Let $P$ be the product of the first $50$ nonzero square numbers. Find the largest integer $k$ such that $7^k$ divides $P$.

[i]2018 CCA Math Bonanza Individual Round #2[/i]
1. First, we need to express \( P \) as the product of the first 50 nonzero square numbers:
   \[
   P = 1^2 \cdot 2^2 \cdot 3^2 \cdots 50^2 = (1 \cdot 2 \cdot 3 \cdots 50)^2 = (50!)^2
   \]

2. Next, we need to determine the largest integer \( k \) such that \( 7^k \) divides \( (50!)^2 \). This requires finding the highest power of 7 that divides \( 50! \).

3. To find the highest power of a prime \( p \) (in this case, \( p = 7 \)) that divides \( n! \), we use the formula:
   \[
   \sum_{i=1}^{\infty} \left\lfloor \frac{n}{p^i} \right\rfloor
   \]
   For \( n = 50 \) and \( p = 7 \):
   \[
   \left\lfloor \frac{50}{7} \right\rfloor + \left\lfloor \frac{50}{7^2} \right\rfloor + \left\lfloor \frac{50}{7^3} \right\rfloor + \cdots
   \]

4. Calculate each term:
   \[
   \left\lfloor \frac{50}{7} \right\rfloor = \left\lfloor 7.142857 \right\rfloor = 7
   \]
   \[
   \left\lfloor \frac{50}{49} \right\rfloor = \left\lfloor 1.020408 \right\rfloor = 1
   \]
   \[
   \left\lfloor \frac{50}{343} \right\rfloor = \left\lfloor 0.14577 \right\rfloor = 0
   \]
   Since \( 7^3 = 343 \) is greater than 50, all higher terms will be zero.

5. Summing these values gives the highest power of 7 in \( 50! \):
   \[
   7 + 1 = 8
   \]

6. Since \( P = (50!)^2 \), the highest power of 7 in \( (50!)^2 \) is:
   \[
   2 \times 8 = 16
   \]

Conclusion:
\[
\boxed{16}
\]