how to find mod 100 in the following case what  is the remainder when $3^{999}$ divided by 100?
Is there is any short method to find remainder ?
 i tried 
$3^{1}$ divided by 100=3
$3^{2}$ divided by 100=9
$3^{3}$ divided by 100=27
$3^{4}$ divided by 100=81
but unable to proceed further
 A: HINT:
$$3^{4n+3}=3^3(10-1)^{2n}=3^3(1-10)^{2n}$$
Now $\displaystyle(1-10)^{2n}\equiv1-\binom{2n}110\pmod{100}$
A: I would note that $3\cdot 33 \equiv 99 \equiv -1 (\bmod{100})$ and conclude that $3^{-1} \equiv -33\equiv 67 (\bmod{100}).$  Then I would compute $\phi(100) = 40$ so that I knew $3^{40} \equiv 1 (\bmod{100})$.  Now rightly equipped:
$$3^{999} \equiv 3^{-1}3^{1000} \equiv 67\cdot (3^{40})^{25} \equiv 67 (\bmod{100}).$$
A: ${\rm mod}\ {10}\!:\ \ \ 3^4\equiv 1\,\Rightarrow\, 3^{100}\!\equiv{ 1}\ $ so $\ \color{#0a0}{3^{100} = 1\! +\! 10n}$
$ {\rm mod}\ 100\!:\ \color{#c00}{3^{1000}}\!\equiv (\color{#0a0}{3^{100}})^{10}\equiv  (\color{#0a0}{1\!+\!10n})^{10}\equiv 1^{10}\equiv\color{#c00}{\bf 1}\ $ by Binomial Theorem [or this Lemma]
therefore $\ \ 3^{999}\equiv\dfrac{\color{#c00}{3^{1000}}}3\equiv  \dfrac{\color{#c00}{\bf 1}}3\equiv \dfrac{-99}3\equiv -33$
A: There are some good more-direct methods in the other answers, but let me just help you "proceed further".
Note that "$x \bmod 100$" effectively means the remainder of $x$ when divided by $100$, and $x \equiv y \bmod 100$ means that $x$ and $y$ have the same remainder when divided by $100$.
So you already have:
\begin{align}
3^1 &\equiv 3 \bmod 100\\
3^2 &\equiv 9 \bmod 100\\
3^3 &\equiv 27 \bmod 100\\
3^4 &\equiv 81 \bmod 100\\
\end{align}
but at this point we're not really finding remainders because the values are less than $100$. So the next step $3^5=243$ means we actually start doing that.
$243 \equiv 43 \bmod 100$. Now the step after that gets interesting, because we can ignore that $200$ we just threw away. Even when we multiply it by $3$ (or $9$, etc), it's still a multiple of $100$ and makes no difference to the remainder, so we can just forget it in subsequent steps:
\begin{align}
3^5 &\equiv 243 \equiv 43 \bmod 100\\
3^6 &\equiv 43\cdot 3 \equiv 129 \equiv 29 \bmod 100\\
3^7 &\equiv 29\cdot 3 \equiv 87  \bmod 100\\
3^8 &\equiv 87\cdot 3 \equiv 261 \equiv 61 \bmod 100\\
3^9 &\equiv 61\cdot 3 \equiv 183 \equiv 83 \bmod 100\\
3^{10} &\equiv 83\cdot 3 \equiv 249 \equiv 49 \bmod 100\\
\end{align}
And so on. What you will find on continuing this process is that $3^{20}\equiv 1 \bmod 100$, and this means that the remainder values cycle from there onwards, allowing look up your result from where the $999$ exponent falls in that cycle.
