Find the last $3$ digits of ${{7^{7}}^7}^7$.
I want to find the last $3$ digits, so I'll work with (mod $1000$).
With Euler's Theorem, $7^{\varphi(1000)} \equiv 1$(mod $1000)$.
We have $\varphi(1000)=\varphi(2^35^3)=400$. Therefore, $7^{400} \equiv 1$(mod $1000)$.
With some calculation, I find $7^{7} \equiv 343$(mod $400)$.
What do I have to do now? Thanks in advance.
Best Answer
You need $7^{7^7}$ modulo $400$. To calculate this, you can reduce $7^7$ modulo $\phi(400)=160$. This gives you $7^{7^7}\equiv 7^{23}\equiv 343\mod 400$
Hence, $7^{7^{7^7}}\equiv 7^{343}\equiv 343\mod 1000$.
So, the last three digits are $343$.