12. Let $[x]$ denote the greatest integer not exceeding $x$. Find the last two digits of $\left[\frac{1}{3}\right]+\left[\frac{2}{3}\right]+\left[\frac{2^{2}}{3}\right]+\cdots+\left[\frac{2^{2014}}{3}\right]$.
(2 marks)
Let $[x]$ denote the greatest integer not exceeding $x$. Find the last two digits of $\left[\frac{1}{3}\right]+\left[\frac{2}{3}\right]+\left[\frac{2^{2}}{3}\right]+\cdots+\left[\frac{2^{2014}}{3}\right]$.
(2 marks)
12. 15
12. Note that the remainder when \(2^n\) is divided by 3 is 1 when \(n\) is even, and is 2 when \(n\) is odd. Hence \(\left[\frac{2^n}{3}\right]=\frac{2^n-1}{3}\) when \(n\) is odd, and \(\left[\frac{2^n}{3}\right]=\frac{2^n-2}{3}\) if \(n\) is even. It follows that
\[
\begin{aligned}
S & =\left[\frac{1}{3}\right]+\left[\frac{2}{3}\right]+\left[\frac{2^2}{3}\right]+\cdots+\left[\frac{2^{2014}}{3}\right] \\
& =0+\left(\frac{2-2}{3}+\frac{2^2-1}{3}\right)+\left(\frac{2^3-2}{3}+\frac{2^4-1}{3}\right) \cdots+\left(\frac{2^{2013}-2}{3}+\frac{2^{2014}-1}{3}\right) \\
& =\left(\frac{2}{3}+\frac{2^2}{3}-1\right)+\left(\frac{2^3}{3}+\frac{2^4}{3}-1\right)+\cdots+\left(\frac{2^{2013}}{3}+\frac{2^{2014}}{3}-1\right) \\
& =\left(\frac{2}{3}+\frac{2^2}{3}+\frac{2^3}{3} \cdots+\frac{2^{2014}}{3}\right)-1007 \\
& =\frac{2^{2015}-2}{3}-1007
\end{aligned}
\]

The last two digits of powers of 2 are listed as follows: \(02,04,08,16,32,64,28,56,12,24\), \(48,96,92,84,68,36,72,44,88,76,52,04,08, \ldots\) 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, say \(k=3m\). Thus the last two digits of \(S\) are the same as those of \(100m+22-7\), i.e., 15.