Find the least nonnegative residue of $3^{1442}$ mod 700 So I have that $700=7\cdot2^2\cdot5^2$ and I got that $3^2\equiv1\pmod2$ so then $3^{1442}\equiv1\pmod2$ also $3^2\equiv1\pmod{2^2}$ so $3^{1442}\equiv1\pmod{2^2}$ which covers one of the divisors of $700$. Im not sure if I'm supposed to use $2$ or $2^2$ and I was able to find that $3^2\equiv-1\pmod5$ so $3^{1442}\equiv-1\pmod5$, For mod $7$ I wasn't able to come up with an answer in a way like the other two, and I'm not really sure how to do this to find the least non negative residue 

Go $\pmod4$, $\pmod7$ and $\pmod{25}$. We have
\begin{align}
3^2 \equiv 1\pmod4\\
3^6 \equiv 1\pmod7\\
3^{20} \equiv 1\pmod{25}
\end{align}
This gives us that
\begin{align}
3^{60} \equiv 1\pmod4\\
3^{60} \equiv 1\pmod7\\
3^{60} \equiv 1\pmod{25}
\end{align}
This means
$$3^{60} \equiv 1\pmod{700}$$
Note that $3^{1442} = 3^{24\cdot60+2} = \left(3^{60}\right)^{24} \cdot 3^2$. Hence, we obtain
$$3^{1442} \equiv 3^2\pmod{700} \equiv9\pmod{700}$$
