master > master: Kritzelei
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| $13^{1003}\mod 5$ | ||||
| 
 | ||||
| $=3^{1003}\mod 5$ | ||||
| 
 | ||||
| $=3^{1003}\mod 5$ | ||||
| 
 | ||||
| $3^0=\boxed{1}$ | ||||
| $3^1=\boxed{3}$ | ||||
| $3^2=9\equiv \boxed{4}$ | ||||
| $3^3=3^2\cdot 3\equiv 4\cdot 3 =12\equiv \boxed{2}$ | ||||
| $3^4=3^3\cdot 3\equiv 2\cdot 3 =6\equiv \boxed{1}$ | ||||
| 
 | ||||
| ---- | ||||
| $3^5=3^4\cdot 3\equiv 1\cdot 3=3$ | ||||
| 
 | ||||
| ---- | ||||
| 
 | ||||
| $3^{4k+r}=3^{4k}\cdot 3^{r}=(3^{4})^{k}\cdot 3^{r}\equiv 1^{k}\cdot 3^{r}=1\cdot 3^{r}=3^{r}$ | ||||
| 
 | ||||
| $1003=4\cdot 250 + \boxed{3}$ | ||||
| 
 | ||||
| $\Longrightarrow$ $3^{1003}\equiv 3^{3}\mod 5=2$ | ||||
| 
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| 
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| Wenn $n=p\in\mathbb{P}$, dann für $k\in\{1,2,\ldots,n-1\}$ gilt $k^{e}\not\equiv 0$. (Wegen Ex. von Inversen: ist $a$ das Inverse von $k$, dann ist $a^{e}$ das Inverse von $k^{e}$.) | ||||
| 
 | ||||
| Wenn $n\notin\mathbb{P}$, dann kann es sein, dass es $k\in\{1,2,\ldots,n-1\}$ gibt, so dass $k^{e}\equiv 0$. | ||||
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