<s>[INST] 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?
 [/INST] 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
$3^{k+1}$.
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