Help me in solving MXFACS problem

help
,
1

My issue

i dont knew why my code is wrong. can anybody explain it.
My code is failing on the test case
input:
1
102107304

my output:
2

expected output:
3

i thought the lowest divisor orther than 1 is 2. but why is it saying the lowest divisor is 3?

My code

for i in range(int(input())):
    a=int(input())
    for i in range(2,a+1):
        if(a%i==0):
            print(i)
            break;

Learning course: 2000 to 2500 difficulty problems
Problem Link: Maximum Factors Problem Practice Problem in 2000 to 2500 difficulty problems - CodeChef

2

@anon27703279
yeah the answer would be 3 because in the prime factorization of n 3 comes more than 2 that’s y the answer would be 3
plzz refer my c++ code for better understanding of the logic

#include <bits/stdc++.h>
using namespace std;

int main() {
	// your code goes here
    int t;
    cin>>t;
    while(t--)
    {
        int n;
        cin>>n;
        map<int,int> mp;
        while(n%2==0)
        {
            n=n/2;
            mp[2]++;
        }
        for(int i=3;i<=sqrt(n);i+=2)
        {
            if(n%i==0)
            {
                while(n%i==0)
                {
                    n=n/i;
                    mp[i]++;
                }
            }
        }
        if(n>2)
        mp[n]++;
        int mx=0;
        for(auto x:mp)
        {
            mx=max(mx,x.second);
        }
        for(auto x:mp)
        {
            if(x.second==mx)
            {
                cout<<x.first<<endl;
                break;
            }
        }
    }
}
3

The lowest divisor is not necessarily the answer. Consider N = 18. If you choose K = 2 then M = 9, which has three divisors. However if you choose K = 3, M = 6 which has 4 divisors and is the answer.

More generally, if prime factorization of N is:
N = p_1^{a_1} \cdot p_2^{a_2} \cdot ... \cdot p_k^{a_k}, the number of divisors of N would be
\phi(N) = (1+a_1)\cdot(1+a_2)\cdot ... (1+a_k).
If you choose a prime p_i to divide, the difference in number of divisors would be:
\phi(N)-\phi({\frac{N}{p_i}}) = \frac{\phi(N)}{1+a_i}
To maximize \phi(\frac{N}{p_i}), we would want to minimize the difference. This would occur when we choose such a prime p_i that a_i is maximum.