4-digit 5's complement of a negative number
Let me demystify $b$'s complement. 

Let $b \geq 2$ be an integer. Given a $d$-digit integer $x$ is base $b$, we want to find a $d$-digit integer $y$ such that $y \equiv -x \pmod{b^d}$. Why is this useful? Suppose that $z$ is also a $d$-digit integer, and assume that $z-x \geq 0$. Then $z-x \equiv z+y \pmod{b^d}$, and so if we add $z$ and $y$ and ignore the carry, we will get $z-x$ to achieve subtraction.

How do we find $y$, the $b$'s complement of $x$? We take $y =k\times b^d-x= b^d-x = (b^d-1-x)+1$ (with $k\in \mathbb{Z}$, let $k = 1$). Now the representation of $b^d-1$ in base $b$ consists of $d$ digits of $b-1$, so $b^d-1-x$ is obtained by inverting all the digits of $x$ (subtracting them from $b-1$). This is $b$'s complement: the $b$'s complement $y$ of $x$ is obtained by inverting all digits and adding 1.