Proof $13 \mid (k\cdot 2^n+1)$ if $n\equiv2 \pmod{12}$ and $k\equiv3 \pmod{13}$? 
Proof $13 \mid (k \cdot 2^n + 1)$ if $n\equiv2 \pmod{12}$ and $k\equiv3 \pmod{13}$
Hint: for $k$ odd: $2^n \equiv-k' \pmod p$ and $kk' \equiv1 \pmod p$

My thoughts:
$13\mid(k-3) \Rightarrow k=13a+3$
and 
$12|(n-2)  \Rightarrow n=12b+2$
so
$\begin{align}k\cdot 2^n+1 &=(13a+3)2^{12b+2}+1 \\
&=4(13a+3)(2^{b})^{12}+1\\
\textrm{ or }
&=(k-3+3)\cdot 2^{n-2+2}+1\\
&=4\cdot 2^{n-2}(k-3+3)+1\\
&=4\cdot 2^{n-2}(k-3)+3\cdot2^n+1\end{align}$
I don't kow how to use the hint :(

By Fermat we have $2^{12} \equiv 1 \pmod{13}$ so $2^{12b}=1+13c$. 
So
\begin{eqnarray*}
k2^n+1=(3+13a)(4+4 \times 13 c)+1=13(1+4a+12c+52ac)
\end{eqnarray*}
