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Find all positive integer solutions to $24x+18y=6420$.

Here's my work.

Simplifying the equation gives $4x+3y=1070$. Note that this equation has solutions because $\gcd (4,3)=1\mid 1070$.

We will use the Euclidean Algorithm to solve $4x+3y=1$.

We have that $$4=3(1)+1\\ 3=1(3)$$ Hence $$1=4-3(1)=4(1)-3(1).$$ Hence one solution $(x_0,y_0)$ to the equation is $(1070,-1070)$. We know that all solutions to the equation $4x+3y=1070$ are of the form $(x_0 + \dfrac{b}{d}k, y_0-\dfrac{a}{d}k),$ where $b=3$, $a=4$, $d=\gcd (4,3)=1$, and $k\in\mathbb{Z}$. Hence, to find all positive integer solutions, we need to solve $1070+4k> 0\;(1)$ and $-1070-3k > 0\;(2)$. Simplifying $(1)$ gives $k >-\dfrac{1070}{4}=-267.5$ and simplifying $(2)$ gives $k<-\dfrac{1070}{3}=-356\dfrac{2}{3}$. Hence, since there is no intersection between the set of solutions to $(1)$ and $(2)$, the equation has no positive solutions.

Edit: The problem was updated.

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  • $\begingroup$ Even if you only require $x,y$ to be non-negative, it is clear that there are no solutions. $x$ clearly has to be $0$, $y=1$ doesn't work, and $y=2$ is too big. $\endgroup$
    – lulu
    Nov 25, 2019 at 0:59
  • $\begingroup$ There is most likely something wrong with the question. I got it from a university textbook. It should not have such an elementary solution. $\endgroup$
    – user726063
    Nov 25, 2019 at 1:00
  • $\begingroup$ The problem was updated. $\endgroup$
    – user726063
    Nov 25, 2019 at 17:31

3 Answers 3

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Your solution seems correct. However, it'd be much faster to simply notice that if $x,y\geq1$, $$154x+24y\geq178>30.$$

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  • $\begingroup$ I think there might be something wrong with the wording for this question. $\endgroup$
    – user726063
    Nov 25, 2019 at 0:56
  • $\begingroup$ @user23749 Possibly. As it stands, though, there exist integral solutions (as you’ve shown), but not positive integral ones. $\endgroup$
    – ViHdzP
    Nov 25, 2019 at 0:58
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    $\begingroup$ I think whoever asked this question wanted to find integral solutions w/in a certain range or sth. I think it's a little fishy that this question has an answer a grade $1$ student could literally come up with. The real answer should use a similar approach to mine. $\endgroup$
    – user726063
    Nov 25, 2019 at 0:59
  • $\begingroup$ Listen! The question was updated! I used the approach above because I knew I would have to. Sorry about the confusion! $\endgroup$
    – user726063
    Nov 25, 2019 at 17:30
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You're looking at $$ 77𝑥+12𝑦=15 $$ right? For positive $x$ and $y$. Let $u = x-1, v = y-1$, then (1) $u$ and $v$ are nonnegative, and (2)

$$ 77x + 12 y = 77 + 12 + (77u + 12 v) = 89 + (77u + 12v) $$ which is at least $89$, because each of $u$ and $v$ is nonnegative.

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  • $\begingroup$ Sorry. I think there's something wrong with the question. It should not be so simple that a grade $1$ student could solve it. $\endgroup$
    – user726063
    Nov 25, 2019 at 0:59
  • $\begingroup$ $x = 3, y = -18$. $\endgroup$
    – David G. Stork
    Nov 25, 2019 at 1:06
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    $\begingroup$ Much as I usually find you're right on point, @DavidG.Stork, the number $-18$ is definitely not positive as required in the question. :) $\endgroup$
    – John Hughes
    Nov 25, 2019 at 2:40
  • $\begingroup$ Ah yes.... I don't think there's a solution for $\{ x, y \} \in \mathbb{Z}^+$.... at least none I can find. But $x \to 15/154, y \to 5/8$ for $\{ x,y \} \in \mathbb{R}^+$. $\endgroup$
    – David G. Stork
    Nov 25, 2019 at 4:08
  • $\begingroup$ I strongly suspect that the "+" in this problem was supposed to be a "-", which then makes it more interesting. $\endgroup$
    – John Hughes
    Nov 25, 2019 at 12:22
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$x = 266; y = 2$ gives a solution, so evidently your current solution (to the most weirdly edited problem ever!) is wrong.

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