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**Problem :-**Given a natural number, calculate sum of all its proper divisors. A proper divisor of a natural number is the divisor that is strictly less than the number.

For example, number 20 has 5 proper divisors: 1, 2, 4, 5, 10, and the divisor summation is: 1 + 2 + 4 + 5 + 10 = 22.

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**Solution :-**

#include <bits/stdc++.h>

using namespace std;

int main()

{

int t;

cin>>t;

while(t--)

{

int n,i,res=0;

cin>>n;

for(i=1;i<=n/2;i++)

{

if(n%i==0)

{

res+=i;

}

}

cout<<res<<endl;

}

return 0;

}

**Output:-**

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