Check if product of first N natural numbers is divisible by their sum | C ++ implementation of the approach ; Function that returns true if n is prime ; Corner cases ; This is checked so that we can skip middle five numbers in below loop ; Function that return true if the product of the first n natural numbers is divisible by the sum of first n natural numbers ; Driver code
#include <bits/stdc++.h> NEW_LINE using namespace std ; bool isPrime ( int n ) { if ( n <= 1 ) return false ; if ( n <= 3 ) return true ; if ( n % 2 == 0 n % 3 == 0 ) return false ; for ( int i = 5 ; i * i <= n ; i = i + 6 ) if ( n % i == 0 || n % ( i + 2 ) == 0 ) return false ; return true ; } bool isDivisible ( int n ) { if ( isPrime ( n + 1 ) ) return false ; return true ; } int main ( ) { int n = 6 ; if ( isDivisible ( n ) ) cout << " Yes " ; else cout << " No " ; return 0 ; }