TITLE: Last digit of $3^{459}$.
QUESTION [8 upvotes]: I am supposed to find the last digit of the number $3^{459}$. Wolfram|Alpha gives me $9969099171305981944912884263593843734515811805621702621829350243852275145577745\\3002132202129141323227530694911974823395497057366360402382950449104721755086093\\572099218479513977932448616356300654729978057481366551670706\color{red}{\mathbf{7}}$
Surely there's some sort of numerical trick to doing this. I thought maybe modular arithmetic was involved? Any ideas on how to approach this problem?

REPLY [8 votes]: Try out the first few powers of $3$: we have that 
$$
3^1=3, 3^2=9, 3^3=27, 3^4=81, 3^5=243, 3^6=729,\ldots
$$
It seems like the final digit cycles in a pattern, namely
$$
3\to9\to7\to1\to3\to9\to7\to1,
$$
of length $4$. Since
$$
459=4\cdot114+3,
$$
the final digit is the third in the cycle, namely $\color{red}{\mathbf{7}}$, and, sure enough, your Wolfram|Alpha computation confirms this.

REPLY [6 votes]: $$3^{4}=1\pmod{10}\implies (3^{4})^{114}=1 \pmod{10}$$ So $$3^{459}=3^{4 \times 114}\cdot 3^{3}=3^{3}\pmod{10}=7\pmod {10}$$