smc

From $f(1)=0$, we know that $a+b+c=0$. From the first inequality, we get $50<49a+7b+c<60$. Subtracting $a+b+c=0$ from this gives us $50<48a+6b<60$, and thus $\frac{25}{3}<8a+b<10$. Since $8a+b$ must be an integer, it follows that $8a+b=9$. Similarly, from the second inequality, we get $70<64a+8b+c<80$. Again subtracting $a+b+c=0$ from this gives us $70<63a+7b<80$, or $10<9a+b<\frac{80}{7}$. It follows from this that $9a+b=11$. We now have a system of three equations: $a+b+c=0$, $8a+b=9$, and $9a+b=11$. 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 $15000<19305<20000\to 5000(3)<19305<5000(4)$, we find that $k=3\to \boxed{3}$.