Simple solution || beats 98%|| 3ms.
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<!-- 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```\nclass Solution {\n    public int countPairs(int[] nums, int k) {\n        int c=0;\n        for(int i=0;i<nums.length;i++)\n        {\n            for(int j=i+1;j<nums.length;j++)\n            {\n\n                if(nums[i]==nums[j] && (i*j)%k==0)\n                    c++;\n            }\n        }\n        return c;\n    }\n}\n```
['Java']