$ \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)$ ?

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 $9$.    
