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Question: For some positive integers $x, y$ let $g=\gcd(x,y)$ and $\ell=lcm(2x,y)$. Given that the equation $xy+3g + 7\ell = 168$ holds, the largest value of $2x+y$ is...

Is my answer below acceptable as a correct answer?

My Answer: Since $\ell=lcm(2x,y)$, $lcm(x,y)=\ell$ or $lcm(x,y)=\frac{1}{2}\ell$

  1. Case 1: $lcm(x,y)=\ell$

\begin{align*} xy+3g + 7\ell &= 168\\ g\ell+3g+7\ell &=168\\ (g+7)(\ell+3)&=189 \end{align*} The last equation gives $g=2$ and $\ell=18$ where no integer $x$ and $y$ satisfy the conditions

  1. Case 2: $lcm(x,y)=\frac{1}{2}\ell$

\begin{align*} xy+3g + 7\ell &= 168\\ \frac{1}{2}g\ell+3g+7\ell &=168\\ gl+6g+14\ell&=336\\ (g+14)(\ell+6)&=420 \end{align*} The last equation gives $g=1$ and $\ell=22$. Then $gcd(x,y)=1$ and $lcm(x,y)=11$. It means that $x=1,y=11$ or $x=11, y=1$.

The largest possible value of $2x+y=23$

How do I know that $x=1,y=11$ or $x=11, y=1$ are the olny solutin?How can we make sure of that?

masrosid
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  • For a solution-verification question to be on topic you must specify precisely which step in the proof you question, and why so. This site is not meant to be used as a proof checking machine. – Bill Dubuque Oct 31 '23 at 20:35
  • Could there be more solutions to Case 2? You've found one solution, how do you know you've gotten them all? EG I believe that $ g = 7, l = 14$ works, leading to $ x = 7, y = 7$, though $2x+y$ is smaller. – Calvin Lin Oct 31 '23 at 20:35
  • @BillDubuque I am sorry if this is wrong way to answer question. I just don't know what to do – masrosid Oct 31 '23 at 20:53
  • @CalvinLin I don't know if there are any other solution. Thats why I asked here. – masrosid Oct 31 '23 at 20:54
  • The factorization method you use is a generalization of completing the square which is known as completing a product or rectangle, which has been discussed here many times. We can't give feedback unless you explain in much more detail how you are finding your solutions after completing the product. Please do so (in the question, not in comments). – Bill Dubuque Oct 31 '23 at 21:15

1 Answers1

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We have $xy+3g+7l=168\Rightarrow\dfrac{xy+3g}{7}+l=24\Rightarrow l=23,22,21$ because $xy+3g=7,14,21$. It follows $l=22$ the only even.

From $lcm(2x,y)=22$ and the largest value of $x$ is $11$ which gives the required largest value is $2\cdot11+1=23$ (the other fitting values are easily discarded because $xy+3g=14$).

Piquito
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