Let $x$ and $y$ be integers such that $2x+3y$ is a multiple of $17$. Show that $9x+5y$ must also be a multiple of $17$.

Question

Let $x$ and $y$ be integers such that $2x+3y$ is a multiple of $17$. Show that $9x+5y$ must also be a multiple of $17$.

So $2x+3y \equiv 0 \pmod{17}$. Adding $7x$ and $2y$ we have that $9x+5y \equiv 7x+2y \pmod{17}$, thus we would need to show that $7x+2y \equiv0 \pmod{17}$.

How should I approach this? I cannot see to be able to show the last congruence...

score 4 · Accepted Answer · answered Oct 25 '20 at 09:22

4

It's easier to just work directly with the original number.

Note that

$$ 17x + 17y - 4(2x+3y) = 9x + 5y.$$

Everything on the left is a multiple on 17, and hence so is the number on the right.

answered Oct 25 '20 at 09:22

asahay

290

$17x$ and $17y$ are clearly multiples of $17$, but $17$ doesn't divide $4(2x+3y)$? – Oct 25 '20 at 09:25
2

$2x+3y$ is a multiple of 17 by assumption, so $4(2x+3y)$ is also a multiple of 17. – Laars Helenius Oct 25 '20 at 09:28
Ah yes! Thanks! – Oct 25 '20 at 09:29
@Daniel The key idea is to scale $2x+3y$ by a factor $,c,$ to get $,9x+5y,,$ For that we need $,2c\equiv 9\equiv -8\iff c\equiv -4.,$ See here for more. At asahay: please try to resist quickly answering FAQs before there is a chance for dupes to be located. – Bill Dubuque Oct 25 '20 at 09:48

Let $x$ and $y$ be integers such that $2x+3y$ is a multiple of $17$. Show that $9x+5y$ must also be a multiple of $17$.

1 Answers1