Again a remainder problem :)))

Let $17^{17}=S$ , then $17^{17^{17}}=17^S$ . To find $17^S \pmod{1729}$ , we can just apply the Chinese Remained Theorem on the results $17^S \pmod{7}$ , $17^S \pmod{13}$ , $17^S \pmod{19}$ , since we have $1729=7*13*19$ .

We first note that:

$S=17^{17}\equiv(-1)^{17}\pmod{6}\equiv5\pmod{6}$

$S=17^{17} \equiv(5)^{17}\pmod{12}\equiv5\pmod{12}$

$S=17^{17}\equiv(-1)^{17}\pmod{18}\equiv17\pmod{18}$

These congruences help in the application of Fermat's Little Theorem to figure out the congruences later modulo the various primes. The second congruence is an extension from the result $5^4\equiv1\pmod{12}$ as taken from Euler's totient theorem.

We then have:

$17^S\pmod{7}\equiv3^S\pmod{7} \equiv3^5\pmod{7}\equiv5\pmod{7}$

$17^S\pmod{13}\equiv4^S\pmod{13} \equiv4^5\pmod{13}\equiv10\pmod{13}$

$17^S\pmod{19}\equiv(-2)^S\pmod{19} \equiv-2^{17}\pmod{19}\equiv9\pmod{19}$

Then applying CRT on these congruences give us $17^S \equiv 712 \pmod{1729}$ , hence our answer is $\boxed{712}$

@Calvin Lin Can you correct this question properly?