From $f(1)=0$, we know that $a+b+c=0$.  From the first inequality, we get $50\textless49 a+7 b+c\textless60$.  Subtracting $a+b+c=0$ from this gives us $50\textless48 a+6 b\textless60$, and thus $\frac{25}{3}\textless8 a+b\textless10$. Since $8 a+b$ must be an integer, it follows that $8 a+b=9$.  Similarly, from the second inequality, we get $70\textless64 a+8 b+c\textless80$. Again subtracting $a+b+c=0$ from this gives us $70\textless63 a+7 b\textless80$, or $10\textless9 a+b\textless\frac{80}{7}$.  It follows from this that $9 a+b=11$. We now have a system of three equations: $a+b+c=0,8 a+b=9$, and $9 a+b=11$. Solving gives us $(a, b, c)=(2,-7,5)$ and from this we find that $f(100)=2(100)^{2}-7(100)+5=19305$.  Since $15000\textless19305\textless20000 \rightarrow 5000(3)\textless19305\textless5000(4)$, we find that $k=3 \rightarrow(\mathbf{C}) 3$. 
Let $f(x)=a x^{2}+b x+c$, where $a, b$, and $c$ are integers. Suppose that $f(1)=0,50\textless f(7)\textless60,70\textless f(8)\textless80$, $5000 k\textless f(100)\textless5000(k+1)$ for some integer $k$. What is $k$? (2011 AMC 12A Problems, Question 20) 
Options:
1) $$1$$ 
2) $$2$$ 
3) $$3$$ 
4) $$4$$ 
5) $$5$$ 