# 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?

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The solution according to what? – Thomas Andrews Nov 1 '12 at 16:40
You have shown $11^{13}-5$ can be divided by 6; so too can $11^{13}+1$. So $11^{13}-1$ cannot; $11^{1}-1=10$ cannot either. – Henry Nov 1 '12 at 16:40
$11^{13} - 5 \equiv 0 \mod 6$ so $11^3 - 1 \equiv 4 \mod 6$. – Graphth Nov 1 '12 at 16:41
$11 \equiv -1 \mod 6$ should help copmputations like this. – Arthur Nov 1 '12 at 16:41
Probably $\:11^{13}\!-1\:$ is a typo for $\:11^{13}\!+1\equiv (-1)^{13}\!+1\equiv 0\pmod 6.\ \$ – Bill Dubuque Nov 1 '12 at 18:45

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 $11 \equiv 2 \equiv -1 \, (\text{mod} \, 3)$.

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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 $6$.

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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 .

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