# 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?.

It's always even, and it has digits $999\dots996$ so it's always divisible by $3$: it's $3$ times $333\dots332$.

• Good answer! You got there 43 sec before me! Apr 7, 2017 at 23:58

Because it is even and $\;10^n-4\equiv1^n-1\equiv0\mod3$, so it's divisible by $2$ and $3$.

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 $3$.

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. $10^n\equiv 4\mod 6$

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 $n\in \mathbb{N}$.