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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?

  • 3
    Probably $\:11^{13}\!-1\:$ is a typo for $\:11^{13}\!+1\equiv (-1)^{13}\!+1\equiv 0\pmod 6.\ \ $2012-11-01

3 Answers 3

1

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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4

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 $6$.

4

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) $.