$100$ stones, each weighs $1$ kg or $10$ kgs or $50$ kgs, weighs $500$ kgs in total. How many values can the number of stones weighing $10$ kgs take?

$ 
\textbf{(A)}\ 0
\qquad\textbf{(B)}\ 1
\qquad\textbf{(C)}\ 2
\qquad\textbf{(D)}\ 3
\qquad\textbf{(E)}\ 4
$
To solve this problem, we need to determine how many different values the number of stones weighing 10 kg can take, given the constraints. Let's denote:
- \( x \) as the number of stones weighing 1 kg,
- \( y \) as the number of stones weighing 10 kg,
- \( z \) as the number of stones weighing 50 kg.

We are given the following conditions:
1. The total number of stones is 100:
   \[
   x + y + z = 100
   \]
2. The total weight of the stones is 500 kg:
   \[
   1x + 10y + 50z = 500
   \]

We need to find the possible values of \( y \).

Step 1: Express \( x \) in terms of \( y \) and \( z \) using the first equation:
\[
x = 100 - y - z
\]

Step 2: Substitute \( x \) into the second equation:
\[
1(100 - y - z) + 10y + 50z = 500
\]

Step 3: Simplify the equation:
\[
100 - y - z + 10y + 50z = 500
\]
\[
100 + 9y + 49z = 500
\]

Step 4: Rearrange to solve for \( y \):
\[
9y + 49z = 400
\]

Step 5: Solve for \( y \) in terms of \( z \):
\[
y = \frac{400 - 49z}{9}
\]

Step 6: Determine the integer values of \( z \) such that \( y \) is a non-negative integer:
- \( 400 - 49z \) must be divisible by 9.
- \( 400 - 49z \geq 0 \) implies \( z \leq \frac{400}{49} \approx 8.16 \), so \( z \) can be 0, 1, 2, 3, 4, 5, 6, 7, or 8.

Step 7: Check each value of \( z \) to see if \( 400 - 49z \) is divisible by 9:
- \( z = 0 \): \( 400 - 49(0) = 400 \), \( \frac{400}{9} \) is not an integer.
- \( z = 1 \): \( 400 - 49(1) = 351 \), \( \frac{351}{9} = 39 \), which is an integer.
- \( z = 2 \): \( 400 - 49(2) = 302 \), \( \frac{302}{9} \) is not an integer.
- \( z = 3 \): \( 400 - 49(3) = 253 \), \( \frac{253}{9} \) is not an integer.
- \( z = 4 \): \( 400 - 49(4) = 204 \), \( \frac{204}{9} = 22.67 \), which is not an integer.
- \( z = 5 \): \( 400 - 49(5) = 155 \), \( \frac{155}{9} \) is not an integer.
- \( z = 6 \): \( 400 - 49(6) = 106 \), \( \frac{106}{9} \) is not an integer.
- \( z = 7 \): \( 400 - 49(7) = 57 \), \( \frac{57}{9} = 6.33 \), which is not an integer.
- \( z = 8 \): \( 400 - 49(8) = 8 \), \( \frac{8}{9} \) is not an integer.

From the above, the only valid value of \( z \) is 1, which gives \( y = 39 \).

Conclusion:
The number of stones weighing 10 kg can only take one value, which is 39.

The final answer is \(\boxed{1}\)