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Describe the inverse of $5$ modulo $18$ as a positive power of $5\pmod{18}$.


I've got that the inverse of $5$ is $11$, but is this question asking to find a $t$ such that $$ 11=5^t\pmod{18}?$$

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A positive $t$ specifically, yes. –  anon May 24 '13 at 3:04
    
Oh, OK. Thank you! –  Trancot May 24 '13 at 3:04
    
$=$ not $\equiv$, yes? –  Trancot May 24 '13 at 3:05
    
When learning congruences one typically restricts to using $\equiv$, but oftentimes we can just write "$=$" for the same idea without confusion. I would definitely not say "$=$ not $\equiv$," though. –  anon May 24 '13 at 3:07

3 Answers 3

$$5^2=25\equiv 7\equiv-11 \pmod {18}, \quad 5^3\equiv 5\cdot7\equiv-1 \pmod {18}$$

$$\implies 11=(-11)(-1)\equiv 5^2\cdot5^3\equiv5^5 \pmod {18}$$

Now, as $5^3\equiv-1 \pmod {18}, 5^6=(5^3)^2\equiv(-1)^2\equiv1$

$\implies 5^{6u}=(5^6)^u\equiv1^u\equiv1\pmod{18}$ for any integer $u$

So, the general solution will be $t=6u+5$ as $5^{5+6u}\equiv1\cdot5^5\equiv1\cdot11 \pmod {18}$

If we need $5+6u>0, u\ge 0$

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We usually put the $\pmod{18}$ only at the end, otherwise it just looks weird. –  Patrick Da Silva May 24 '13 at 3:35

$$5^t \equiv 11 \pmod{18}$$

Notice that $\phi(18)=6$

$$5^{6} \equiv1\pmod{18}$$

$$5^5 \cdot 5 \equiv 1\pmod{18} \implies 5x \equiv 1\pmod{18}$$

$$\implies x \equiv 11\pmod{18}$$

Therefore, $$5^5 \equiv 11\pmod{18}$$

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You can see that your modulos are all rendered inproperly. Use the \pmod command instead of the \mod command, it renders the parenthesis right. I edited your answer to do so. –  Patrick Da Silva May 24 '13 at 3:37
    
Yes. Alright. But when is \mod command served.? –  Inceptio May 24 '13 at 3:39

mod $\,18\!:\ 5^3\! \equiv 5(25)\equiv 5(7) = -1 \,\Rightarrow\, 5^6\! \equiv (-1)^2\!\equiv 1,\ $ i.e. $\ 5\cdot \color{#c00}{5^5} \equiv 1,\,$ so $\, 5^{-1}\! \equiv \color{#c00}{5^5}$

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