find the last two digits of $2^{250}$. Suppose we want the last two digits of $3^{250}$, one can use the theorem $a^{\phi(n)}\cong 1(\mod n)$ whenever $(3,n)=1$. But instead, if i have $2^{250}$, how do i solve this problem, because here $(2,100)\neq 1$.
notation
$(a,b)=$the gcd of $a$ and $b$
 A: As $(2^{250},100)=2^2,\left(2^{250-2},\dfrac{100}{2^2}\right)=1$
Let us find $2^{248}\pmod{25}$
As $(2,25)=1$ and $\phi(25)=20,2^{20}\equiv1\pmod{25}$ and $248\equiv8\pmod{20}$
$\implies2^{248}\equiv2^8\pmod{25}\equiv6$ as $2^8=256$
$\implies2^{248}\cdot2^2\equiv6\cdot2^2\pmod{25\cdot2^2}$ 
Done!
A: Well, the last two digits are divisible for $4$, so
$$2^{250}\equiv 4a \pmod{100}\iff 2^{248}\equiv a \pmod{25}$$
and $(2,25)=1$
A: Nowhere near as neat as Exodd's solution but I thought I'd post this because I think it gives a handy insight into the behaviour of some powers of $2$ modulo $100$.
I exploited the fact that $2^{14} = 16384 \equiv -16 = -2^4 \pmod{100}$ and $2^{16} = 65536 \equiv -64 = -2^6 \pmod{100}$, giving $2^{30} = 2^{14} \cdot 2^{16} \equiv (-2^4)\cdot (-2^6) = 2^{10} \pmod{100}$.
So $2^{250} = 2^{30\cdot 8 + 10} \equiv (2^{10})^8\cdot 2^{10} = 2^{90} = (2^{30})^3 \equiv (2^{10})^3 = 2^{30} \equiv 2^{10} = 1024 \equiv 24 \pmod{100}$.
As I said, not as elegant, and perhaps not very general, but I put it down here in case it gives the original asker or others a little more insight.
