Finding the last two digits of $6543^{210}$ I have to find the last two digits of $6543^{210}$, my strategy is to use the Euler theorem and then some algebra to reduce this to $6543^{10}$, however I can't think of any easy way to proceed after this, any ideas?
 A: This might be the fastest way:  For the last two digits, you want to look modulo $100$.  Notice that your number is relatively prime to $100$, and that $\phi(100)=40$. By Euler's theorem $$6543^{210}\equiv 6543^{10}\equiv 43^{10}\pmod{100}.$$   Here we can try using repeated squaring.  $$43^2\equiv 49\pmod{100}.$$ $$43^4\equiv 49^2\equiv 1\pmod{100}.$$  Since $43^4\equiv 1$, we see that $$43^{10}\equiv 43^2\equiv 49\pmod{100}.$$
A: Binomial Theorem $\Rightarrow\rm mod\ 100\!:\ (-7\!+\!50)^{\large 2+4N}\!\equiv (-7)^{\large 2+4N}\! \equiv 7^{\large 2}\, $ by $\, 7^{\large 4}\! \equiv (-1\!+\!50)^{\large 2}\!\equiv 1$
A: One can use the binomial theorem (thrice). To do so, write $h$ for everything that is a multiple of $100$, possibly varying from line to line.
Since $6543=43+h$, one knows that
$$
6543^{210}=(43+h)^{210}=43^{210}+h.
$$
Now,
$$
43^{210}=(3+40)^{210}=3^{210}+210\cdot3^{209}\cdot40+h=3^{210}+h.
$$
Finally,
$$
3^{210}=9^{105}=(-1+10)^{105}=-1+105\cdot10+h=1049+h=49+h,
$$
that is, $6543^{210}=49+$ some multiple of $100$.
