<s>[INST] Find the unit digit of: $\left\lfloor{10^{20000}\over 100^{100}+3}\right\rfloor$ 
Find the unit digit of:
$$\left\lfloor{10^{20000}\over 100^{100}+3}\right\rfloor$$

I am completely clueless about how to deal with such huge powers.
I noticed that the numerator is $({100^{100}})^{100}$, which brings some sort of similarity with the denominator, but I couldn't get anything useful from it, I tried to add and subtract terms in order to reduce the power:
Let $x = 100^{100}$
${x^{100}-3^{100} + 3^{100}\over x+3 }$
Now the first two terms combined are divisible by $x+3$ , but it still didn't work out to solve for the unit digit.
Could someone please teach me how to solve for the unit digit?
Thanks !
 [/INST] $$\frac{x^{100}-3^{100} + 3^{100}}{ x+3 }$$
"Now the first two terms combined are divisible by $x+3$". In fact, you have done the critical step!
$$\frac{x^{100}-3^{100} + 3^{100}}{ x+3 }= \frac{x^{100}-3^{100}}{ x+3 }+\frac{3^{100}}{ x+3 }$$
Since $\dfrac{3^{100}}{ x+3 }=\dfrac{3^{100}}{ 100^{100}+3 }<1$, we see that the unit digit wanted is the same as the unit digit of $\dfrac{x^{100}-3^{100}}{ x+3 }$.
The unit digit of $x^{100}-3^{100}$ is 9 since the unix digit of $x=100^{100}$ is $0$ and the unit digit of $3^{100}=(81)^{25}$ is $1$. The unit digit of $x+3$ is $3$.
So the unit digit of $\dfrac{x^{100}-3^{100}}{ x+3 }$ is $\dfrac{\cdots9}{\cdots3} = 3$
</s>