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Let $x,y$ be positive integers, $x<y$, and $x+y=667$. Given that $\dfrac{\text{lcm}(x,y)}{\text{gcd}(x,y)}=120,$ find all such pairs $(x,y)$.

The only way I can think of solving this is trying all possibilities where one number is odd and the other even, and testing them all. Using this, I found one solution, $(115,552)$, but I'm wondering if there is a more efficient way to do this problem.

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  • $\begingroup$ Python script gives $(115,552)$ and $(232,435)$. $\endgroup$ Commented Dec 24, 2016 at 19:16
  • $\begingroup$ @barakmanos I want to solve without scripts or calculators $\endgroup$
    – suomynonA
    Commented Dec 24, 2016 at 19:16
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    $\begingroup$ Of course, that's why I wrote this as a comment and not as an answer - in order to provide some more insight on the question (BTW, you've failed to specify the other pair for some reason). $\endgroup$ Commented Dec 24, 2016 at 19:17
  • $\begingroup$ BTW, since the prime factorization of $667$ is $23\cdot29$, there aren't many possibilities to explore (i.e., the $\gcd$ is $\in[1,23,29]$). So you only have $23+29$ pairs to check. $\endgroup$ Commented Dec 24, 2016 at 19:20

2 Answers 2

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If a number divides $x$ and $y$ then the number divides $x+y$. Therefore $\gcd(x,y)$ is a divisor of $667$.

Since $667=23\times 29$ there are only $4$ cases.

If $\gcd(x,y)=1$ we get $\text{lcm}(x,y)=120$, clearly impossible since one of the numbers is larger than $120$.

If $\gcd(x,y)=23$ then $\text{lcm}(x,y)=23\times 120$, dividing everything by $23$ we must find coprime $x'$ and $y'$ that add to $29$ and have product equal to $120$. Solving the quadratic equation leads to $(5,24)$ and $(24,5)$

If $\gcd(x,y)=29$ then we must find coprime $x'$ and $y'$ that add to $23$ and have product $120$. Solving the quadratic leads to $(8,15)$ and $(15,8)$

If $\gcd(x,y)=23\times 29$ then we must find $x',y'$ that add to $1$ and have product $120$, clearly impossible.


So the only solutions are $(5\times 23,24\times 23),(24\times 23, 5\times 23),(8\times 29, 15\times 29),(15\times 29,8\times 29)$

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Let $\dfrac xX=\dfrac yY=(x,y)=d$(say) $\implies(X,Y)=1$

$120=\dfrac{XYd}d=XY$

As $120=1\cdot120,2\cdot60,3\cdot40,4\cdot30,5\cdot24,6\cdot20,8\cdot15,10\cdot12$

The possible set of values of $(X,Y)$ such that $X<Y;(X,Y)=1$ are $\{(1,120);(3,40);(5,24);(8,15)\}$

Now use the fact that $\displaystyle667=x+y=d(X+Y)\implies\dfrac{23\cdot29}{X+Y}=d$ which is a positive integer to find the possible set of values of $(X,Y)$ to be $\{(5,24);(8,15)\}$

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