Let $P$ be the product of the first 100 positive odd integers. Find the largest integer $k$ such that $P$ is divisible by $3^k$.
To find the largest integer \( k \) such that the product \( P \) of the first 100 positive odd integers is divisible by \( 3^k \), we need to count the number of factors of 3 in \( P \). This can be done using the de Polignac's formula (Legendre's formula), which states that the exponent of a prime \( p \) in \( n! \) is given by:

\[
e_p(n!) = \left\lfloor \frac{n}{p} \right\rfloor + \left\lfloor \frac{n}{p^2} \right\rfloor + \left\lfloor \frac{n}{p^3} \right\rfloor + \cdots
\]

However, since we are dealing with the product of the first 100 positive odd integers, we need to adjust our approach. The product of the first 100 positive odd integers can be written as:

\[
P = 1 \cdot 3 \cdot 5 \cdot 7 \cdot \ldots \cdot 199
\]

We need to count the number of factors of 3 in this product. We will count the multiples of 3, 9, 27, 81, etc., among the first 100 positive odd integers.

1. **Count the multiples of 3:**
   The sequence of the first 100 positive odd integers is \( 1, 3, 5, 7, \ldots, 199 \). Every third odd number is a multiple of 3. Therefore, the number of multiples of 3 is:

   \[
   \left\lfloor \frac{100}{3} \right\rfloor = 33
   \]

2. **Count the multiples of 9:**
   Every ninth odd number is a multiple of 9. Therefore, the number of multiples of 9 is:

   \[
   \left\lfloor \frac{100}{9} \right\rfloor = 11
   \]

3. **Count the multiples of 27:**
   Every twenty-seventh odd number is a multiple of 27. Therefore, the number of multiples of 27 is:

   \[
   \left\lfloor \frac{100}{27} \right\rfloor = 3
   \]

4. **Count the multiples of 81:**
   Every eighty-first odd number is a multiple of 81. Therefore, the number of multiples of 81 is:

   \[
   \left\lfloor \frac{100}{81} \right\rfloor = 1
   \]

5. **Count the multiples of 243:**
   Every two hundred forty-third odd number is a multiple of 243. Therefore, the number of multiples of 243 is:

   \[
   \left\lfloor \frac{100}{243} \right\rfloor = 0
   \]

Adding these counts together gives the total number of factors of 3 in \( P \):

\[
33 + 11 + 3 + 1 = 48
\]

The final answer is \(\boxed{48}\).