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Let $x$ and $y$ be integers. Show that $9x + 5y$ is divisible by 19 if and only if $x + 9y$ is divisible by 19.

 

This is a repost, though the last post's solution was removed for some reason.

 

Thanks!

 Aug 3, 2020
edited by new3r  Aug 3, 2020
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Help Please!

\(\text{Let $x$ and $y$ be integers. Show that $9x + 5y$ is divisible by 19 $\\$if and only if $x + 9y$ is divisible by 19.}\)

 

My attempt:

 

\(\begin{array}{|rcll|} \hline x + 9y &\equiv& 0 \pmod{19} \\ x + 9y &=& 19n,\ \qquad n\in \mathbb{Z} \\ \mathbf{x} &=& \mathbf{19n -9y} \\ \hline \end{array} \)

 

\(\begin{array}{|rcll|} \hline 9x + 5y &\overset{\color{red}?}{\equiv}& 0 \pmod{19} \quad | \quad \mathbf{x=19n -9y} \\ 9(19n -9y) + 5y &\overset{\color{red}?}{\equiv}& 0 \pmod{19} \\ 9*19n -81y + 5y &\overset{\color{red}?}{\equiv}& 0 \pmod{19} \\ 9*19n -76y &\overset{\color{red}?}{\equiv}& 0 \pmod{19} \\ 9*{\color{red}19}n -4*{\color{red}19}y & \equiv& 0 \pmod{19} \checkmark \\ \hline \end{array}\)

 

laugh

 Aug 3, 2020

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