Can $11^{13}-1$ be divided exactly by 6? Can $11^{13}-1$ be divided exactly by 6?
My solution:
$$11^2 \equiv 1 \pmod 6$$
$$11^{12} \equiv 1 \pmod 6$$
$$11^{13} \equiv 5 \pmod 6 $$
Hence, $(11^{13}-\mathbf{5})$ can be divided exactly by 6. However, according to the solution on my book, （$11^{13}-\mathbf{1}$）can be divide exactly by 6. What's wrong?
 A: Assume $11^{13}-1$ was divisible by $6$, then we'd have
$$11^{13}-1\equiv 0\pmod 6.$$
In other words, $11^{13}\equiv 1$. However, by your computation $11^{13}\equiv 5$, this is a contradiction because $1$ and $5$ are not congruent modulo $6$. Hence, $11^{13}-1$ is not divisible by $6$.
A: It is already not divisible by $ 3 $; notice that
\begin{align}
11^{13} - 1 &\equiv (-1)^{13} - 1 \\
            &\equiv -1 - 1 \\
            &\equiv -2 \\
            &\equiv 1 \\
            &\not\equiv 0 \, (\text{mod} \, 3).
\end{align}
Note $ 11 \equiv 2 \equiv -1 \, (\text{mod} \, 3) $.
A: As          $\frac {11}{6}  $  gives remainder of $5$   or $-1$ . we take $-1$ as it eases our calculation . 
1) $ \frac{11^{13}}6 $  gives remainder  $-1$ . 
2) $ \frac{1}6 $  gives remainder 1
FIND : RESULT $1$ - RESULT $2$
$-1 -1 = -2$  which is equal to $4 $ as $6-2 =4$ 
So remainder is $4$  . It means it is definitely NOT  DIVISIBLE by  6 .
.You can check your answers in one of the most trusted sites : wolframalpha , where you can do such calculation involving large numbers . 
I have done the following for you in the link below .
http://www.wolframalpha.com/input/?i=11%5E13%E2%88%921+mod+6
Even scientific calculators in computers perform, mod operation of such large numbers with ease . 
These computing sites give confidence on our calculation.
