how to find mod 100 in the following case

Question

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

Well, I'd first compute the order of $3\pmod {100}$. As $\varphi(100)=40$ the only possibilities are $2,4,5,10,20,40$. — lulu, Dec 01 '16 at 16:16
See http://math.stackexchange.com/questions/2038690/least-residue-of-75-dos-50 — lab bhattacharjee, Dec 01 '16 at 16:17
@lulu, We can use http://mathworld.wolfram.com/CarmichaelFunction.html as well — lab bhattacharjee, Dec 01 '16 at 16:23

score 3 · Answer 1 · answered Dec 01 '16 at 16:16

3

HINT:

$$3^{4n+3}=3^3(10-1)^{2n}=3^3(1-10)^{2n}$$

Now $\displaystyle(1-10)^{2n}\equiv1-\binom{2n}110\pmod{100}$

answered Dec 01 '16 at 16:16

lab bhattacharjee

279,016

score 2 · Accepted Answer · answered Dec 01 '16 at 16:25

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:

$$3^{999} \equiv 3^{-1}3^{1000} \equiv 67\cdot (3^{40})^{25} \equiv 67 (\bmod{100}).$$

score 1 · Answer 3 · edited Apr 13 '17 at 12:20

${\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 $\ \ 3^{999}\equiv\dfrac{\color{#c00}{3^{1000}}}3\equiv \dfrac{\color{#c00}{\bf 1}}3\equiv \dfrac{-99}3\equiv -33$

score 1 · Answer 4 · answered Dec 02 '16 at 22:52

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.

how to find mod 100 in the following case

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