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How does one calculate $3^{(29^{10001})} \mod 35$? I'm just not seeing how to start on it, unless someone could give me a hint? Thanks!

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Hint: To use Euler's theorem, you need to compute $29^{10001} \bmod \phi(35)$, noting that $\phi(35)=24$.

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  • $\begingroup$ Thank you! I want to give my calculations but that would require a lot of typing just to say that I think the outcome must be 33. $\endgroup$ – simp Jun 27 '17 at 23:01
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Use first the Chinese remainder theorem and compute first $3^{29^{10001}}$ mod. $5$ and mod.$7$. Note that, by lil' Fermat $$3^{29^{\scriptstyle 10001}}\equiv3^{29^{\scriptstyle10001} \bmod4}\pmod5,\qquad3^{29^{\scriptstyle 10001}}\equiv3^{29^{\scriptstyle10001} \bmod6}\pmod7. $$ Now $29\equiv 1\pmod4$, so $3^{29^{\scriptstyle10001} \bmod4}\equiv 3^1=3\pmod5$.

Similarly, $29\equiv -1\pmod6$, so $3^{29^{\scriptstyle10001} \bmod4}\equiv 3^{-1}\equiv 5\pmod7$.

Finally everything comes down to solving the system of congruences \begin{cases} x\equiv3\pmod5,\\x\equiv 5\pmod 7.\end{cases} As a Bézout's relation between $7$ and $5$ is $\;3\cdot5-2\cdot7=1$, t-he solution is $$x\equiv 5(3\cdot5)-3(2\cdot 7)=33\pmod{35}. $$

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  • $\begingroup$ Thank you for your insights! $\endgroup$ – simp Jun 27 '17 at 23:32
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Using Carmichael Function,

$$3^{29^{2n+1}}\equiv3^{29^{2n+1}\pmod{\lambda(35)}}\pmod{35}$$

Now as $\lambda(35)=12$ and $29\equiv5\pmod{12},29^{2n+1}\equiv5^{2n+1}$

as $\lambda(12)=2,2n+1\equiv1\pmod2\implies5^{2n+1}\equiv5^1\pmod{12}$

$$\implies3^{29^{2n+1}}\equiv3^5\pmod{35}\equiv?$$

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