$ \forall n \in \Bbb N , 18\mid1^n+2^n+\ldots+9^n-3(1+6^n+8^n)$ How to prove:$ \forall n \in \Bbb N , 18\mid1^n+2^n+\ldots+9^n-3(1+6^n+8^n)$ ?
 A: It is enough to check divisibility by $2$ and by $9$. Divisibility by $2$ is trivial, since $1^n+\cdots+9^n$ and $3(1^n+6^n+8^n)$ are both odd. 
To show divisibility by $9$, we first check separately the case $n=1$. This is easy, the expression in that case is $0$. For $n\ge 2$, the terms $3^n$, $6^n$, and $9^n$ are divisible by $9$, so we need to show that
$$1^n+2^n+4^n+5^n+7^n +8^n -3(1^n+8^n)$$
is divisible by $9$. Recall that if $a$ is relatively prime to $9$, then $a^6\equiv 1\pmod{9}$, since $\varphi(9)=6$, where $\varphi$ is the Euler $\varphi$-function.
Thus we will be finished if we can show that 
$1^n+2^n+4^n+5^n+7^n+8^n-3(1^n+8^n)$ is divisible by $9$ for $n=0,1,2,3,4,5$. This is a finite computation, so in principle we are almost finished.
To cut down on the work, note that $8\equiv -1\pmod{9}$, $7\equiv -2\pmod{9}$, and $5\equiv -4\pmod{9}$. So if $n$ is odd, the terms $1^n$ and $8^n$, and $2^n$ and $7^n$, and $4^n$ and $5^n$ "cancel" modulo $9$.
Thus it only remains to check the cases $n=0$, $2$, and $4$. The case $n=0$ is easy. For the other $2$ cases, compute modulo $9$.    
A: $$\sum_{1\le a\le 9}a^n-3(1+6^n+8^n)$$
$$=\sum_{0\le a\le \frac{9-1}2}\{a^n+(9-a)^n\}-3\{1^n+8^n\}-3\cdot6\cdot6^{n-1}\text{ for }n\ge1$$
If $n$ is odd, $9\mid \{a^n+(9-a)^n\}$ and $9\mid \{1^n+8^n\}$
$$\implies 9\mid \{\sum_{1\le a\le 9}a^n-3(1+6^n+8^n)\}\text{ if } n \text{ is odd} $$
If $n$ is even, $8^n\equiv(-1)^n\pmod 9\equiv1$ and $(9-a)^n\equiv(-a)^n\pmod9\equiv a^n$
$$\text{So,}\sum_{0\le a\le \frac{9-1}2}\{a^n+(9-a)^n\}-3\{1^n+8^n\}-3\cdot6\cdot6^{n-1}$$
$$\equiv2(1+2^n+3^n+4^n)-6\pmod 9$$
$$\equiv2(2^n+4^n-2)\pmod 9 \text{ and }9\mid3^n\text{ for }n\ge2$$
Now, $2^n+4^n-2=(3-1)^n+(3+1)^n-2=(1-3)^n+(1+3)^n-2$ as $n$ is even
$(1-3)^n+(1+3)^n-2$
$=\{1-\binom n13+\binom n23^2-\cdots\}+\{1+\binom n13+\binom n23^2+\cdots\}-2$
$\equiv1-3n+1+3n-2\pmod 9\equiv0\pmod 9$
Thomas has already hinted that $\sum_{1\le a\le 9}a^n-3(1+6^n+8^n)$ is always even which can be derived as follows $\sum_{1\le a\le 9}a^n-3(1+6^n+8^n)\equiv1+3+5+7+9-3\pmod2\equiv0$
So, lcm$(2,9)\mid \{\sum_{1\le a\le 9}a^n-3(1+6^n+8^n)\}$ for all natural number.
If $0$ is also considered $$\sum_{1\le a\le 9}a^n-3(1+6^n+8^n)$$
becomes $$\sum_{1\le a\le 9}1-3(1+1+1)=9-9=0\text{ which is divisible by any nonzero number including }18$$
