Prove that for all x, y ∈ Z if 7 divides 8x + 23y, then 7 divides x + 2y.

Question

Can someone help me solve this question?

Prove that for all x, y ∈ Z if 7 divides 8x + 23y, then 7 divides x + 2y.

I know it's about division algorithm. I just started practicing but not being able to solve.

Thanks in advance.

Answer 1

We know that: $$8x+23y\equiv 0\mod 7$$ We can write $8$ as $7+1$ and $23$ as $7\cdot3+2$, substituibg, we have: $$7x+x+7\cdot3y+y\equiv 0\mod 7$$ And so, by the proprieties of modular arithmetic, we have: $$x+2y\equiv 0\mod 7$$ which implies that $x+2y$ is a multiple of $7$.

Answer 2

Observe that If $\; a\; |\; b \;$ and $ \;a \;| \;c\; $ then

$$ a\; | \;(b-c)$$

We know that $$ 7 \;|\; 21 y$$ and $$7\; | \;7x$$

so, if $ \; 7 \;|\; 8x + 23y \; $ then

$$7 \;|\; (8x+23y) - 7x - 21y$$