Remainder in polynomial division The remainder when $x^{50}$ is divided by $(x-3)(x+2)$ is of the form $ax + b$. Find the units digit of $a$.
I tried to tackle the problem using the polynomial remainder theorem but got stuck as the divisor is a quadratic expression. 

Siong Thye Goh has a good idea but we need to work $\bmod 50$.
We can compute the following modular exponentiation using the Square and Multiply Algorithm:
$$
\begin{align}
x^{50}&=(x-3)(x+2)Q(x)+ax+b\\
(-2)^{50}&\equiv24\equiv-2a+b\pmod{50}\\
3^{50}&\equiv49\equiv\phantom{-}3a+b\pmod{50}
\end{align}
$$
Therefore,
$$
25\equiv5a\pmod{50}
$$
which means that
$$
\bbox[5px,border:2px solid #C0A000]{a\equiv5\pmod{10}}
$$

Exponentiation Using The Square and Multiply Algorithm
$$
\begin{array}{}
&\bmod{50}\\
(-2)^1&\equiv-2\\
(-2)^2&\equiv4&\text{square}\\
(-2)^3&\equiv-8&\text{multiply}\\
(-2)^6&\equiv14&\text{square}\\
(-2)^{12}&\equiv-4&\text{square}\\
(-2)^{24}&\equiv16&\text{square}\\
(-2)^{25}&\equiv-32&\text{multiply}\\
(-2)^{50}&\equiv24&\text{square}
\end{array}
$$
$$
\begin{array}{}
&\bmod{50}\\
3^1&\equiv3\\
3^2&\equiv9&\text{square}\\
3^3&\equiv27&\text{multiply}\\
3^6&\equiv29&\text{square}\\
3^{12}&\equiv41&\text{square}\\
3^{24}&\equiv31&\text{square}\\
3^{25}&\equiv-7&\text{multiply}\\
3^{50}&\equiv49&\text{square}
\end{array}
$$
