Best method for proving that $7\times11^{2n+1}-3^{4n-1}$ is divisible by $10$ I am asked to prove by induction that $7\times11^{2n+1}-3^{4n-1}$ is divisible by $10$.
I wonder whether there is a more direct method, for example factorizing by $10$.
If an expression is divisible by $10$, does this mean that I can factorize it by $10$?
Thanks in advance
 A: A direct and intuitive way. Every integer power of $11$ has $1$ as last digit, so every number of the form $7 \cdot 11^{2n+1}$ ends with the digit $7$.
The last digits of the integer powers of $3$ follows the four-step cycle $(3,9,7,1)$, so that every number of the form $3^{4n-1}$ has $7$ as last digit.
Therefore, subtracting this second number from the first one we get a number ending with the digit $0$. This implies that this last number is divisible by $10$.
A: Below are four proofs using various methods.

You seem to seek a direct proof showing a factor of $10$ so let's do that first.
$$\begin{align} x &\,=\ \ \ \ \, 7\cdot 11^{\large 2n+1}\ -\ 3^{\large 4n-1}\\[.2em]
\Rightarrow\ \   3x &\,=\ \ \, 21\cdot 11\cdot 121^n - 81^n\\[.2em]
 &\,=\, (21\cdot11\!-\!1)121^n+ \color{#0a0}{121^n-81^n}\\[.2em] 
  &\,=\, \color{#c00}{10}\,(23\cdot 121^n + \color{#c00}4(121^{n-1} + \cdots + 81^{n-1}))\
\end{align}\qquad$$
where we used the Factor Theorem to deduce $\,\color{#c00}{10\cdot 4} = 121\!-\!81\,$ divides $\,\color{#0a0}{121^n-81^n}.\,$
Thus $\,10\mid 3x\Rightarrow 10\mid x\,$ by Euclid (or directly $\,10\mid 7(3x)\!-\!20x = x,\,$ or cancel $3$ from  $121$'s and $81$'s).

It's much easier by modular arithmetic (congruences)
$$\begin{align}\bmod 10\!:\ \  3x &\equiv 21\cdot 11^{\large 2n+1} - 81^{\large n}\\
            \iff\      3x  &\equiv \ \ 1\ \cdot\ 1^{\large 2n+1}\ -\ 1^{\large n} \equiv\color{#0a0} 0\\
            \iff\ \ \       x  &\equiv\,3^{-1}\cdot\color{#0a0} 0\equiv 0
\end{align}\qquad$$
by basic congruence laws. We used the fact that scaling by an invertible (here $3$) yields an equivalent congrence (recall by Bezout that $3$ is invertible being coprime to the modulus $10)$

By induction: base case $\,n=1\,$ is $\!\bmod 10\!:\ 7\cdot 11^3\equiv 3^3\,$ (or $\,7\cdot 11\equiv 1/3\,$ for $\,n=0)\,$  which are both true, and the induction step follows conceptually by simply by multiplying the first two congruences below using  $\rm\color{#0a0}{CPR} =$ Congruence Product Rule,
$$\begin{align}\bmod 10\!:\qquad\ \ \  \color{#c00}{11^{\large 2}}\ &\equiv\ \color{#c00}{3^{\large 4}}\\[.2em]
{\rm times}\ \ \ \ \ \ \ 7\cdot 11^{\large 2n+1}&\equiv3^{\large 4n-1}\quad \  P(n)_{\phantom{|}}\\[.2em]
\hline
\Longrightarrow\ \ \ \ \ 7\cdot 11^{\large 2n+\color{#c00}3}&\equiv 3^{\large 4n+\color{#c00}3}\quad\ P(n\!+\!\color{#c00}1),
\ \ \rm by \ \,\color{#0a0}{CPR}^{\phantom{|^|}}\end{align}\qquad  $$
If congruences are unfamiliar we can preserve the arithmetical essence of this simple proof by using an analogous product rule for divisibility (DPR), as explained here.

Or as here use Binomial Theorem  on $\,(1\!+\!10)^{2n+1}$ and $(-1\!+\!10)^{4n-1}$ (or $\,(1\!+\!80)^n\,$ in $3x)$

Remark $ $ All these methods do in fact use induction (on $n),\,$ but it may be hidden (encapsulated) in the proof of a theorem that is invoked, e.g. the Factor Theorem or Binomial theorem, or the Congruence Power Rule $\,a\equiv b\Rightarrow\, a^n\equiv b^n$.
A: we have $$7{(10+1)}^{2n+1}-3{(10-1)}^{2n-1}$$
indeed by binomial theorem it is equivalent to:
$$7(10k+1)-3(10m-1)=10+70k-30m$$
which is div by $10$
A: Using modular arithmetic:
\begin{align}3(7\cdot 11^{2n+1}-3^{4n-1})&=21\cdot 11^{2n+1}-9^{2n}\\&\equiv 1\cdot (1)^{2n+1}-(-1)^{2n}\\&=0 \pmod {10}.\end{align}
A: We have $$3(3^3)=81 \equiv 1 \pmod{10}$$
$$3^{-1}\equiv 3^3 \equiv 7 \pmod{10}$$
\begin{align}
7(11^{2n+1})-3^{4n-1} &\equiv 7 (1^{2n+1}) - (3^4)^n3^{-1}\\
&\equiv 7(1)-(1)(7) \\
&\equiv 0 \pmod{10}
\end{align}
A: Best method?  Depends on what you know.
If you know modular arithmetic and Eulers Theorem then
$7\cdot 11^{2n+1} - 3^{4n-1}\equiv 7\cdot 1^{2n+1} - (3^{-1})\cdot 3^{4n}\equiv 7-(3^{-1})\pmod{10}$ and as $3\cdot 7 \equiv 1 \pmod {10}$ then $3^{-1}\equiv 7 \pmod {10}$ and $7-7\equiv 0\pmod {10}$ so $10|7\cdot 11^{2n+1} - 3^{4n-1}$.
But that assumes you are comfortable with many concepts
If you don't know any modular arithmetic:
$11^k = (10+1)^k = 10^k + k10^{k-1} + ...... + k\cdot 10 + 1 = 10 M + 1$.
And $3^{4n-1} = 3\times 3^{4n-2}= 3\times 9^{2n-1}$.  And if $k=2n-1$ is odd the $9^k = (10-1)^k = 10^k - k10^{k-1} + .......   + k\cdot 10 - 1= 10N-1$.
So $7\cdot 11^{something} - 3^{1+2\times something\ odd} = 7(10M + 1)- 3(10N-1)= 70M -30N +10$.
A: Since
$\quad 11^k \equiv 1 \pmod{10} \quad \forall k \in \Bbb Z$
and
$\quad 3^{4n-1} \equiv {(3^2)}^{2n} \times 3^{-1} \equiv (-1)^{2n} \times 7 \equiv 7 \pmod{10}$
we can write
$\quad 7\times11^{2n+1}-3^{4n-1} \equiv 7 - 7 \equiv 0 \pmod{10}$

Elementary Number Theory:
The invertible elements in $\mathbb{Z}/{10}\mathbb{Z}$ are
$\quad [1], [3], [7] \text{ and } [9]$
allowing one to quickly determine that $3^{-1} \equiv 7 \pmod{10}$.
A: The OP is looking for a factorization argument and so we give a problem/hint that can be directly solved without elementary number theory using simple algebra; this is the step case of the induction proof.
Problem: Show that if $k \ge 1$ and
$\tag 1  \displaystyle 7\times11^{2k+1}-3^{4k-1} = 10q$
then
$\tag 2  \displaystyle 7\times11^{2(k+1)+1}-3^{4(k+1)-1} = 10 \times (11^2 q + 4 \times 3^{4k-1})$
A: Multiply $n=7\times11^{2n+1}-3^{4n-1}$  by $3$ to get $3n=21\times11^{2n+1}-81^{n}$, which is divisible by $10$
(ones digit is $0$, because it is the difference of two numbers whose ones digit is $1$).
Now if $3n$ is divisible by $10$, for an integer $n$, then $n$ is divisible by $10$.  QED.
