3\), so the smallest possible value of \((x + y)^2=3^2=9\) Answer B



Intern
Joined: 26 Feb 2014
Posts: 7
GPA: 4

If x and y are integers and 2x + 9y = 57, then the least possible valu
[#permalink]
Show Tags
10 May 2017, 19:06
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 = 576/7 = 49/7 = 7. Clearly (x+y)^2 = 9 is the right answer
Hope this helps some people.



Retired Moderator
Status: Long way to go!
Joined: 10 Oct 2016
Posts: 1344
Location: Viet Nam

Re: If x and y are integers and 2x + 9y = 57, then the least possible valu
[#permalink]
Show Tags
10 May 2017, 19:30
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 \(x_1=3t+2 \implies (3t+2)+3y_1=8 \implies 3t+3y_1=6 \implies t+y_1 =2 \implies y_1 = 2t\) Hence we have \(x=3x_1=3(3t+2)=9t+6\) and \(y=2y_1+1=2(2t)+1=42t+1=52t\) The roots of this equation is \((x,y)=(9t+6,52t)\) with \(t\) is an integer. Now we have \((x+y)^2=(9t+6+52t)^2=(7t+11)^2\) 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.
_________________



Retired Moderator
Status: Long way to go!
Joined: 10 Oct 2016
Posts: 1344
Location: Viet Nam

Re: If x and y are integers and 2x + 9y = 57, then the least possible valu
[#permalink]
Show Tags
10 May 2017, 20:01
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 \(\Big\{ \begin {array}{lr} 2x + 9y = 57 \\ x+y=1 \end{array} \implies \Big\{ \begin {array}{lr} 2x + 9(1x) = 57 \\ x+y=1 \end{array} \implies \Big\{ \begin {array}{lr} 7x = 48 \\ x+y=1 \end{array} \implies \Big\{ \begin {array}{lr} 7x = 48 \\ x+y=1 \end{array} \implies \Big\{ \begin {array}{lr} x = \frac{48}{7} \\ x+y=1 \end{array}\) Eliminate this case because \(x\) is not integer. With \(x+y=1\) we have \(\Bigg\{ \begin {array}{lr} 2x + 9y = 57 \\ x+y=1 \end{array} \implies \Big\{ \begin {array}{lr} 2x + 9(1x) = 57 \\ x+y=1 \end{array} \implies \Big\{ \begin {array}{lr} 7x = 66 \\ x+y=1 \end{array} \implies \Big\{ \begin {array}{lr} 7x = 66 \\ x+y=1 \end{array} \implies \Big\{ \begin {array}{lr} x = \frac{66}{7} \\ x+y=1 \end{array}\) 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 \(\Big\{ \begin {array}{lr} 2x + 9y = 57 \\ x+y=3 \end{array} \implies \Big\{ \begin {array}{lr} 2x + 9(3x) = 57 \\ x+y=3 \end{array} \implies \Big\{ \begin {array}{lr} 7x = 30 \\ x+y=3 \end{array} \implies \Big\{ \begin {array}{lr} 7x = 30 \\ x+y=3 \end{array} \implies \Big\{ \begin {array}{lr} x = \frac{30}{7} \\ x+y=3 \end{array}\) Eliminate this case because \(x\) is not integer. With \(x+y=3\) we have \(\Bigg\{ \begin {array}{lr} 2x + 9y = 57 \\ x+y=3 \end{array} \implies \Big\{ \begin {array}{lr} 2x + 9(3x) = 57 \\ x+y=3 \end{array} \implies \Big\{ \begin {array}{lr} 7x = 84 \\ x+y=3 \end{array} \implies \Big\{ \begin {array}{lr} x = 12 \\ y=9 \end{array}\) This one satisfies the given conditions. Hence this choice is the correct answer.
_________________



Intern
Joined: 17 Jul 2017
Posts: 4

Re: If x and y are integers and 2x + 9y = 57, then the least possible valu
[#permalink]
Show Tags
03 Jan 2018, 08:28
spence11 wrote: 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 \(x+y\)  \(x=5+2=7,y=69=3,x+y=4\)
 \(x=7+2=9,y=39=12,x+y=3\)
Testing solutions in either direction only produces \(x+y>3\), so the smallest possible value of \((x + y)^2=3^2=9\) Answer B spence11I 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 ?



NonHuman User
Joined: 09 Sep 2013
Posts: 11649

Re: If x and y are integers and 2x + 9y = 57, then the least possible valu
[#permalink]
Show Tags
23 Feb 2019, 10:27
Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up  doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________




Re: If x and y are integers and 2x + 9y = 57, then the least possible valu
[#permalink]
23 Feb 2019, 10:27






