FASTEST way with java(beats 99.5%)
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 count = 0;\n        for(int a = 0; a < nums.length-1; a++){\n            for(int b = a+1; b < nums.length; b++){\n                if(nums[a]==nums[b]){\n                    if((a*b)%k==0){\n                        count++;\n                    }\n                }\n            }\n        }\n        return count;\n    }\n}\n```
['Java']