number theory - 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.

Answer

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.