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If x and y are integers and 2x + 9y = 57, then the least possible valu
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21 Mar 2017, 05:01
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25% (02:43) correct 75% (02:39) wrong based on 328 sessions
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Re: If x and y are integers and 2x + 9y = 57, then the least possible valu
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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.




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



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



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



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



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



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



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



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



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




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