This question is related to number theory and i am getting stuck on it it is : 2^98 when divided by 33 what is the remainder?
i am pretty sure you solve it using fermat theorem I think , you first take mod 3 and mod 11 and , make a linear equation (this is where i am getting stuck )
By Euler's totient theorem we know that as 2 and 33 are coprime so:- $$2^{\varphi(33)} \equiv 1(\text{mod }33)$$ $$\Rightarrow 2^{20} \equiv 1(\text{mod }33)$$ $$\Rightarrow 2^{80} \equiv 1(\text{mod }33)$$ Thus:- $$2^{80}\cdot2^{18} \equiv 1\cdot(2^{5})^{3}\cdot2^{3} \equiv (32)^{3}\cdot8 \equiv (-1)^{3}\cdot8 \equiv -8 \equiv 25(\text{mod }33) $$
