Simple solution 100% beat
Complexity
Time complexity:O(n*n)

Space complexity:O(1)
Code

# Complexity
- Time complexity:O(n*n)
<!-- Add your time complexity here, e.g. $$O(n)$$ -->

- Space complexity:O(1)
<!-- Add your space complexity here, e.g. $$O(n)$$ -->

# Code
```python3 []
class Solution:
    def numberOfPairs(self, nums1: List[int], nums2: List[int], k: int) -> int:
        for i in range(len(nums2)):
            nums2[i]=nums2[i]*k
        
        cnt=0
        for i in nums1:
            for j in nums2:
                if i%j==0:
                    cnt+=1
        return cnt
```
['Array', 'Hash Table', 'Python3']