What should the 'efficient' way of finding the last two digits of $2^{2016}$ be? The way I found them was by multiplying the powers of $2$ because $2016=1024+512+256+128+64+32$. I heard that one way would be with the Chinese Remainder Lemma but I don't really know how I should start?
Number Theory – How to Find Last Two Digits of $2^{2016}$
decimal-expansionelementary-number-theorymodular arithmetic
Best Answer
By brute force:
Powers of $2$ end in
$$01,02,\color{blue}{04,08,16,32,64,28,56,12,24,48,96,92,84,68,36,72,44,88,76,52},04,08,16\cdots$$ and so on with a period of $20$.
Hence $$2^{2016}\to2^{16}\to36.$$