- #1
Math100
- 797
- 221
- 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 ##.