Find last two digits of a number $A=(2016^{2015^{2014}}+2014^{2015^{2016}}+2017)^{2017}$ Find last two digits of a number $A=(2016^{2015^{2014}}+2014^{2015^{2016}}+2017)^{2017}$.
I have tried to write $A=100k+r$ and find r in $A$ but I stuck at it. Any solution will be aprreciated. Thank you.

Consider the equation mod 100, then we get $(16^{2015^{2014}}+14^{2015^{2016}}+17)^{2017}$. Now $16^6 = 16 mod(100)$, and power of 14 recycle after 10 times. so we have to consider the powers of 16 mod 5 and powers of 14 mod 10 and taking considerations of the first terms. 
by $(16^5+14^5+17)^{2017}(mod100)=(76+24+17)^{2017}=17^{2017}$. Now order of 17 is 20 in mod 100 and 2017 is 17 mod 100 so $17^{17}=77$ mod 100
