The Powers of 2014 Unleashed!

$2014^{2014}-1\equiv 0 \; \textrm{mod} \; (2014^{2014}-1)$ $\Rightarrow 2014^{2014}\equiv 1 \; \textrm{mod} \; (2014^{2014}-1)$ $\Rightarrow \left ( 2014^{2014} \right )^{1007} \equiv 1^{1007} \equiv 1 \; \textrm{mod} \; (2014^{2014}-1)$ $\Rightarrow 2014^{2028098}\equiv 1 \; \textrm{mod} \; (2014^{2014}-1)$ $\Rightarrow 2014^{2028098} \times 2014^{1007} \equiv 1 \times 2014^{1007} \; \textrm{mod} \; (2014^{2014}-1)$ $\Rightarrow 2014^{2029105} \equiv 2014^{1007} \; \textrm{mod} \; (2014^{2014}-1)$ As, $2014^{1007} < 2014^{2014}-1, \; \; 2014^{1007}$ is the desired reminder. Thus the answer is $2014+1007=\boxed{3021}$

$2014^{2029105} = 2014^{1007\times 2014 + 1007} = 2014^{1007} 2014^{1007\times 2014}$

$\quad \quad \quad \quad \quad = 2014^{1007} \left[ (2014^{2014}-1) \displaystyle \sum _{n=0} ^{1006} {2014^{2014n}}-1 \right]$

$\quad \quad \quad \quad \quad = 2014^{1007} (2014^{2014}-1) \displaystyle \sum _{n=0} ^{1006} {2014^{2014n}}-2014^{1007}$

Therefore, $\quad 2014^{2029105} \equiv 2014^{1007} \pmod {2014^{2014}}$

$\quad \quad \quad \quad \quad \Rightarrow a = 2014$ and $b = 1007\quad \Rightarrow a+b = \boxed {3021}$

Just observe a series for small nos. first

$\large{\Rightarrow (3^{ 5 })mod(3^{ 3 }-1)=(3^{ 5~ mod~ 3 })\\ \Rightarrow (4^{ 7 })~ mod~ (4^{ 4 }-1)=(4^{ 7~ mod~ 4 })\\ ..\\ ..\\ ..\\ \Rightarrow a^{ b }~ mod~ a^{ a }=a^{ b~ mod~ a }\\ \Rightarrow \{ 2014^{ 2029105 }\} ~ mod~ \{ (2014^{ 2014 })-1\} =2014^{ 2029105~ mod~ 2014 }\\ \Rightarrow i.e\quad 2014^{ 1007 }\\ \Rightarrow \boxed { a+b=3021 } }$