Group Event 8
Given that the equation $x^{2}+(m+1) x-2=0$ has 2 integral roots $(\alpha+1)$ and $(\beta+1)$ with $\alpha<\beta$ and $m \neq 0$. Let $d=\beta-\alpha$.
G8.1 Find the value of $m$.
G8.2 Find the value of $d$.
$$
\begin{array}{l}
(\alpha+1)(\beta+1)=-2 \\
\Rightarrow \alpha+1=-1, \beta+1=2 \text { or } \alpha+1=-2, \beta+1=1 \\
\Rightarrow(\alpha, \beta)=(-2,1),(-3,0)
\end{array}
$$

When $(\alpha, \beta)=(-3,0)$, sum of roots $=(\alpha+1)+(\beta+1)=-(m+1) \Rightarrow m=0$ (rejected)
When $(\alpha, \beta)=(-2,1)$, sum of roots $=(\alpha+1)+(\beta+1)=-(m+1) \Rightarrow m=-2$
$$
d=\beta-\alpha=1-(-2)=3
$$

Let $n$ be the total number of integers between 1 and 2000 such that each of them gives a remainder of 1 when it is divided by 3 or 7 .