Remainder of division by $23$ Find
$$(2014^{16}+2016^{32})^9 \ \ \ (\text{mod }23)$$
I know that 
$$2014=2\times 19 \times 53$$
$$2016=2^5\times 3^2\times 7$$
This means that $2014 \equiv 2 \times (-4)\times (-3)$. I've tried solving this in multiple ways but I can't get the same answer as Wolframalpha, which is $(2014^{16}+2016^{32})^9 \equiv 15$ (mod $23$).
I could use some help.
 A: First of all, $2014 = 87\cdot 23 + 13 \equiv 13$ so you have miscalculated there. Then $2016 = 2014 + 2 \equiv 15$. Now use square-and-multiply.
Check your results here:
$$13^{16} \equiv 4 \pmod{23}\\
15^{32} \equiv 3 \pmod{23}\\
7^9 \equiv 15 \pmod{23}$$
A: $2014\equiv 13\pmod{23}$.  So $2014^{16}\equiv 13^{16}\pmod{23}$.  
Similarly, $2016^{32}\equiv 15^{32}\pmod{23}$.
You can evaluate these by successive squaring, reducing $\pmod{23}$ as you go.  (I.e. for instance do $15^2, 15^4, 15^8, 15^{16}, 15^{32}$, where each is the square of its predecessor).  
Once you've evaluated the parentheses, you should be able to finish it off.
A: This can be done simply and quickly with purely mental arithmetic mod $\,23.\,$
We transform powers to powers of $\,2\,$ and use $\,\color{#c00}{a^{22}\equiv 1}\,$ if $\,23\nmid a\,$ by little Fermat. 
$2016 = 20(100)+16 \equiv -3(8)+16 \equiv -8 \ \Rightarrow\ 2014\equiv 2016-2\equiv -10$
$2016^{32}\equiv (-8)^{32} \color{#c00}\equiv 8^{10}\equiv \color{#0a0}{2^{30}}\color{#c00}\equiv  2^8 = (2^4)^2 = (-7)^2 = 3$
$2014^{16}\equiv (-10)^{16} = 100^8 \equiv 8^8\equiv  2^{24}\color{#c00}\equiv 2^2\equiv 4$
$(2016^{32}\! + 2014^{16})^9\equiv (3\!+\!4)^9\equiv (-16)^9 \equiv -(2^4)^9\equiv -\color{#0a0}{2^{30}} 2^6 \equiv -3(-5) \equiv 15$
