This is the solution of the first problem given in the Coding Contest of Reload22

LCD_GCD.cpp

#include <bits/stdc++.h>
using namespace std;
 
int main() {
    int t; cin>>t;
    while(t--) {
        long long n; cin>>n;
        long long ans = 0;
        // pairs in which lcm/gcd=1
        ans+=n;
        
        //pairs in which lcm/gcd=2
        ans+=(n/2)*2;
        
        //pairs in which lcm/gcd=3
        ans+=(n/3)*2;
        
        cout<<ans<<'\n';
    }
    return 0;
}