Find the last two digits of the given number Problem:

Find the last $2$ digits of $7^{2008}$.

Unfortunately I have no idea how to solve this problem. I know that for the last digit, we have to apply the concept of cyclicity, but I'm not aware of how to extend this to the last $2$ digits. I would be truly grateful for any help. Many thanks in advance!

Since $7$ and $100$ are coprime with respect to each other, you can use Euler's theorem:


*

*$7^{\phi(100)}\equiv1\pmod{100}$

*$\phi(100)=\phi(2^2\cdot5^2)=2\cdot4\cdot5=40$

*$7^{40}\equiv1\pmod{100}$

*$7^{2008}=7^{40\cdot50+8}=(7^{40})^{50}\cdot7^8$

*$(7^{40})^{50}\cdot7^8\equiv1^{50}\cdot7^8\equiv1\cdot5764801\equiv5764801\equiv01\pmod{100}$

