C++ || Brute Force ✔✔
Complexity\n- Time complexity: O(n^2)\n Add your time complexity here, e.g. O(n) \n\n- Space complexity: O(1)\n Add your space complexity here, e.g. O(n) \n\n# 
# Complexity\n- Time complexity: $$O(n^2)$$\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: $$O(1)$$\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\npublic:\n    int countPairs(vector<int>& nums, int k) {\n        int ans=0;\n        for (int i=0; i<nums.size(); i++){\n            for (int j=i+1; j<nums.size(); j++){\n                if (nums[i]==nums[j] && (i*j)%k==0)ans++;\n            }\n        }\n        return ans;\n    }\n};\n```
['Array', 'Greedy', 'C++']