Serie(ou)s Sum

$4^{5^6}$ is $0$ mod $8$ and $-1$ mod $5$ , hence $24$ mod $40$ .

$3^{24}$ is $3^{24} \equiv 1$ mod $4$ , and $3^{24} \equiv 3^4 \equiv 6$ mod $25$ , so it's $81$ mod $100$ . Since $\phi(100) = 40$ , $3^{4^{5^6}} \equiv 3^{24} \equiv 81 \, ({\rm mod} \, 100).$

Finally, we want $2^{3^{4^{5^6}}}$ mod $1000$ . This is $0$ mod $8$ , and it's congruent to $2^{81}$ mod $125$ by the above computations, since $\phi(125) = 100$ .

Some easy computations yield $2^{81} \equiv 102$ mod $125$ , so we get a final answer of $\fbox{352}$ by the Chinese Remainder Theorem.

I am not quite familiar with Euler's totient function and Chinese Remainder Theorem but this is pretty much the way I solved (with all due respect to sir Patrick Corn):

$5^6 \equiv 1 \pmod{2} \rightarrow 4^{5^6}=4^{2k+1} \equiv 0 \pmod{4}$

$\rightarrow 3^{4^{5^{6}}} = 81k \rightarrow 2^{3^{4^{5^6}}} =2^{81k}$

$=A=({2^{10} \times 2^{10} \times 2^{10} \times 2^{10} \times 2^{10} \times 2^{10} \times 2^{10} \times 2^{10} \times 2})^k$

$\rightarrow A \equiv 24^4 \times 24^4 \times 2 \equiv 776 \times 776 \times 2 \equiv (352) \pmod{1000}$

I know my solution is not even close as elegant to sir Patrick's solution but I just wanted to share other solutions.Peace out!