Beat's 87 % Easy solution without using inbuilt functions
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    def countPairs(self, nums: List[int], k: int) -> int:\n        count = 0\n        for i in range(len(nums)):\n            for j in range(i+1,len(nums)):\n                if nums[i]==nums[j] and i*j%k==0:\n                    count +=1\n        return count\n        \n```
['Python3']