Easy Solution || -beats 100%
Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time 
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\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        \n        int pairs = 0;\n\n        for( int i = 0 ; i < nums.size()-1 ; i++ ){\n\n            for( int j  = i+1 ; j < nums.size() ; j++ ){\n\n                if ( ( ( i * j ) % k == 0 ) && ( nums[i] == nums[j]) ) pairs++; \n            }\n        }\n\n        return pairs;\n    }\n};\n```
['C++']