Prove 27 divides 2^(5n+1) + 5^(n+2) for n ≥ 1

[Math100]

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 $n\geq 1$.

 

[SammyS]

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 $n\geq 1$.

You could simply establish that $2^5=32$ and that $32\equiv 5 \pmod {27}$ . Then the actual proof is a snap

 

[fresh_42]

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 $n\geq 1$.

Correct. And thanks for the $()$.

Last time I missed to explain better why the parentheses around $27\mid (2^{5n+1}+5^{n+2})$ are better: The "divides" symbol belongs to multiplication and the plus sign belongs to addition. That's why the parentheses around the addition are reasonable; the same as the distributive law.