Find the last three digits of $17^{256}$

Question

Find the last three digits of $ 17 ^{256} $

We have to check mod $1000$ I tried to check some patterns but in vain.!

Answer 1

As $1000=8\times 125$ comptute first $17^{256}$ modulo $8$ and modulo $256$, then use the Chinese Remainder Theorem to recover $17^{256}\mod 1000$.

Modulo $8$: $\enspace 17\equiv 1\mod 8$, hence $17^{256}\equiv 1 \mod 8$.

Modulo $125$:

By Euler's theorem, $n^{\varphi(125)}\equiv 1\mod125$ for all $n$. As $\varphi(125)=100$, we have $17^{256}=17^{56} \mod 125$. More over we can check $17^{50}\equiv -1 \mod 125$, hence: $$17^{56}\equiv -17^6\equiv -69\equiv 56\mod 125.$$

$17^{256}\bmod 1000\,$ is the solution of the system of congruences: $$\begin{cases}x\equiv 1\mod 8\\x\equiv 56\mod 125\end{cases}$$ The extended euclidean algorithm yields Bézout's identity: \begin{array}[t]{c@{\qquad}r@{\qquad}r@{\qquad}c} r_i & u_i & v_i & q_i\\ \hline 125 & 1 & 0 & \\ 8 & 0 & 1 & 15\\ \hline 5 & 1 & -15 & 1 \\ 3 & -1 & 16 & 1\\ 2 & 2 & -31 & 1 \\ 1 & -3 & 47 \\ \hline \end{array} \begin{align*}&-3\times 125+47\times 8=1,\\ \text{whence}\quad x\equiv-3\times 125&\times\color{red}{1}+47\times 8 \times\color{red}{56}=20681\equiv \color{red}{681} \mod 1000.\end{align*}