How can I find the last digit of $17^{68}$ and the last both digits of $14^{200}$? I have to compute the last digit of $17^{68}$ and the last both digits of $14^{200}$.
For the first exercise, I get
$$17^4=...1 \Rightarrow  17^{68}=(17^4)^{17} = (...1)^{17}=\ .... 1.$$
For the last exercise, I don't know how I can do it. 
 A: $100=4\times25$ and $gcd(4,25)=1$
$14^{200}\equiv2^{200}\times7^{200}\equiv0$ mod $4$ (since $4=2^2$)
$14^{200}\equiv(-11)^{200}\equiv11^{200}\equiv121^{100}\equiv(-4)^{100}\equiv4^{100}\equiv(-1)^{20}\equiv1$ mod $25$
(since $4^5=1024\equiv-1$ mod $25$)
So we have, for some integer $k$ :
$14^{200}=1+25k\equiv k+1\equiv0$ mod $4$
Therefore, for some integer $k'$ such that $k=-1+4k'$:
$14^{200}=1+25k=1+25\times(4k'-1)=-24+100k'\equiv 76$ mod $100$
Edit : your answer is okay for the first exercise, by the way.
A: Hint: For the first use modulo 10. For the second use modulo $100$.
Modulo $10$ means the remainder after a division with 10. 
E.g. $$17 \equiv 7 \pmod {10}, 17^2\equiv49\pmod {10}=-1 \pmod {10}$$ and$$ 17^{68}=(-1)^{34}\pmod {10}\equiv1\pmod {10}$$
A: Hint for the second one:
$$14^2 = 196 = -4 \text{ mod } 100$$
Then successive powers of $-4$ mod $100$ are $-4, 16, -64, 56, -24, 96$.  (Aha!)
A: \begin{align}
  & {{17}^{\varphi (10)}}={{17}^{4}}\overset{10}{\mathop{\equiv }}\,1\,\,\,\Rightarrow \,\,\,{{7}^{68}}\overset{10}{\mathop{\equiv }}\,1 \\ 
 & {{2}^{22}}\overset{100}{\mathop{\equiv }}\,4\,\,\,\,\Rightarrow \,\,\,\,{{2}^{198}}\overset{100}{\mathop{\equiv }}\,{{4}^{9}}\overset{100}{\mathop{\equiv }}\,44\,\,\,\,\,\Rightarrow \,\,\,\,{{2}^{200}}\overset{100}{\mathop{\equiv }}\,76 \\ 
 & {{14}^{2}}\overset{100}{\mathop{\equiv }}\,4\,\,\,\,\Rightarrow \,\,\,\,{{14}^{200}}\overset{100}{\mathop{\equiv }}\,{{2}^{200}}\overset{100}{\mathop{\equiv }}\,76 \\ 
\end{align}
A: You should start learning a bit of modular arithmetic to solve this kind of problems. To calculate the n last numbers of a p number we just calculate the:  $$ p \equiv {x}\ mod\ 10^n $$
In this case we got:
$$ 17^{68} \equiv 7^{68} \equiv (7^2)^{34} \equiv (-1)^{34} \equiv 1\ mod\ 10$$
The other case:
$$14^{200} \equiv (14^2)^{100} \equiv (-4)^{100} \equiv (-24)^{20} \equiv 76^5 \equiv 76\ mod\ 100$$
