Easiest soln u will ever find :)
IntuitionApproachComplexity
Time complexity:

Space complexity:
Code
# Intuition
<!-- Describe your first thoughts on how to solve this problem. -->

# Approach
<!-- Describe your approach to solving the problem. -->

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

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

# Code
```cpp []
class Solution {
public:
    int maxDivScore(vector<int>& nums, vector<int>& divisors) {

        int n = divisors.size();
        int maxcount=INT_MIN;
        int minelement = INT_MAX;
        
        for(int i=0;i<n;i++){
            
            int count=0;
            for(int j=0;j<nums.size();j++){

                if(nums[j]%divisors[i]==0){
                    count++;

                }
            }
           
   // find max count, if count are equal check for the smallest divisor availbe         
            if(count>maxcount || (count==maxcount && divisors[i]<minelement) ){
                maxcount = count;
                minelement=divisors[i];


            }



        }
        return minelement;
        
    }
};
```
![d9a30a34-57a0-40f1-adab-43bec86a259a_1723795301.0363479.png](https://assets.leetcode.com/users/images/b5fc3135-163f-4dc4-9586-2c62087b4997_1735234033.965795.png)

['C++']