6. determine all $k$ for which a natural number $n$ exists so that $1^{n}+2^{n}+3^{n}+4^{n}$ ends with exactly $k$ zeros.

## Solution
Sei $A(n)=1^{n}+2^{n}+3^{n}+4^{n}$. The examples $A(1)=10, A(3)=100$ and $A(4)=354$ show that $k=0,1,2$ is possible. Now assume that $A(n)$ ends with at least three zeros, then $8 \mid A(n)$ applies. For $n \geq 3$, however, $A(n) \equiv 1+0+3^{n}+0=4$ or 2 $(\bmod 8)$, contradiction.