<s>[INST] Prove that $2^{5n + 1} + 5^{n + 2} $ is divisible by 27 for any positive integer My question is related to using mathematical induction to prove that $2^{5n + 1} + 5^{n + 2} $ is divisible by 27 for any positive integer.
Work so far: 
(1) For n = 1:  
$2^{5(1) + 1} + 5^{(1) + 2} = 26 + 53 = 64 + 125 = 189$
Check if divisible by $27$:  $189$ mod $27$ = $0$
As no remainder is left, the base case is divisible by $27$.  
(2) Assume $n = k$, then $2^{5k + 1} + 5^{k + 2} = 27k$  
(3) Prove that this is true for n = k + 1:  
$$2^{5(k + 1) + 1} + 5^{(k + 1) + 2} $$
$$= 2^{5k + 5 + 1} + 5^{k + 1 + 2} $$
                $$  = 32 * 2^{5k + 1} + 5 * 5^{k + 2}$$
$$=  ? $$
I know I am supposed to factor out 27 somehow, I just cant seem to figure it out. Any help would be appreciated.
 [/INST] \begin{eqnarray*}
=\color{red}{27} \times 2^{5n+1}+5 \color{red}{(2^{5n+1}+5^{n+1})}.
\end{eqnarray*}
</s>