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. 
 A: 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
$$
\begin{array}{}
&\bmod{50}\\
(-2)^1&\equiv-2\\
(-2)^2&\equiv4&\text{square}\\
(-2)^3&\equiv-8&\text{multiply}\\
(-2)^6&\equiv14&\text{square}\\
(-2)^{12}&\equiv-4&\text{square}\\
(-2)^{24}&\equiv16&\text{square}\\
(-2)^{25}&\equiv-32&\text{multiply}\\
(-2)^{50}&\equiv24&\text{square}
\end{array}
$$
$$
\begin{array}{}
&\bmod{50}\\
3^1&\equiv3\\
3^2&\equiv9&\text{square}\\
3^3&\equiv27&\text{multiply}\\
3^6&\equiv29&\text{square}\\
3^{12}&\equiv41&\text{square}\\
3^{24}&\equiv31&\text{square}\\
3^{25}&\equiv-7&\text{multiply}\\
3^{50}&\equiv49&\text{square}
\end{array}
$$
A: Edit: robjohn's solution is super awesome.
$$x^{50}=A(x-3)(x+2)+ax+b$$
Substitute $x=-2$ and $3$ and take $\mod 10$.
$$ (-2)^{50}= -2a+b $$
$$(3)^{50}= 3a+b $$
$$3^{50}-(-2)^{50}=5a$$
$$(3-(-2))\left( \sum_{i=0}^{49} 3^i(-2)^{49-i} \right)=5a$$
A: Hint
$x^{50}=(x-3)(x+2)q(x)+ax+b$ (Division algorithm)
Now, take $x=3$ and $x=-2$ to get two equations. Two variables and two equations. Hope you get it.
This is the complete problem. (Use the hint and try yourself first)
Taking $x=3$,
$3^{50}=3a+b,$
Taking $x=-2$
$(-2)^{50}=2^{50}=-2a+b$
Subtracting, we get
$3^{50}-2^{50}=5a\implies 9^{25}-4^{25}=5a$
Firstly, observe that the LHS is divisible by $(9-4)=5$(Why?). So, you get an integer value of $a$. (Just for a check)
$\therefore a= 9^{24}+9^{23}\cdot 4+ 9^{22}\cdot 4^2+...+4^{24}$
Now, you may use modular arithmetic. 
$a\equiv 1-4+6-4+6-4+...+6 \equiv 1+12\times 2\equiv 5\pmod{10}$ (Why?)
Hope you get it.
A: Below we solve it simply - with purely mental arithmetic. By polynomial division with remainder followed by evaluation at $\,x=3,$ and $\,x=-2\,$ we obtain
$$\begin{align} x^{50} &= (x\!-\!3)(x\!+\!2)\, q(x) + ax + b\\
\Rightarrow\quad  3^{50} &= \ \ \ 3a + b\\
 (-2)^{50} &= -2a + b\\
\Rightarrow\  \ 3^{50}\!-(-2)^{50} &=\ \ \  5a
\end{align}\ \ \ $$
But $\ 3\equiv -2\pmod{5}\,\Rightarrow\, 3^{50}\equiv (-2)^{50}\,\Rightarrow\, 5a\equiv 0\pmod{25},\,$ by Lemma below.
So ${\rm mod}\ 50\,$ either  $5a\equiv 0$ or $\,5a\equiv 25.\,$ But $\,5a = 3^{50}-(-2)^{50}\,$ is odd, so it must be $\,5a\equiv 25\pmod{50}.\,$
Hence  $\,a\equiv 5\pmod{10},\,$ by cancelling $\,5$.

Lemma $\ \ c \equiv d \pmod n\,\Rightarrow\, c^{nk} \equiv d^{nk} \pmod{n^2}$.
Proof $\ $ By hypothesis $\ c = d+nj\,$ for some integer $\,j\,$ so by the Binomial Theorem
$$  c^{nk} = (d+nj)^{nk} = d^{nk} + (\color{#c00}nk)(\color{#c00}nj) d^{nk-1} + (\color{#c00}nj)^{\color{#c00} 2}(\cdots) \equiv d^{nk}\!\! \pmod{\!\color{#c00}{n^2}}$$
Remark $ $ This method of solving for $\,a\,$ may be viewed as Lagrange (or Newton) interpolation, which is a special case of CRT = Chinese remainder theorem.
