I have tried to reduce it to $(6^3)^{171} \equiv 16^{171} \pmod {100}$. I am not completely familiar with Euler' theorem but I have done this:$16^{171}\equiv (16^{40})^4 \cdot16^{11}\equiv 16^{11} \pmod {100}$. How do I continue from here?
Moreover is there a more systematic method for solving this? Thanks in advance!
Use the Chinese remainder theorem: $$\mathbf Z/100\mathbf Z\simeq \mathbf Z/4\mathbf Z\times \mathbf Z/25\mathbf Z.$$ Now $6^2\equiv 0\bmod 4$, so $6^n\equiv 0$ for any $n\ge 2$.
On the other hand, $6$ has order $5\bmod 25$, so $$6^{513}\equiv 6^{513\bmod 5}=6^3\equiv 16\mod 25.$$ Also, a Bézout's relation between $4$ and $25$ is $\:25-6\cdot 4=1$, so by the inverse isomorphism, $$6^{513}\equiv0\cdot 25-16\cdot 6\cdot 4=-384\equiv 16\mod 100.$$
Hint: $6^{n+5} \equiv 6^n \bmod 100$ for $n\ge2$
Like How to find last two digits of $2^{2016}$
as $(6^m,100)=4$ for $m\ge2$
$$6^{5n+1}=(1+5)^{5n+1}\equiv1+\binom{5n+1}15\pmod{25}\equiv6$$
$$\implies6^{5m+3}\equiv6\cdot6^2\pmod{25\cdot6^2}\equiv216\pmod{100}\text{ as }100 \text{ divides }25\cdot6^2$$
Hints:
$6^n\equiv0\bmod4 $ for $n\ge2$
$6^3=216\equiv16\bmod25$
$6^5=7776\equiv1\bmod25$
Recall that
$\quad 2^9 = 512$
We define a recursive sequence as follows,
$\quad a_0 \equiv 6 \pmod{100}$
$\quad a_{n+1} \equiv {a_{n}}^2 \pmod{100}$
so that the answer we are seeking is $6 \cdot a_9 \pmod{100}$.
Without using any computing assistance (and with limited ability to square numbers)
we apply the algorithm,
$\quad a_1 \equiv 36 \pmod{100}$
$\quad a_2 \equiv (25+11)^2 \equiv 25^2 + 11 \cdot 50 + 11^2 \equiv 25 + 50 + 21 \equiv 96 \pmod{100}$
$\quad a_3 \equiv (-4)^2 \equiv 16 \pmod{100}$
$\quad a_4 \equiv 16^2 \equiv 56 \pmod{100}$
$\quad a_5 \equiv (50+6)^2 \equiv 50^2 + 6 \cdot 100 + 6^2 \equiv 36 \pmod{100}$
$\quad a_6 \equiv a_2 \pmod{100}$
$\quad a_7 \equiv a_3 \pmod{100}$
$\quad a_8 \equiv a_4 \pmod{100}$
$\quad a_9 \equiv a_5 \pmod{100}$
So the answer is $6 \cdot a_9 \equiv 6 \cdot 36 \equiv 16 \pmod{100}$.
We have
$\displaystyle {(1+5)^{512} =}$ $$ \sum_{k=0}^{512} \binom{512}{k}5^k = \sum_{k\in \{0,1,511,512\}}^{ } \binom{512}{k}5^k + \sum_{k=2}^{255} \binom{512}{k}(5^k + 5^{512-k}) + \binom{512}{256}5^{256}$$
If we apply $\text{(mod 100)}$ to the rhs we get
$$1 + 12 \cdot 5 + 12 \cdot 25 + 25 + 50\sum_{k=2}^{255} \binom{512}{k} + 25\binom{512}{256} \text{ (mod 100)}$$
As a consequence of Lucas's theorem or using Vandermonde's identity, we know that
$$ \binom{2^n}{m} \text{ is even when } 0 \lt m \lt 2^n$$
So $2$ divides $\binom{512}{k}$ for $2 \le k \le 255$.
In addition, using the theory found in
$\quad$ Middle binomial coefficient mod 4
we can now write
$$6^{513} \text{ (mod 100)} \equiv 6 \cdot (1 + 60 + 25 + 50) \text{ (mod 100)} \equiv 6 \cdot 36 \text{ (mod 100)} \equiv 16 \text{ (mod 100)} $$
