Problem (Algebra):
Let $P(x)=x^{2}+T x+800$, and let $r_{1}$ and $r_{2}$ be the roots of $P(x)$. The polynomial $Q(x)$ is quadratic, it has leading coefficient 1, and it has roots $r_{1}+1$ and $r_{2}+1$. Find the sum of the coefficients of $Q(x)$.

Solution:
Let $Q(x)=x^{2}+A x+B$. Then $A=-\left(r_{1}+1+r_{2}+1\right)$ and $B=\left(r_{1}+1\right)\left(r_{2}+1\right)$. Thus the sum of the coefficients of $Q(x)$ is $1+\left(-r_{1}-r_{2}-2\right)+\left(r_{1} r_{2}+r_{1}+r_{2}+1\right)=r_{1} r_{2}$. Note that $T=-\left(r_{1}+r_{2}\right)$ and $800=r_{1} r_{2}$, so the answer is $\mathbf{8 0 0}$ (independent of $T$ ). [Note: With $T=108,\left\{r_{1}, r_{2}\right\}=\{-8,-100\}$.

Final answer: 800