Proving $1+2^n+3^n+4^n$ is divisible by $10$ How can I prove 

$$1+2^n+3^n+4^n$$ is divisible by $10$  if $$n\neq 0,4,8,12,16.....$$

 A: $$3\equiv1\pmod2\implies 1+2^n+3^n+4^n\equiv0\pmod2$$
If $n$ is odd, $1+4^n=1^n-(-4)^n$ is divisible by $1-(-4)=5$
Similarly, $2^n+3^n\equiv0\pmod5,$
For $n=4k+2,2^{4k+2}\equiv4\pmod5,3^{4k+2}\equiv-1,4^{4k+2}\equiv1$
A: After checking that $S(n)=1+2^n+3^3+4^n\equiv0$ mod $10$ for $n=1$, $2$, and $3$ (i.e., computing $S(1)=10$, $S(2)=30$ and $S(3)=100$), induction does the rest, thanks to the difference
$$\begin{align}
S(n+4)-S(n)&=2^n(2^4-1)+3^n(3^4-1)+4^n(4^4-1)\\
&=15\cdot2^n+80\cdot3^n+255\cdot4^4\\
&\equiv5\cdot2^n+0\cdot3^n+5\cdot4^n\mod 10\\
&=5\cdot2(2^{n-1}+2^{2n-1})\\
&\equiv0\mod 10
\end{align}$$
Remark:  This also shows that $S(4n)\equiv4$ mod $10$, since $S(0)=1+1+1+1=4$.
A: HINT : You can consider each right-most digit.
$$1:1,1,1,1,1,\cdots$$
$$2^n:2,4,8,6,2,\cdots$$
$$3^n:3,9,7,1,3,\cdots$$
$$4^n:4,6,4,6,4,\cdots$$
So, we have
$$1+2+3+4=1\color{red}{0},1+4+9+6=2\color{red}{0},1+8+7+4=2\color{red}{0},1+6+1+6=14.$$
A: We consider it $(\mathrm{mod}\;2)$ and $(\mathrm{mod}\;5)$ separately. 
Clearly $1$ and $3^n$ are odd, while $2^n$ and $4^n$ are even, so their sum is even. 
Now by Fermat's Little Theorem, when $a$ is not divisible by $5$, $a^5 \equiv a \mod 5$, so $a^4 \equiv 1 \mod 5$. Thus we only need to check $n=1$, $2$, and $3$. 
(note that $1$, $2$, $3$, and $4$ are not divisible by $5$. 
If $n=1$, we get $1+2+3+4=10$, which is divisible by $5$
If $n=2$, we get $1+4+9+16=30$, which is divisible by $5$. 
If $n=3$, we get $1+8+27+64=100$, which is divisible by $5$.
Thus we have proved it $(\mathrm{mod}\;5)$ and we are done. 
