<s>[INST] To find a remainder, when divided by $101$ 
There is a large number given in an exam. It is said to find a remainder when it is divided by $101$. The number is
  $$1+11^{{(1+11+11^2+11^3+...+11^{100}~~)}^{(11^{101}~~+11^{102}~~+\cdots+11^{1000}~~)}}$$

We know $11$ and $101$ are itself prime then $(101,11)=1$ and $\phi(101)=100$    
Now,
$$1+11+11^2+11^3+....+11^{100}\equiv r\mod 100 \\ \Rightarrow1+11+21+21*11+21^2+21^2*11+.....+(21^2)^{50}\equiv r \mod 100 \\ \Rightarrow (1+11+21+31+41+51+61+71+81-9)*10 \equiv r\mod 100 \\ \Rightarrow 0\equiv r\mod 100 \\ \therefore 0^{{(11^{101}+11^{102}+.....+11^{1000}~~)}}\equiv0 \mod 100 \\ \therefore 1+11^0\equiv1+1\equiv2 \mod 101$$ 
Is it correct? Any help is appreciated.
 [/INST] Let's show $1+11+11^2+11^3+....+11^{100} \equiv 1 \mod 100$.
Easy to see that $11^{101}-1$ is divisible by $10$.
Not trivial, but not that hard to work that $11^{10} \equiv 1 \mod 100$.
With above two results we have
$$10(1+11+11^2+11^3+....+11^{100}) = 11^{101}-1 \equiv 11-1  \equiv 10 \mod 100$$
This yields $$1+11+11^2+11^3+....+11^{100} \equiv 1 \mod 100$$

$$1+11^{{(1+11+11^2+11^3+...+11^{100})}^{(11^{101}+11^{102}+.....+11^{1000})}} \\ \equiv 1 + 11^{1\mod 100} \\ \equiv 1 + 11  \\ \equiv 12 \mod 101$$
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