# 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.

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.

• This is brilliant – Singh Chief Feb 8 '18 at 20:00

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 $27.$

It is: $$2^{5n+1}+5^{n+2}=2\cdot 32^n+25\cdot 5^n=$$ $$2\cdot (27+5)^n+25\cdot 5^n=2\cdot (27^n+\cdots +5^n)+25\cdot 5^n=$$ $$27m+2\cdot 5^n +25\cdot 5^n=27(m+5^n).$$

\begin{eqnarray*} =\color{red}{27} \times 2^{5n+1}+5 \color{red}{(2^{5n+1}+5^{n+1})}. \end{eqnarray*}

$$32\cdot2^{5k+1}+5\cdot5^{k+2}=5(2^{5k+1}+5^{k+2})+27(2^{5k+1})$$ Both terms now have factors in them divisible by 27:

• First has $$2^{5k+1}+5^{k+2}$$
• Second has $$27$$

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.