For ## n\geq 1 ##, use congruence theory to establish....

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For natural numbers n≥1, it is established that 27 divides the expression 2^(5n+1) + 5^(n+2) using congruence theory. The proof demonstrates that the expression simplifies to 0 modulo 27 through a series of congruences. Specifically, it shows that 2^(5n) and 5^n can be manipulated to reveal that their sum is congruent to 0 modulo 27. The discussion also emphasizes the importance of proper notation, particularly the use of parentheses to clarify the relationship between addition and multiplication in the divisibility statement. Overall, the proof effectively confirms the divisibility condition for all natural numbers n≥1.
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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 ##.
 
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Math100 said:
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
 
Last edited:
Math100 said:
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.
 
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