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.
 A: Because $$2^{5n+1}+5^{n+2}=2\cdot32^n-2\cdot5^n+27\cdot5^n=2(32-5)(32^{n-1}+...+5^{n-1})+27\cdot5^n,$$
which is divided by $27.$
A: Use the fact that $32=5+27$ to get
$$\begin{align}2^{5(k+1)+1}+5^{k+1+2}&=2^{5k+6}+5^{k+3}\\&=32\cdot2^{5k+1}+5\cdot5^{k+2}\\&=5\cdot2^{5k+1}+27\cdot2^{5k+1}+5\cdot5^{k+2}\\&=5(2^{5k+1}+5^{k+2})+27\cdot2^{5k+1}\\&=27(5t+2^{5k+1})\end{align}$$ for some positive integer $t$. The result follows.
A: It is:
$$2^{5n+1}+5^{n+2}=2\cdot 32^n+25\cdot 5^n=$$
$$2\cdot (27+5)^n+25\cdot 5^n=2\cdot (27^n+\cdots +5^n)+25\cdot 5^n=$$
$$27m+2\cdot 5^n +25\cdot 5^n=27(m+5^n).$$
A: \begin{eqnarray*}
=\color{red}{27} \times 2^{5n+1}+5 \color{red}{(2^{5n+1}+5^{n+1})}.
\end{eqnarray*}
A: $$32\cdot2^{5k+1}+5\cdot5^{k+2}=5(2^{5k+1}+5^{k+2})+27(2^{5k+1})$$ Both terms now have factors in them divisible by 27:


*

*First has $2^{5k+1}+5^{k+2}$

*Second has $27$
A: 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.
