Hong Kong Preliminary Selection Contest

Note that the remainder when $2^n$ is divided by $3$ is $1$ when $n$ is even, and $2$ when $n$ is odd. Hence $\left[\frac{2^n}{3}\right]=\frac{2^n-1}{3}$ when $n$ is even, and $\left[\frac{2^n}{3}\right]=\frac{2^n-2}{3}$ when $n$ is odd. It follows that $$\begin{array}{rl}S& =\left[\frac{1}{3}\right]+\left[\frac{2}{3}\right]+\left[\frac{2^2}{3}\right]+\cdots +\left[\frac{2^{2014}}{3}\right]\\ & =0+\left(\frac{2}{3}-\frac{2}{3}+\frac{2^2}{3}-\frac{2}{3}\right)+\left(\frac{2^3}{3}-\frac{2}{3}+\frac{2^4}{3}-\frac{2}{3}\right)+\cdots +\left(\frac{2^{2013}}{3}-\frac{2}{3}+\frac{2^{2014}}{3}-\frac{2}{3}\right)\\ & =\left(\frac{2}{3}+\frac{2^2}{3}-1\right)+\left(\frac{2^3}{3}+\frac{2^4}{3}-1\right)+\cdots +\left(\frac{2^{2013}}{3}+\frac{2^{2014}}{3}-1\right)\\ & =\left(\frac{2}{3}+\frac{2^2}{3}+\frac{2^3}{3}+\cdots +\frac{2^{2014}}{3}\right)-1007\\ & =\frac{2^{2015}-2}{3}-1007\end{array}$$ The last two digits of powers of $2$ are listed as follows: 02, 04, 08, 16, 32, 64, 28, 56, 12, 24, 48, 96, 92, 84, 68, 36, 72, 44, 88, 76, 52, 04, 08, ... The pattern repeats when the exponent is increased by $20$. So the last two digits of $2^{2015}$ are the same as those of $2^{15}$, i.e. $68$. Now write $2^{2015}-2=100k+66$. Since $\frac{2^{2015}-2}{3}$ is an integer, $k$ is a multiple of $3$, and so we write $k=3m$. Thus the last two digits of $S$ are the same as those of $100m+22-7$, i.e. $15$.