Given a <strong>0-indexed</strong> integer array <code>nums</code> of length <code>n</code> and an integer <code>k</code>, return <em>the <strong>number of pairs</strong></em> <code>(i, j)</code> <em>where</em> <code>0 &lt;= i &lt; j &lt; n</code>, <em>such that</em> <code>nums[i] == nums[j]</code> <em>and</em> <code>(i * j)</code> <em>is divisible by</em> <code>k</code>.
<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [3,1,2,2,2,1,3], k = 2
<strong>Output:</strong> 4
<strong>Explanation:</strong>
There are 4 pairs that meet all the requirements:
- nums[0] == nums[6], and 0 * 6 == 0, which is divisible by 2.
- nums[2] == nums[3], and 2 * 3 == 6, which is divisible by 2.
- nums[2] == nums[4], and 2 * 4 == 8, which is divisible by 2.
- nums[3] == nums[4], and 3 * 4 == 12, which is divisible by 2.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,2,3,4], k = 1
<strong>Output:</strong> 0
<strong>Explanation:</strong> Since no value in nums is repeated, there are no pairs (i,j) that meet all the requirements.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 100</code></li>
	<li><code>1 &lt;= nums[i], k &lt;= 100</code></li>
</ul>

from collections import defaultdict

class Solution:
    def countPairs(self, nums: List[int], k: int) -> int:
        # Create a dictionary to store the indices of each number
        index_dict = defaultdict(list)
        for i, num in enumerate(nums):
            index_dict[num].append(i)
        
        # Initialize the count of valid pairs
        pair_count = 0
        
        # Iterate through each list of indices in the dictionary
        for indices in index_dict.values():
            n = len(indices)
            # Check all pairs of indices
            for i in range(n):
                for j in range(i + 1, n):
                    # If the product of indices is divisible by k, count the pair
                    if (indices[i] * indices[j]) % k == 0:
                        pair_count += 1
        
        return pair_count

# Example usage:
solution = Solution()
nums = [3, 1, 2, 2, 2, 1, 3]
k = 2
print(solution.countPairs(nums, k))  # Output: 4
55