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

Question

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.

Answer 1

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 $14^{14^{14}}\equiv 36\pmod{100}$.

Answer 2

Hints:

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

Answer 3

Review the theory and techniques found here.

We write as true (no calculator is necessary),

$\; 14^{14} \equiv (14^2)^7 \equiv 16 \times 16^2 \times 16^2 \times 16^2 \equiv 16 \times 16 \times 16 \times 16 \equiv 16 \pmod{20}$

So (again using mental calculations),

$\; \displaystyle{ {14}^{{14}^{14}} \equiv 76 \times {14}^{16} \equiv 76 \times ({14}^{2})^8 \equiv 76 \times ({96}^{2})^4 \equiv ({16}^{2})^2 \equiv 36 \pmod{100}}$