Why is $10^n-4$ always a multiple of 6? I don't have any idea on how to prove it, and I need it for one of my questions which is still unanswered: What is the largest number smaller than 100 such that the sum of its divisors is larger than twice the number itself?.
 A: It's always even, and it has digits $999\dots996$ so it's always divisible by $3$: it's $3$ times $333\dots332$.
A: Because it is even and $\;10^n-4\equiv1^n-1\equiv0\mod3$, so it's divisible by $2$ and $3$.
A: We see $$10^n - 4 = 6 + \sum^{n-1}_{k=1} 9\cdot 10^k$$ and $6$ divides $9\cdot 10^k$ for any $k\ge 1$ since each of those is an even number divisible by $3$. 
A: Note that $4^n\equiv 4\mod 2$ and $4^n\equiv 4\mod 3$ so by the Chinese Remainder theorem $4^n\equiv 4\mod 6$ i.e. $10^n\equiv 4\mod 6$ 
A: Here is the standard proof for the above claim:
Prove $10^n-4$ is always a multiple of 6, for $n\in \mathbb{N}$.
That is: $10^n-4=6m$, for $m\in \mathbb{N}$.
We prove the above via induction:
Consider the base case, that is, when n = 1, we thus have:
$$10^{1}-4=10-4=6=6m \ for \ m=1 \ \checkmark$$ 
Consider the $n^{th}$ case, that is: $$10^n-4=6m \Rightarrow10^n=6m+4, \ for \ n,m\in \mathbb{N}$$
Then we want to prove the $(n+1)^{th}$.
So consider the following:
$$10^{n+1}-4=(10^n\cdot 10)-4=10(6m+4)-4=60m+36=6(10m+6).$$
Indeed we see $10^{n+1}-4$ is a multiple of $6$ given the $n^{th}$ case.
Therefore by induction, the claim holds for all $n\in \mathbb{N}$.
