Apples cost $2$ dollars. Bananas cost $3$ dollars. Oranges cost $5$ dollars. Compute the number of distinct baskets of fruit such that there are $100$ pieces of fruit and the basket costs $300$ dollars. Two baskets are distinct if and only if, for some type of fruit, the two baskets have differing amounts of that fruit.
To solve the problem, we need to find the number of distinct baskets of fruit such that there are $100$ pieces of fruit and the basket costs $300$ dollars. We are given the following costs for each type of fruit:
- Apples cost $2$ dollars each.
- Bananas cost $3$ dollars each.
- Oranges cost $5$ dollars each.

Let:
- $x$ be the number of apples,
- $y$ be the number of bananas,
- $z$ be the number of oranges.

We need to solve the system of equations:
\[
x + y + z = 100
\]
\[
2x + 3y + 5z = 300
\]

1. **Express one variable in terms of the others:**
   From the first equation, we can express $x$ in terms of $y$ and $z$:
   \[
   x = 100 - y - z
   \]

2. **Substitute $x$ into the second equation:**
   Substitute $x = 100 - y - z$ into the second equation:
   \[
   2(100 - y - z) + 3y + 5z = 300
   \]
   Simplify the equation:
   \[
   200 - 2y - 2z + 3y + 5z = 300
   \]
   Combine like terms:
   \[
   200 + y + 3z = 300
   \]
   Subtract 200 from both sides:
   \[
   y + 3z = 100
   \]

3. **Express $y$ in terms of $z$:**
   From the equation $y + 3z = 100$, we can express $y$ as:
   \[
   y = 100 - 3z
   \]

4. **Determine the range of $z$:**
   Since $y$ must be a nonnegative integer, $100 - 3z \geq 0$:
   \[
   100 \geq 3z
   \]
   \[
   z \leq \frac{100}{3}
   \]
   Since $z$ must be an integer, the maximum value of $z$ is 33. Therefore, $z$ can take any integer value from 0 to 33, inclusive.

5. **Count the number of possible values for $z$:**
   The number of possible values for $z$ is:
   \[
   33 - 0 + 1 = 34
   \]

Thus, there are 34 distinct baskets of fruit that satisfy the given conditions.

The final answer is $\boxed{34}$.