The average of the numbers $ 1,2,3,...,98,99$, and $ x$ is $ 100x$.  What is $ x$?

$ \textbf{(A)}\ \frac{49}{101} \qquad\textbf{(B)}\ \frac{50}{101} \qquad\textbf{(C)}\ \frac12 \qquad\textbf{(D)}\ \frac{51}{101} \qquad\textbf{(E)}\ \frac{50}{99}$
1. First, we need to find the sum of the numbers from 1 to 99. The sum of the first \( n \) natural numbers is given by the formula:
   \[
   S = \frac{n(n+1)}{2}
   \]
   For \( n = 99 \):
   \[
   S = \frac{99 \cdot 100}{2} = 4950
   \]

2. Let the unknown number be \( x \). The average of the numbers \( 1, 2, 3, \ldots, 99, x \) is given to be \( 100x \). The total number of terms is \( 100 \) (99 numbers plus \( x \)).

3. The average of these 100 numbers can be expressed as:
   \[
   \text{Average} = \frac{\text{Sum of all numbers}}{\text{Number of terms}}
   \]
   Therefore:
   \[
   100x = \frac{4950 + x}{100}
   \]

4. To solve for \( x \), we multiply both sides of the equation by 100 to clear the fraction:
   \[
   100 \cdot 100x = 4950 + x
   \]
   \[
   10000x = 4950 + x
   \]

5. Next, we isolate \( x \) by subtracting \( x \) from both sides:
   \[
   10000x - x = 4950
   \]
   \[
   9999x = 4950
   \]

6. Finally, we solve for \( x \) by dividing both sides by 9999:
   \[
   x = \frac{4950}{9999}
   \]

7. Simplify the fraction by dividing the numerator and the denominator by their greatest common divisor (GCD), which is 50:
   \[
   x = \frac{4950 \div 50}{9999 \div 50} = \frac{99}{199.98} \approx \frac{50}{101}
   \]

Thus, the value of \( x \) is:
\[
x = \frac{50}{101}
\]

The final answer is \(\boxed{\frac{50}{101}}\)