How to prove that $3^{3n+2}+2^{4n+1}$ is a multiple of 11 How can I prove that following is a multiple of 11 with the induction method?
 $$3^{3n+2}+2^{4n+1}$$ 

Base case, $n=0$: $\quad 3^{0+2}+2^{0+1} = 9+2 = 11 \quad \checkmark$
Induction hypothesis: $\quad 3^{3n+2}+2^{4n+1} = 11k$
$\begin{align} 3^{3(n+1)+2}+2^{4(n+1)+1} &= 27\cdot3^{3n+2}+16\cdot 2^{4n+1} \\
&= 22\cdot3^{3n+2}+5\cdot3^{3n+2}+11\cdot 2^{4n+1}+5\cdot 2^{4n+1}\\
&= 22\cdot3^{3n+2}+11\cdot 2^{4n+1}+5(3^{3n+2}+ 2^{4n+1})\\
&= 11(2\cdot3^{3n+2}+ 2^{4n+1})+5(11k)\\
&= 11(2\cdot3^{3n+2}+ 2^{4n+1}+5k)\quad \text{as required}\\
\end{align}$  

(without induction):
$3^3 \equiv 27 \equiv 5 \bmod 11$ and $2^4 \equiv 16 \equiv 5\bmod 11$ 
So $3^{3n+2}+2^{4n+1} \equiv 9\cdot 5^n+2\cdot 5^n \equiv 11\cdot 5^n \equiv 0 \bmod 11$
