- #1

Math100

- 756

- 204

- Homework Statement
- For ## n\geq 1 ##, use congruence theory to establish the following divisibility statement:

## 27\mid (2^{5n+1}+5^{n+2}) ##.

- Relevant Equations
- None.

Proof:

Let ## n\geq 1 ## be a natural number.

Then \begin{align*} 2^{5n+1}+5^{n+2}&\equiv (2^{5n}\cdot 2+5^{n}\cdot 5^{2})\pmod {27}\\

&\equiv [(2^{5})^{n}\cdot 2+5^{n}\cdot 25]\pmod {27}\\

&\equiv (32^{n}\cdot 2+5^{n}\cdot 25)\pmod {27}\\

&\equiv (5^{n}\cdot 2+5^{n}\cdot 25)\pmod {27}\\

&\equiv (5^{n}\cdot 27)\pmod {27}\\

&\equiv 0\pmod {27}.

\end{align*}

Therefore, ## 27\mid (2^{5n+1}+5^{n+2}) ## for ## n\geq 1 ##.

Let ## n\geq 1 ## be a natural number.

Then \begin{align*} 2^{5n+1}+5^{n+2}&\equiv (2^{5n}\cdot 2+5^{n}\cdot 5^{2})\pmod {27}\\

&\equiv [(2^{5})^{n}\cdot 2+5^{n}\cdot 25]\pmod {27}\\

&\equiv (32^{n}\cdot 2+5^{n}\cdot 25)\pmod {27}\\

&\equiv (5^{n}\cdot 2+5^{n}\cdot 25)\pmod {27}\\

&\equiv (5^{n}\cdot 27)\pmod {27}\\

&\equiv 0\pmod {27}.

\end{align*}

Therefore, ## 27\mid (2^{5n+1}+5^{n+2}) ## for ## n\geq 1 ##.