1498A - GCD Sum - CodeForces Solution


brute force math *800

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Python Code:

import sys
from math import gcd
def f(n):
    s=str(n)
    x=0
    for i in range(len(s)):
        x=x+int(s[i])
    return x
for _ in range(int(sys.stdin.readline())):
    n=int(sys.stdin.readline())
    s1=f(n)
    s2=f(n+1)
   
    if gcd(s1,n)>1:
        print(n)
    elif gcd(s2,n+1)>1:
        print(n+1)
    else:
        print(n+2)
   

C++ Code:

/**
 *    author:  Mohammed-issa
 *    created: 00.00.00       
 **/
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define endl "\n"

int main() {
    int t;
    cin >> t;
    for (int i = 0; i < t; i++) {
        long long n;
        cin >> n;
        while (true) {
            long long n2 = n;
            int sum = 0;
            while (n2 > 0) {
                sum += n2 % 10;
                n2 /= 10;
            }
            if (gcd(n, sum) > 1) {
                cout << n << endl;
                break;
            }
            n++;
        }
    }
}


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