Showing that a $3^n$ digit number whose digits are all equal is divisible by $3^n$ 
Let $c$ be a $3^n$ digit number whose digits are all equal. Show that $3^n$ divides
  $c$.

I have no idea how to solve these types of problems. Can anybody help me please?
 A: A number whose digits are all equal and of length $3^n$ is thus of the form $c = \sum\limits_{i=0}^{3^n-1} a \cdot 10^i = a \dfrac {10^{3^n}-1}{10-1}$ by the geometric series.
Since we have to account for the possibility that $a = 1$ we need to show that $3^n \mid \dfrac{10^{3^n}-1}9$, i.e. $3^{n+2} \mid 10^{3^n}-1$.
This we can do by induction: Base case is trivial, suppose $3^{n+2} \mid 10^{3^n}-1$.
Then we have $10^{3^n} \equiv 1 \pmod {3^{n+2}}$, and so: $10^{3^n} \equiv 1 + k 3^{n+2} \pmod {3^{n+3}}$. Now compute $10^{3^{n+1}} = \left(10^{3^n}\right)^3$ modulo $3^{n+3}$.

Addendum: In general, when considering repdigits it is useful to have them in the form obtained through the geometric series. Since the length of the repdigit is in the exponent, and exponents are well-behaved under modulo calculations, it is one of the most informative forms for discovering properties of repdigits; in particular those pertaining to divisibility.
A: Because the digits sum to a multiple of $3$, a block of $3$ identical digits is divisible by $3$.
$3$ of those blocks of $3$ digits, having been divided by $3$, is also divisible by $3$.
$3$ of those blocks of $3^2$ digits, having been divided by $3^2$, is also divisible by $3$.
$3$ of those blocks of $3^3$ digits, having been divided by $3^3$, is also divisible by $3$.
and so on. For example,
$\begin{align}
&111/3\\
=&037
\end{align}$
$\begin{align}
&111\ 111\ 111/3^2\\
=&037\ 037\ 037/3\\
=&012\ 345\ 679
\end{align}$
$\begin{align}
&111111111\ 111111111\ 111111111/3^3\\
=&037037037\ 037037037\ 037037037/3^2\\
=&012345679\ 012345679\ 012345679/3\\
=&004115226\ 337448559\ 670781893
\end{align}$
A: first of all, you may suppose that your number is composed by all digits 1. 
Then use induction: case n=1 is easy, in general try and write the number as the product of 111 and something else.
A: Hint $\ $ Below is the key induction step
$$\begin{eqnarray} \rm \overbrace{11\ \cdots\ 11}^{\large\color{#C00}{ 3^{\large N}}} &=&\rm a\,\color{#C00}{3^N} \\[0.5em] 
\rm \Rightarrow\ \ \ \rm \overbrace{11\ \cdots\ 11}^{\color{#C00}{\large 3^{\large N+1}}} &=&\rm \overbrace{11\, \cdots\, 11}^{\large 3^{\large N}}\ \overbrace{11\ \cdots\ 11}^{\large 3^{\large N}}\,\overbrace{11\ \cdots\ 11}^{\large 3^{\large N}}\\[0.5em]
 &\,=\,&\rm a\,3^N 10^{2k}\! +\, a\,3^N 10^k\! +\, a\, 3^N,\ \ \ k = 3^N\\[0.5em]
 &=&\rm a\,3^N (10^{2k}+\,10^k+\,1) \\[0.5em]
 &=&\rm b\,\color{#C00}{3^{N+1}}\ by \ \ mod\ 3\!:\ 10^{2k}\!+10^k+\!1\equiv 1^{2k}\!+1^{k}\!+1\equiv 3\equiv 0
\end{eqnarray}$$
A: Let $a_1a_2 \dots a_{3^n}$(Call it $Z$) be a $3^n$ digit number with a common digit $a$.
You have $10^{3n}\cdot a+10^{3n-1}\cdot a+10^{3n-2}\cdot a+\dots a$
You can right that in a different way,
$\sum_{i=1}^{3^n}(10^{3i}-1)\cdot a+\sum_1^3^na=Z$
$\sum_1^{3^n}a=3^n\cdot a$ , Use Lord Farin's method of GM on $\sum_{i=1}^{3^n}(10^{3i}-1)$.
A: We shall prove this by induction on $n$.
Base Case: For $n = 1$, we note that any $3$-digit integer with $3$ identical digits is divisible by $3$.
Since, for any $k \in \{1, \dots, 9\}$, $kkk = k \cdot (111)$. Further, $111$ is divisible by $3$. Therefore, $kkk$ is
divisible by $3$.
Hypothesis: Assume that the statement is true for $n = k$, $k \geq 1$.
Induction Step: For $n = k + 1$, $k \geq 1$. Let $x$ be an integer composed of $3^{k+1}$ identical digits.
We note that $x$ can be written as
$x = y \times z$
where $y$ is an integer composed of $3^k$ identical digits, and $z = 10^{2\cdot 3^k}
+ 10^{3^k}+ 1$.
For example, $x = 666666666 = 666 \times 1001001 = y × (10^{2\cdot 3}+ 10^3+ 1)$. $y$ is divisible by $3^k$ by the hypothesis and $z$ is divisible by $3$ (sum of the digits is divisible by $3$). Thus $x$ is divisible by
$3^{k+1}$.
