different language solutions
Intuition\njavascript []\n/**\n * @param {number[]} nums1\n * @param {number[]} nums2\n * @param {number} k\n * @return {number}\n */\nvar numberOfPairs = funct
# Intuition\n```javascript []\n/**\n * @param {number[]} nums1\n * @param {number[]} nums2\n * @param {number} k\n * @return {number}\n */\nvar numberOfPairs = function(nums1, nums2, k) {\n    let count=0;\n    for(let i=0;i<nums1.length;i++){\n        for(let j=0;j<nums2.length;j++){\n            if(nums1[i]%(nums2[j]*k)==0){\n                count++;\n            }\n        }\n    }\n return count;\n};\n```\n```python []\nclass Solution:\n    def numberOfPairs(self, nums1: List[int], nums2: List[int], k: int) -> int:\n        result=0 \n        for i in range(len(nums1)):\n            for j in range(len(nums2)):\n                if nums1[i]%(nums2[j]*k)==0:\n                    result+=1\n        return result            \n\n```\n```java []\nclass Solution {\n    public int numberOfPairs(int[] nums1, int[] nums2, int k) {\n        int result=0;\n        for(int i=0;i<nums1.length;i++){\n            for(int j=0;j<nums2.length;j++){\n                if(nums1[i]%(nums2[j]*k)==0){\n                    result++;\n                }\n            }\n        }\n        return result;\n    }\n}\n```\n```Go []\nfunc numberOfPairs(nums1 []int, nums2 []int, k int) int {\n    count :=0\n    for i := 0; i < len(nums1); i++{\n        for j := 0; j < len(nums2); j++{\n           if nums1[i] % (nums2[j] * k) == 0{\n            count++;\n           }\n        } \n    }\n    return count;\n}\n```\n``` C []\n\nint numberOfPairs(int* nums1, int nums1Size, int* nums2, int nums2Size, int k) {\n    int count=0;\n    for(int i=0;i<nums1.Size;i++){\n        for(int j=0;j<nums2.Size;j++){\n            if(nums)\n        }\n    }\n}\n```\n``` Dart []\nclass Solution {\n  int numberOfPairs(List<int> nums1, List<int> nums2, int k) {\n    int result=0;\n    for(int i=0;i<nums1.length;i++){\n        for(int j=0;j<nums2.length;j++){\n            if(nums1[i] % (nums2[j]*k)==0){\n                result++;\n            }\n        }\n    }\n    return result;\n  }\n}
['Java', 'Go', 'Python3', 'JavaScript', 'C#', 'Dart']