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

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.

• 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.
– lulu
Commented Nov 25, 2019 at 0:59
• There is most likely something wrong with the question. I got it from a university textbook. It should not have such an elementary solution.
– user726063
Commented Nov 25, 2019 at 1:00
• The problem was updated.
– user726063
Commented Nov 25, 2019 at 17:31

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

• I think there might be something wrong with the wording for this question.
– user726063
Commented Nov 25, 2019 at 0:56
• @user23749 Possibly. As it stands, though, there exist integral solutions (as you’ve shown), but not positive integral ones. Commented Nov 25, 2019 at 0:58
• 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.
– user726063
Commented Nov 25, 2019 at 0:59
• Listen! The question was updated! I used the approach above because I knew I would have to. Sorry about the confusion!
– user726063
Commented Nov 25, 2019 at 17:30

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.

• Sorry. I think there's something wrong with the question. It should not be so simple that a grade $1$ student could solve it.
– user726063
Commented Nov 25, 2019 at 0:59
• $x = 3, y = -18$. Commented Nov 25, 2019 at 1:06
• Much as I usually find you're right on point, @DavidG.Stork, the number $-18$ is definitely not positive as required in the question. :) Commented Nov 25, 2019 at 2:40
• 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 $\{ x,y \} \in \mathbb{R}^+$. Commented Nov 25, 2019 at 4:08
• I strongly suspect that the "+" in this problem was supposed to be a "-", which then makes it more interesting. Commented Nov 25, 2019 at 12:22

$$x = 266; y = 2$$ gives a solution, so evidently your current solution (to the most weirdly edited problem ever!) is wrong.