Remainder when $11^{2402}$ is divided by $3000$? What is the remainder when $11^{2402}$ is divided by $3000$?
I just came across this question. I am a beginner in number theory. Your help would mean a lot.Thanks!! 
 A: Use Euler theorem
$$x^{\varphi(n)}\equiv 1 \pmod n$$
in this case $\varphi(3000)=800$ so 
$$11^{2402}\equiv 11^2 \pmod{3000}$$ 
A: As $(3,1000)=1$
$$11^{2402}=(1+10)^{2402}\equiv1+\binom{2402}110^1+\binom{2402}210^2\pmod{1000}$$
$$\equiv1+2\cdot10+\frac{2402\cdot2401}2100\equiv1+20+100$$
$$\implies11^{2402}\equiv121\pmod{1000}\ \ \ \ (1)$$
$$11\equiv-1\pmod3\implies11^{2402}\equiv(-1)^{2402}\equiv1\ \ \ \ (2)$$
$$\implies11^{2402}\equiv1\equiv121\pmod3\ \ \ \ (3)$$
Combining $(1),(3)$ we get $$\implies11^{2402}\equiv121\pmod{3\cdot1000}$$
or apply CRT on $(1),(2)$
A: If you don't know Euler/Carmichael's theorem then you can instead use the Binomial theorem.   
First $\,{\rm mod}\ 1000\!:\ 11^n = (1+10)^n \equiv 1 + 10 n + 100 n(n\!-\!1)/2,\ $ which is $\equiv 1$ if $\,1000$ divides the last two terms, e.g. if $\,100\mid n.\,$ Therefore $\,11^{100}\equiv 1.\,$ Next $\,{\rm mod}\ 3\!:\ 11^2\equiv 2^2\equiv 1,\,$ so $\,11^{2n}\equiv 1.\,$   
So $\, 3,1000\mid 11^{100}\!-1\,\Rightarrow\, 3000\mid \color{#0a0}{11^{100}}\!-\color{#c00}1,\,$ so ${\rm mod}\ 3000\!:\ 11^{2402} = 11^2(\color{#0a0}{11^{100}})^{24} \equiv 11^2 \color{#c00}1^{24}\equiv 11^2$
A: Usage of Carmichael function is often more beneficial for composite numbers  as $$\lambda(3000)=\cdots=100\implies 11^{100}\equiv1\pmod{3000}$$
and $\displaystyle2402\equiv2\pmod{100}$
