Find all the numbers $n$ such that $3\cdot 2^n+2\cdot 3^n\equiv 1 \pmod 7$ 
Find all the numbers $n$ such that $3\cdot 2^n+2\cdot 3^n\equiv 1 \pmod 7$

Attempt:
$\star$ denotes $3\cdot 2^n+2\cdot 3^n$
$$\text{for  }n=1:\quad\star\equiv 5\not\equiv 1\\
\text{for  }n=2:\quad\star\equiv 5\not\equiv 1\\
\text{for  }n=3:\quad\star\equiv 3\not\equiv 1\\
\text{for  }n=4:\quad\star\equiv 0\not\equiv 1\\
\text{for  }n=5:\quad\star\equiv \color{red}1\\
\text{for  }n=6:\quad\star\equiv 5\not\equiv 1\\
\vdots$$
this is how should I aprroach this? or maybe there is smarter way?
 A: As $2^3\equiv1,3^3\equiv-1\pmod7$
let us start with $n=3a,3a+1,3a+2$
Case$\#1:$  $n=3a$
$$1\equiv3\cdot2^{3a}+2\cdot3^{3a}\equiv3+2(-1)^a$$
$\iff2(-1)^a\equiv-2\iff(-1)^a\equiv-1$ as $(2,7)=1$
$\implies a$ is odd $=2b+1$(say) $\implies n\equiv3\pmod6$
Case$\#2:$ $n=3a+1$
$$1=3\cdot2^{3a+1}+2\cdot3^{3a+1}\equiv6(1+(-1)^a)$$
$\iff1+(-1)^a\equiv6^{-1}\equiv6$ as $6^2\equiv1\pmod7$
$\iff(-1)^a\equiv5\pmod7$ which is impossible as $5\not\equiv\pm1\pmod7$
Case$\#3:$ $n=3a+2$
Left for you!
A: You have the right idea. Working mod $7,\,$ by little Fermat $\,2^6\equiv 1\equiv 3^6\pmod 7\,$ so the sequence $\,f_n \equiv 3\cdot 2^n + 2\cdot 3^n \,$  has period $\,6,\,$ i.e $\, f_{n+6}\equiv f_n.\,$ So we need only check the first cycle for $\,n = 0,\ldots, 5.\,$ Using $\,2^3\equiv 1,\ 3^3\equiv -1\,$ we compute the cycle as
$$\begin{eqnarray} 3\ (1&,&2&,&4&,&\ \ \,1&,&\,\ \ 2&,&\,\ \ 4)\\
+\ 2\ (1&,&3&,&9&,&{-}1&,&{-}3&,&{-}9)\\
\equiv \ \  (5&,& 5&,& 2&,&\,\ \ \color{#c00}1&,&\,\ \ 0 &,&\,\ \ \color{#0a0}1)\end{eqnarray}$$
Therefore, we conclude that $\ f_n \equiv 1\iff $  $\,\color{#c00}{n\equiv 3}\ $ or $\,\ \color{#0a0}{n \equiv  5}\, \pmod 6$ 
