What is the largest integer $n$ that satisfies $(100^2-99^2)(99^2-98^2)\dots(3^2-2^2)(2^2-1^2)$ is divisible by $3^n$?

$ 
\textbf{(A)}\ 49
\qquad\textbf{(B)}\ 53
\qquad\textbf{(C)}\ 97
\qquad\textbf{(D)}\ 103
\qquad\textbf{(E)}\ \text{None of the above}
$
To solve the problem, we need to find the largest integer \( n \) such that the product \((100^2 - 99^2)(99^2 - 98^2) \cdots (2^2 - 1^2)\) is divisible by \( 3^n \).

1. **Simplify each term in the product:**
   \[
   k^2 - (k-1)^2 = k^2 - (k^2 - 2k + 1) = 2k - 1
   \]
   Therefore, the product can be rewritten as:
   \[
   (100^2 - 99^2)(99^2 - 98^2) \cdots (2^2 - 1^2) = (199)(197)(195) \cdots (3)(1)
   \]

2. **Identify the terms divisible by 3:**
   We need to count how many terms in the sequence \( 1, 3, 5, \ldots, 199 \) are divisible by 3. This sequence is an arithmetic sequence with the first term \( a = 1 \) and common difference \( d = 2 \).

3. **Find the number of terms in the sequence:**
   The general term of the sequence is given by:
   \[
   a_n = 1 + (n-1) \cdot 2 = 2n - 1
   \]
   Setting \( 2n - 1 = 199 \) to find the number of terms:
   \[
   2n - 1 = 199 \implies 2n = 200 \implies n = 100
   \]
   So, there are 100 terms in the sequence.

4. **Count the terms divisible by 3:**
   We need to find the number of terms \( 2k - 1 \) that are divisible by 3. This means:
   \[
   2k - 1 \equiv 0 \pmod{3} \implies 2k \equiv 1 \pmod{3} \implies k \equiv 2 \pmod{3}
   \]
   Therefore, \( k \) can be \( 2, 5, 8, \ldots \).

5. **Find the number of such \( k \) values:**
   The sequence \( 2, 5, 8, \ldots \) is an arithmetic sequence with the first term \( a = 2 \) and common difference \( d = 3 \). The general term is:
   \[
   k_m = 2 + (m-1) \cdot 3 = 3m - 1
   \]
   Setting \( 3m - 1 \leq 100 \):
   \[
   3m - 1 \leq 100 \implies 3m \leq 101 \implies m \leq \frac{101}{3} \implies m \leq 33.67
   \]
   So, \( m \) can be at most 33. Therefore, there are 33 terms in the sequence \( 1, 3, 5, \ldots, 199 \) that are divisible by 3.

6. **Count the powers of 3 in the product:**
   Each of these 33 terms contributes at least one factor of 3. Additionally, we need to check for higher powers of 3. For each term \( 3k \), we need to check if it is divisible by \( 3^2 \), \( 3^3 \), etc.

   - For \( 3^2 \) (i.e., 9), the terms are \( 9, 27, 45, \ldots \). This sequence has the first term \( 9 \) and common difference \( 18 \). The number of terms is:
     \[
     9 + (m-1) \cdot 18 \leq 199 \implies 18m - 9 \leq 199 \implies 18m \leq 208 \implies m \leq \frac{208}{18} \implies m \leq 11.56
     \]
     So, there are 11 terms divisible by \( 9 \).

   - For \( 3^3 \) (i.e., 27), the terms are \( 27, 81, 135, \ldots \). This sequence has the first term \( 27 \) and common difference \( 54 \). The number of terms is:
     \[
     27 + (m-1) \cdot 54 \leq 199 \implies 54m - 27 \leq 199 \implies 54m \leq 226 \implies m \leq \frac{226}{54} \implies m \leq 4.18
     \]
     So, there are 4 terms divisible by \( 27 \).

   - For \( 3^4 \) (i.e., 81), the terms are \( 81, 243, \ldots \). This sequence has the first term \( 81 \) and common difference \( 162 \). The number of terms is:
     \[
     81 + (m-1) \cdot 162 \leq 199 \implies 162m - 81 \leq 199 \implies 162m \leq 280 \implies m \leq \frac{280}{162} \implies m \leq 1.73
     \]
     So, there is 1 term divisible by \( 81 \).

7. **Sum the contributions:**
   \[
   33 + 11 + 4 + 1 = 49
   \]

Therefore, the largest integer \( n \) such that the product is divisible by \( 3^n \) is \( 49 \).

The final answer is \( \boxed{49} \)