<s>[INST] 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.
 [/INST] 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. 
$2016 = 20(100)+16 \equiv -3(8)+16 \equiv -8 \ \Rightarrow\ 2014\equiv 2016-2\equiv -10$
$2016^{32}\equiv (-8)^{32} \color{#c00}\equiv 8^{10}\equiv \color{#0a0}{2^{30}}\color{#c00}\equiv  2^8 = (2^4)^2 = (-7)^2 = 3$
$2014^{16}\equiv (-10)^{16} = 100^8 \equiv 8^8\equiv  2^{24}\color{#c00}\equiv 2^2\equiv 4$
$(2016^{32}\! + 2014^{16})^9\equiv (3\!+\!4)^9\equiv (-16)^9 \equiv -(2^4)^9\equiv -\color{#0a0}{2^{30}} 2^6 \equiv -3(-5) \equiv 15$
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