💚Easy java solution using two loops.
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        \n     for(int i=0 ; i<nums.length ; i++){\n         for(int j=i+1;j<nums.length ; j++){\n             if(nums[i] == nums[j] && i*j % k == 0){\n                count++;\n             }\n         }\n     }   \n     return count;\n    }\n}\n```
['Array', 'Java']