simple c solution
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:n^2\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nint countPairs(int* nums, int n, int k){\n    int c=0;\n    for(int i=0;i<n-1;i++){\n        for(int j=i+1;j<n;j++){\n            if(nums[i]==nums[j]){\n                if((i*j)%k==0){\n                    c++;\n                }\n            }\n        }\n        \n    }\n    return c;\n\n\n}\n```
['C']