Last 2 digits of $\displaystyle 2014^{2001}$ How to find the last 2 digits of $2014^{2001}$? What about the last 2 digits of $9^{(9^{16})}$?
 A: Finding the last two digits it equivalent to finding
$2014^{2001}\pmod{100}$.
Note $2014\equiv 14\pmod{100}$
Finding this requires using $100=25\times 4$ (coprime factors)
Is there a common solution to the system below?
$x\equiv 2014^{2001}\pmod{25}\equiv 14^{2001}\pmod{25}$
$x\equiv 2014^{2001}\pmod{4}\equiv 2^{2001}\pmod{4}$
A: Hint:
For finding last two digits you need to reduce this modulo $100$. That is you ne need to find
$$2014^{2001} \equiv ? \pmod{100}.$$
This is the  same as asking
$$14^{2001} \equiv ? \pmod{100}.$$
Now in order to facilitate computation, you need to use Euler's Theorem. But keep in mind that $\gcd(14,100) = 2.$ So you need to adjust things a bit.
A: By Euler's Theorem
$2014^{\phi(25)}=2014^{20}\equiv 1\pmod{25}\implies2014^{2000}\equiv 1\pmod{25}\implies2014^{2001}\equiv 2014\pmod{25}\implies2014^{2001}\equiv 14\pmod{25}$
Obviously $2014^{2001}\equiv 0\pmod{4}$
Let $2014^{2001}=25n+14\implies25n+14\equiv 0\pmod{4}\implies n\equiv 2\pmod{4}$
$2014^{2001}=25n+14=25(4k+2)+14=64+100k$
A: $9^{\phi(100)}=9^{40}\equiv 1\pmod{100}\implies9^{40k}\equiv 1\pmod{100}$
Now consider $9^{16}=40k+?$ , that is the same as finding $x\equiv 9^{16}\pmod{40}$
$9^{\phi(40)}=9^{16}\equiv 1\pmod{40}\implies 9^{16}=40k+1$
$9^{40k}\equiv 1\pmod{100}\implies 9^{40k+1}\equiv 9\pmod{100}$
