Beats 99%
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 divisibleNumbersQty = 0;\n\n        for(int i = 0; i < nums.length; i++) {\n\n\n            for(int j = i+1; j < nums.length; j++) {\n                if(nums[i] == nums[j] ) {\n                    int number = i*j;\n                    int dividedNumber = number/k;\n                    if(dividedNumber*k == number)\n                        divisibleNumbersQty++;\n                }\n            }\n        }\n        return divisibleNumbersQty;\n    }\n}\n```
['Java']