Find The Remainder When $100^{100}$ Is Divided By 7 Find the remainder when $100^{100}$ is divided by $7$.
I tried finding a pattern with the residues but it took a lot of time and I haven't found anything. Any answers? 
 Thanks!
 A: \begin{align}
100  &=2 \pmod 7 \\ 
 2^1 &=2 \pmod 7  \\
2^2  &=4 \pmod 7  \\ 
2^3  &=1 \pmod 7 
\end{align}

$$ 2^{100} = 2^{ 3*33+1 }={ 2 }^{ 99 }\cdot 2 = 2 \pmod 7$$

A: Note that 100 has remainder 2 when divided by 7. So the problem is equivalent to determining the remainder of $2^{100}$. Hint: $2^3=8$, which has remainder 1. So $2^{100}=(2^3)^{33}\cdot2$...
A: Using base $7$, $100_{10}=202$, and
$${202}^{202}\equiv{2}^{202}\equiv{2}^4\equiv2.$$ 

Indeed, as the base is the modulus, you can replace a number by its last digit
$$\boxed{abc}=7^2a+7b+c\equiv c$$
and by the little theorem of Fermat, you can replace a power by the sum of its digits
$$n^{\boxed{abc}}=(n^a)^{7^2}(n^b)^7n^c\equiv n^{a+b+c}.$$
A: Since $200\equiv 2\pmod{6}$ and $3$ is a generator of $\mathbb{F}_7^*$ (or, simply, by FLT) we have
$$ 100^{100} \equiv 10^{200}\equiv 3^{200}\equiv 3^{2}\equiv 9\equiv \color{red}{2}\pmod{7}.$$
A: We have $100=7\cdot 14+2=6\cdot 16+4$ Hence $$100^{100}\equiv 2^{100}\pmod7$$ Furthermore by Fermat's little theorem, $2^6\equiv 1\pmod7$ then
$$2^{100}\equiv(1)^{16}\cdot2^4\equiv2\pmod7$$
A: A simple-minded solution is to just start dividing $7$ into $100^{100}$, or $1$ followed by $200$ zeroes.
The remainders for each successive zero in the continuing long division are $3, 2, 6, 4, 5, 1$;  this sequence then repeats, since the remainder of $1$ restarts the original division.
The $33$rd cycle of these 6 remainders ends at the $198$th zero. The next two zeroes leaves the remainder at $2$.
A: $${100}^{100}=10^{200}=(7 + 3)^{200}$$
Using Binomial expansion
$$(7 + 3)^{200}=7k +3^{200} \space \text{where k is fixed integer}$$ 
