# Power tower last digits

Can anyone solve the following problem of finding the 6th last digit from the right of the decimal representation of the following number:

$$6^{6^{6^{6^{6^{6}}}}}$$

Essentially it means reducing this modulo $$10^6$$ and supposedly Chinese Remainder Theorem should be used, but I have no idea how to solve this. Can anyone help?

We're interested in $$6^x \mod 10^6$$; this is determined by $$6^i \mod 2^6$$ and $$6^i \mod 5^6$$. Mod $$2^6$$ it's easy: $$6^x$$ is divisible by $$2^6$$ if $$x \ge 6$$. $$6$$ is coprime to $$5^6$$ with multiplicative order $$3125 = 5^5$$ mod $$5^6$$. In fact, the multiplicative order of $$6$$ mod $$5^m$$ seems to be $$5^{m-1}$$ (prove!). So \eqalign{6 &\equiv 1 \mod 5\cr 6^6 &\equiv 6^1 = 6 \mod 5^2\cr 6^{6^6} &\equiv 6^6 \equiv 31 \mod 5^3\cr 6^{6^{6^6}} &\equiv 6^{31} \equiv 531 \mod 5^4\cr 6^{6^{6^{6^6}}} &\equiv 6^{531}\equiv 1156 \mod 5^5\cr 6^{6^{6^{6^{6^6}}}} &\equiv 6^{1156} \equiv 4281 \mod 5^6\cr } and using Chinese Remainder, $$6^{6^{6^{6^{6^6}}}} \equiv 238656 \mod 10^6$$
• How did you do this mod $5^n$ without calculator? For example $6^{1156}$ Jan 13, 2019 at 21:14
• Also I am interested in how did you reach the conclusion that $6^x$ is $0$ in mod $2^6$. Jan 14, 2019 at 10:44
• $6^x = 2^x \cdot 3^x$ is divisible by $2^x$. Jan 14, 2019 at 12:47