# Last Two digits of ${14}^{{14}^{14}}$

How to calculate the last two digits of ${14}^{{14}^{14}}$? With the help of any method. I have tried and have got the last digit to be $6$. But not sure.

• you have to find x $\equiv$ 14^14^14 (mod 100). And you are correct the last digit is 6, but whats the second last digit? – stackErr May 6 '14 at 11:55

Clearly, $14^{14^{14}}$ is a multiple of $4$. To compute $14^n\pmod{25}$ we should know $n\pmod {\phi(25)}$, i.e. $14^{14}\pmod{20}$. Again, $14^{14}$ is a multiple of $4$, and it is $\equiv (-1)^{14}\equiv 1\pmod 5$. Hence $14^{14}\equiv 16\pmod {20}$. Thus $14^{14^{14}}\equiv 14^{16}\pmod {25}$. This can me computed by repeated squareing: $$14^{16}=(14^2)^8=196^8\equiv (-4)^8=16^4=256^2\equiv 6^2=36\pmod{25}.$$ Since $36$ is already a multiple of $4$, we have immediately that $14^{14^{14}}\equiv 36\pmod{100}$.

Hints:

1. Work modulo $100$.
2. Split $14$ into $7\times 2$.
3. Note that $7^2\times2=98=-2\pmod{100}$.

Using twice the mod Distributive Law $$\ ab\bmod ac = a(b\bmod c)\$$ we have

$$\!\bmod 25\!:\,\ 14^{\large\color{#0a0}{10}}\! \equiv\!\!\overbrace{(8^{\large 2})^{\large 10}\! \equiv 1}^{\large\ \ \ 2\cdot 10\ =\ \phi(25)}\!\!,\$$ & $$\, \ \color{#c00}{14^{\large 14}}\, \bmod\,\color{#0a0}{10}\ =\ 2\!\!\!\!\!\!\overbrace{(14^{\large 14}/2 \bmod 5)}^{\ \ \ \ \ \large (-1)^{\LARGE 14}/2\ \equiv\ 1/2\ \equiv\ 6/2}\!\!\!\!\!\!\! =\color{#c00}6$$

therefore $$\ \,14^{\large 14^{\LARGE 14}}\!\!\bmod 100 = 4\, (14^{\large\color{#c00}{14^{\LARGE 14}}}\!\!/4 \bmod 25) =\! \underbrace{4\,(14^{\large\color{#c00}6}/4\ \bmod\, 25)\, =\, 36}_{\large\ (-11)^{\LARGE \color{#c00}6}\equiv\ (121)^{\LARGE 3} \equiv\ (-4)^{\LARGE 3} \equiv\ 36 \ }$$

Or $$\bmod 25\!:\ (-1\!+\!15)^{\large 14^{\LARGE 14}}\overset{\rm\color{#c00}{BT}}\equiv 1- 15\,(\!\!\underbrace{14^{\large 14}}_{\large (-1)^{\LARGE 14}\ \ }\!\!\!\bmod 5)\equiv 36\$$ via $$\,\rm\color{#c00}{BT} =$$ Binomial Theoem