Finding the last two digits of the given number.

We need to evaluate

$$2^5+2^{{5}^{2}}+2^{{5}^{3}}+2^{{5}^{4}}... 2^{{5}^{2015}} \mod {100}$$

and

• $2^5=32 \mod {100}$
• $2^{5^2}=2^{10}2^{10}2^{5}\equiv24\cdot 24\cdot 32\equiv32 \mod {100}$
• $2^{5^3}=(2^{10}2^{10}2^{5})^5\equiv 32^5 \equiv 32\mod {100}$
• ...

therefore

$$2^5+2^{{5}^{2}}+2^{{5}^{3}}+2^{{5}^{4}}... 2^{{5}^{2015}} \equiv2015\cdot 32\equiv 80\mod {100}$$

Note that: $$02,\underbrace{04,08,16,32,64,\cdots,76,52}_{20}, \underbrace{04,08,16,32,64,\cdots,76,52}_{20},\cdots$$ To find the last two digits of $n=5,25,125,...$, you must consider: $n-1$ mod $20$: $$\begin{align}5-1\equiv4 \pmod{20} \Rightarrow 32\\ 25-1\equiv4 \pmod{20} \Rightarrow 32\\ 125-1\equiv4 \pmod{20} \Rightarrow 32\end{align}$$ Alternatively, you can consider the last two digits of $(2^5)^m=32^m:$ $$\color{red}{32};24;68;76;\color{red}{32};24;68;76;\color{red}{32};24;68;76;...$$ Since $5m=5^n \Rightarrow m=5^{n-1}=1;5;25;125;... \Rightarrow m\equiv 1 \pmod 4$, then: $$\underbrace{32+32+\cdots+32}_{2015}=32\cdot 2015=64480\equiv 80 \pmod{100}.$$