Prove that if $23 ∣ (3a + 5b + 7c)$, then $23 ∣ (14a + 8b + 2c)$. 
Prove that if $23 ∣ (3a + 5b + 7c)$, then $23 ∣ (14a + 8b + 2c)$.

My attempt: I am trying to use the fact that $$gcd(23,3)=gcd(23,5)=gcd(23,7) = 1,$$ but I don't know how to proceed from there. 
 A: Hint: $\bmod 23\!:\,\  \color{#c00}{14}a\!+\!8b\! +\!2c\,\equiv \color{#0a0}{{-}3}\,(\color{#c00}3a\!+\!5b\!+\!7c)\equiv -3(0)\equiv 0$
Scale factor $\,\color{#0a0}{-3}\,$ comes from $\ \color{#c00}{14/3}\equiv -9/3\equiv -3$ 
A: $15(14a+8b+2c)=210a+120b+30c$ and 
$210=3$ mod $23$ since $23\times 9=207$
$120=5$ mod $23$ since $23\times 5=115$
$30=7$ mod $23$ 
This implies that $15(14a+8b+2c)=3a+5b+7c$ mod $23$ since
 $15$ is invertible mod $23$.
A: Using row reduction mod $23$ we get
$$
\pmatrix{
3 & 5 & 7 \\
14 & 8& 2
}
\to
\pmatrix{
3 & 5 & 7 \\
0 & 0 & 0
}
$$
Therefore, $(14,8,2)$ is a non-zero multiple of $(3,5,7)$ mod $23$ and the claim follows. Actually, the converse of the claim also follows.
A: We have in the field $\mathbb F_{23}$ $$\begin{cases}3a+5b+7c=0\\14a+8b+2c=x\end{cases}\Rightarrow\begin{cases}6a+10b+14c=0\\98a+56b+14c=7x\end{cases}$$ $$92a+46b=7x\Rightarrow x=0$$ We are done.
A: Note that:
$(14a + 8b + 2c) + 3(3a + 5b + 7c) = 23(a+b+c)$
So if $23$ divides either of the first two brackets it must divide the other.
