Beats 100.00% of users with Python3
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![image.png](https://assets.leetcode.com/users/images/71858ef4-5038-4659-b45d-7b6fd8dda635_1716829848.1517875.png)\n\n\n# Code\n```\nclass Solution:\n    def numberOfPairs(self, nums1: List[int], nums2: List[int], k: int) -> int:\n        c=0\n        n1=len(nums1)\n        n2=len(nums2)\n        for i in range(n1):\n            for j in range(n2):\n                if nums1[i] % ( nums2[j] *k ) ==0:\n                    c = c+1\n        return c \n\n```
['Python3']