Find the last three digits of $17^{102}$.

A good approach is to use the Chinese Remainder Theorem. Instead of looking at $17^{102}$ modulo $1000$, we'll look at it modulo $2^3 = 8$ and $5^3 = 125$.

Modulo $8$, this is very simple, since $17 \equiv 1 \bmod 8$. Modulo $125$, we have $\phi(125) = 100$, so $17^{102} \equiv 17^2 \equiv 39 \bmod 125$.

Now, to raise these solutions back up to a single solution modulo $1000$, we consider the integers $39 + 125s$ and check which ones are congruent to $1$ modulo $8$. We get that $38 + 125 s \equiv 30 + 125s \equiv 5(6+s) \equiv 0 \bmod 8$. So $s \equiv 2 \bmod 8$. Therefore those integers that are congruent to $39 \bmod 125$ and congruent to $1 \bmod 8$ are of the form $39 + 125(2+8s') = 289 + 1000s'$. So $17^{102} \equiv 289 \bmod 1000.$

You want to use Carmichael's theorem to see that $17^{100} \equiv 1 \mod 1000$.

Carmichael's theorem states that if $a$ and $n$ are coprime, then $a^{\lambda(n)} \equiv 1 \mod n$, where $\lambda(n)$ is the Carmichael function.

$\lambda(n)$ can be calculated recursively for $n = p_1^{a_1} p_2^{a_2} \dots p_k^{a_k}$ as $\lambda(n) = LCM( \lambda(p_1^{a_1}), \lambda(p_2^{a_2}), \dots, \lambda(p_k^{a_k})) $, where $\lambda(n) = \phi(n)$ for $n$ a power of an odd prime or $2,4$ and $\lambda(n) = \frac{1}{2} \phi(n)$ for $n = 2^k$ for $k > 2$.

For $n = 1000 = 2^3 5^3$, we get that $\lambda(n) = LCM(\frac{1}{2} \phi(2^3), \phi(5^3) ) = LCM(2,100) = 100$.

$102$ in binary is $1100110$. Denoting equality mod $1000$ by $\equiv$ one finds by successively computing the last three digits of the squares: $$17^2\equiv289, \ 17^4\equiv521, \ 17^8\equiv441, \ 17^{16}=481, \ 17^{32}\equiv361, \ 17^{64}\equiv321\ .$$ It follows that $$17^{102}\equiv17^2\cdot17^4\cdot 17^{32}\cdot 17^{64}\equiv 289\cdot 521\cdot 361\cdot 321\equiv289\ ,$$ where at each stage in the last multiplication we just compute only the last three digits.