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Order of [11] in $\mathbb{Z_{335}^x}$.
What I did is:
Since, GCF(11, 335) = 1
So, I thought
$11^x \equiv 1 \pmod {335} \Rightarrow 11^{\phi(335)} \equiv 1 \pmod {335}$
$x = \phi(335) = 264$
But when I tried it in Wolfram alpha x = 66(n).

William Elliot
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Naruto Uzumaki
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  • Your work establishes that the order of $11$ will divide $264$. So you need to figure out the appropriate divisors first. You may want to consider $$11^x \equiv 1 \pmod{335} \iff \begin{align}11^x & \equiv 1 \pmod{5}\ 11^x &\equiv 1 \pmod{67}\end{align}$$ – Anurag A Nov 05 '18 at 20:53
  • $\begin{cases} 11^x \equiv 1 \pmod 5 \ 11^x \equiv 1 \pmod {67} \ \end{cases} \Rightarrow \begin{cases} 11^{\phi(5)} \equiv 1 \pmod 5 \ 11^{\phi(67)} \equiv 1 \pmod {67} \ \end{cases} \Rightarrow \begin{cases} x = \phi(5) =4\ x=\phi(67)=66 \ \end{cases}$
    Since 4 doesn't satisfy the second expression but 66 does satisfy both, so 66 is the answer, right?     – Naruto Uzumaki Nov 05 '18 at 21:41
  • Note that modulo $5$ we have $11 \equiv 1$ so that offers no constraint on the exponent. So we can now work modulo $67$, and the order will be a factor of $66$. – Mark Bennet Nov 05 '18 at 22:19
  • See the Order Test. – Bill Dubuque Nov 05 '18 at 22:40

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