Prime tower part 2

Let $x=2^{3^{5^{7^{11^{13}}}}}$ . We have that $x \equiv 0 \pmod{32}$ , now let's find the remainder when $x$ is divided by $3125$ . Let $y=3^{5^{7^{11^{13}}}}$ , we have $\lambda(3125)=2500$ , so $x \equiv 2^{y \mod 2500} \pmod {3125}$ .

Now let's find the remainder when $y$ is divided by $2500$ . Let $z=5^{7^{11^{13}}}$ , we have $\lambda(2500)=500$ , so $y \equiv 3^{z \mod 500} \pmod{2500}$ .

We have $z \equiv 0 \pmod{125}$ and $z \equiv 1 \pmod{4}$ , so, by the Chinese Remainder Theorem , $z \equiv 125 \pmod{500}$ .

Working backwards we have $y \equiv 3^{125} \equiv 443 \pmod{2500}$ , then $x \equiv 2^{443} \equiv 458 \pmod{3125}$ , and finally, using the Chinese Remainder Theorem again, $x \equiv \boxed{94208} \pmod{10^5}$ .

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Note: $\lambda$ denotes the Carmichael function .