Modular exponentiation

Question

How do you solve: $$5^{{9}{^{13}}^{17}} \equiv x\pmod {11}$$

I've been trying with this but no luck. I get to ${{9}{^{13}}^{17}} \equiv x\pmod {11}$ from $5^3 * 5^3 * 5^3 = 64 \equiv 9\pmod {11}$. But volfram alpha disagrees with all that. So just a nudge in the right direction please.

thanks!

Answer 1

By the little Fermat theorem we have

$$5^{10}\equiv 1\pmod{11}$$ and since

$$9^{{13}^{17}}\equiv-1\pmod{10}$$ then we see that

$$5^{9^{{13}^{17}}}\equiv 5^{-1}=9\pmod{11}$$

Answer 2

As an alternate way $$5^5=3125$$ $3124\equiv 0\pmod {11}$ (use the criterion for divisibility with $11$), so $$5^5\equiv 1 \pmod {11}$$ On the other hand, $9\equiv 4\pmod 5$ and $13\equiv 1\pmod 2\Rightarrow 13^{17}\equiv 1\pmod 2$, so $9^{13^{17}}\equiv 4\pmod 5$. It follows that $$5^{9^{13^{17}}}\equiv 5^4\pmod {11}$$ $5^4=625,\ 625\equiv 9\pmod {11}$ (apply again the divisibility criterion by $11$), so the conclusion is $$5^{9^{13^{17}}}\equiv 9\pmod {11}.$$