- #1
Math100
- 797
- 221
- Homework Statement
- Find the remainder when ## 823^{823} ## is divided by ## 11 ##.
- Relevant Equations
- Euler-Fermat theorem.
Assume ## (a, m)=1 ##.
Then we have ## a^{\phi(m)}\equiv 1\pmod {m} ##.
Observe that ## 823\equiv 9\pmod {11}\equiv -2\pmod {11} ##.
This implies ## 823^{823}\equiv (-2)^{823}\pmod {11} ##.
Applying the Euler-Fermat theorem produces:
## gcd(-2, 11)=1 ## and ## (-2)^{\phi(11)}\equiv 1\pmod {11} ##.
Since ## \phi(p)=p-1 ## where ## p ## is any prime, it follows that ## \phi(11)=10 ##.
Now we have ## (-2)^{10}\equiv 1\pmod {11} ##.
Thus
\begin{align*}
&(-2)^{823}\equiv [((-2)^{10})^{82}\cdot (-2)^3]\pmod {11}\\
&\equiv [(1)^{82}\cdot (-8)]\pmod {11}\\
&\equiv -8\pmod {11}\\
&\equiv 3\pmod {11}.\\
\end{align*}
Therefore, ## 823^{823}\equiv 3\pmod {11} ## and the remainder when ## 823^{823} ## is divided by ## 11 ## is ## 3 ##.
This implies ## 823^{823}\equiv (-2)^{823}\pmod {11} ##.
Applying the Euler-Fermat theorem produces:
## gcd(-2, 11)=1 ## and ## (-2)^{\phi(11)}\equiv 1\pmod {11} ##.
Since ## \phi(p)=p-1 ## where ## p ## is any prime, it follows that ## \phi(11)=10 ##.
Now we have ## (-2)^{10}\equiv 1\pmod {11} ##.
Thus
\begin{align*}
&(-2)^{823}\equiv [((-2)^{10})^{82}\cdot (-2)^3]\pmod {11}\\
&\equiv [(1)^{82}\cdot (-8)]\pmod {11}\\
&\equiv -8\pmod {11}\\
&\equiv 3\pmod {11}.\\
\end{align*}
Therefore, ## 823^{823}\equiv 3\pmod {11} ## and the remainder when ## 823^{823} ## is divided by ## 11 ## is ## 3 ##.