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

Question

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.

Answer 1

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

Answer 2

You're looking at $$ 77+12=15 $$ 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.

Answer 3

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