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.

Here is a different take, for fun.
Let $x_n = 2^{5n + 1} + 5^{n + 2} = 2\cdot 32^n+25\cdot 5^n$.
Since $32$ and $5$ are the roots of an equation $x^2=ax+b$ (*), we have $x_{n+2}=ax_{n+1}+bx_n$.
The result follows by induction because the base cases $27 \mid x_0 = 27$ and $27 \mid x_1 = 189$ are easily checked. 
(*) where $a=32+5, b=32\cdot 5$. These values are not important. What matters is that the recurrence is linear with integer coefficients.
