# Use $\operatorname{ord}_{11}3$ to find remainder when..

Find $\operatorname{ord}_{11}3$. Then use what you found to find the remainder when you divide $3^{82}$ by $11$.

Work thus far:

$$\operatorname{ord}_{11}3=\ ?$$ $$3^1\equiv3\pmod{11}$$ $$3^2\equiv9\pmod{11}$$ $$3^3\equiv5\pmod{11}$$ $$3^4\equiv4\pmod{11}$$ $$3^5\equiv1\pmod{11}$$ $$\operatorname{ord}_{11}3=5$$

Not sure how to use this to find the remainder through.

Thanks

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If by $\,ord_{11}3\,$ you mean the order of $\,3\,$ in the group $\,\Bbb F_{11}^*=\left(\Bbb Z/11\Bbb Z\right)^*\,$ , then working modulo $\,11\,$ in the following we get

$$ord_{11}3=5\Longrightarrow 3^{82}=\left(3^5\right)^{16}\cdot 3^2=1\cdot 9=9$$

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How does $(3^5)^{16}$ become $1$? Because $ord_{11}3=5$? Are we using the fact that $3^5\equiv1 \pmod{m}$? – student.llama Oct 18 '12 at 3:09
Yes, of course. – DonAntonio Oct 18 '12 at 3:20

Hint $\rm\ \color{#0A0}{3^5}\equiv \color{#C00}1\:\Rightarrow\: 3^{2+5K}\!\equiv\, 3^2 (\color{#0a0}{3^5})^K\!\equiv\, 3^2(\color{#C00}1)^K\!\equiv\, 9$

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We know that $3^{10} \equiv 1 \pmod{11}$ So $\operatorname{ord}_{11} 3 \in \{2,5, 10\}$.

• $3^2 \equiv 9 \equiv -2 \pmod{11}$
• $3^4 \equiv 4 \pmod{11}$
• $3^5 \equiv 1 \pmod{11}$

So $\operatorname{ord}_{11} 3 = 5$.

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