Let $f(x)=ax^2+bx+c$, where $a$, $b$, and $c$ are integers. Suppose that $f(1)=0$, $50 < f(7) < 60$, $70 < f(8) < 80$, and $5000k < f(100) < 5000(k+1)$ for some integer $k$. What is $k$?
 
 $ \textbf{(A)}\ 1 \qquad
\textbf{(B)}\ 2 \qquad
\textbf{(C)}\ 3 \qquad
\textbf{(D)}\ 4 \qquad
\textbf{(E)}\ 5
$
1. Given the quadratic function \( f(x) = ax^2 + bx + c \), we know that \( f(1) = 0 \). This implies:
   \[
   a(1)^2 + b(1) + c = 0 \implies a + b + c = 0
   \]

2. We are also given the inequalities:
   \[
   50 < f(7) < 60
   \]
   \[
   70 < f(8) < 80
   \]
   \[
   5000k < f(100) < 5000(k+1)
   \]

3. Substituting \( x = 7 \) into \( f(x) \):
   \[
   f(7) = 49a + 7b + c
   \]
   Using \( a + b + c = 0 \), we can rewrite \( c \) as \( c = -a - b \):
   \[
   f(7) = 49a + 7b - a - b = 48a + 6b
   \]
   Therefore:
   \[
   50 < 48a + 6b < 60
   \]

4. Substituting \( x = 8 \) into \( f(x) \):
   \[
   f(8) = 64a + 8b + c
   \]
   Using \( c = -a - b \):
   \[
   f(8) = 64a + 8b - a - b = 63a + 7b
   \]
   Therefore:
   \[
   70 < 63a + 7b < 80
   \]

5. We now have two inequalities:
   \[
   50 < 48a + 6b < 60
   \]
   \[
   70 < 63a + 7b < 80
   \]

6. To solve these inequalities, we can test integer values for \( a \) and \( b \). Let's start with \( a = 2 \) and \( b = -7 \):
   \[
   48(2) + 6(-7) = 96 - 42 = 54
   \]
   \[
   63(2) + 7(-7) = 126 - 49 = 77
   \]
   Both values satisfy the inequalities:
   \[
   50 < 54 < 60
   \]
   \[
   70 < 77 < 80
   \]

7. Now, we need to check \( f(100) \):
   \[
   f(100) = 100^2a + 100b + c = 10000a + 100b + c
   \]
   Using \( a = 2 \), \( b = -7 \), and \( c = -a - b = -2 + 7 = 5 \):
   \[
   f(100) = 10000(2) + 100(-7) + 5 = 20000 - 700 + 5 = 19305
   \]

8. We need to find \( k \) such that:
   \[
   5000k < 19305 < 5000(k+1)
   \]
   Dividing 19305 by 5000:
   \[
   19305 / 5000 = 3.861
   \]
   Therefore, \( k = 3 \).

The final answer is \( \boxed{3} \).