+4609 Let $x$ and $y$ be integers. Show that $9x + 5y$ is divisible by 19 if and only if $x + 9y$ is divisible by 19.
$9x + 5y$. There are infinite solutions to your problem, both positive and negative:
($9x95) + ($5x171) =$17,100, which is divisible by 19=$900.
+26367 Let x and y be integers.
Show that 9x + 5y is divisible by 19 if and only if x + 9y is divisible by 19.
\(\begin{array}{lcll} \text{Let } x+9y = n\cdot 19 \qquad n \in Z \\ \end{array}\)
\(\begin{array}{|rcll|} \hline 9x + 5y \stackrel{?} \equiv 0 \pmod{19} \qquad & | \qquad x = n\cdot 19-9y \\\\ 9\cdot (n\cdot 19-9y) + 5y \stackrel{?} \equiv 0 \pmod{19} \\ 9n\cdot 19-81y + 5y \stackrel{?} \equiv 0 \pmod{19} \\ 9n\cdot 19 - 76y \stackrel{?} \equiv 0 \pmod{19} \qquad & | \qquad 76 = 4\cdot 19 \\ 9n\cdot 19 - 4\cdot 19 y \stackrel{?} \equiv 0 \pmod{19} \\ {\color{red}{19}}\cdot (9n-4y) \equiv 0 \pmod{19} \ \checkmark \\ \hline \end{array} \)
