How to find last two digits of $2^{2016}$

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 $$2^{2016}\to2^{16}\to36.$$


Essentially we need $2^{2016}\pmod{100}$

As $(2^{2016},100)=4$

let us find $2^{2016-2}\pmod{100/4}$

Now as $2^{10}\equiv-1\pmod{25}$

$2^{2014}=2^{201\cdot10+4}=(2^{10})^{201}\cdot2^4\equiv(-1)^{201}\cdot2^4\equiv9\pmod{25}$

$$\implies2^2\cdot2^{2014}\equiv2^2\cdot9\pmod{2^2\cdot25}$$


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This is an $\ds{\ul{old\ school}}$ proof:

  1. Since $\ds{2^{2016} = \pars{2^{4}}^{2016/4} = 16^{504}}$, it's obvious the $\ds{\ul{2^{2016}\ last\ digit}}$ is $\ds{\color{#f00}{\large 6}}$.

    Namely, the last digit of any $n^{\mathrm{th}}$-powers $\ds{\pars{n = 1,2,3,\ldots}}$ of $\ds{\ 16}$ is $\ds{\color{#f00}{6}}$.
  2. Then, $\ds{{2^{2016} - 6 \over 10}= {16^{504} - 6 \over 10}}$ is an $\ds{\ul{integer}}$ and its last digit is the digit before the $\ds{2^{2016}}$ last digit: \begin{align} \fbox{$\ds{\ {16^{504} - 6 \over 10}\ }$} & = {\pars{16^{504} - 16} + 10 \over 10} = {16\pars{16^{503} - 1} \over 10} + 1 = {16\times 15 \over 10}\,{16^{503} - 1 \over 16 - 1} + 1 \\[3mm] & = \fbox{$\ds{\ 24\sum_{n = 0}^{502}16^{n} + 1\ }$}\tag{1} \end{align} The above sum last digit is the last digit of $\ds{\pars{1 + 6\times 502} = 301\ul{3}.\ }$ The last digit of $\ds{2\ul{4} \times 301\ul{3}}$ is $\ds{\ul{2}}$ such that the last digit of $\ds{\pars{1}}$ is $\ds{\pars{\ul{2} + \ul{1} = \color{#f00}{\large 3}}}$
  3. Then, $\ds{2^{2016}\ \ul{last\ two\ digits}\ \mbox{is}\ \color{#f00}{\large 36}}$.