C++ Program to find GCD of three numbers

August 23, 2021

C++ Program to find GCD of three numbers 

 

Here, in this section we will discuss GCD of three numbers in C++. GCD (Greatest Common Divisor) of two numbers is the largest number that divides both numbers.

Example : The GCD of 20, 45 and 30 will be :

                   Factors of 20 are 2 X 2 X 5

                   Factors of 45 are 3 X 3 X 5

                   Factors of 30 are 2 X 3 X 5

                   Common factor of 20, 45 and 30  : 5 , So the required GCD will be 5

Sum of odd digits of a number in java

Algorithm :

  • Take the input of three numbers from the user.
  • Store them in variable a, b and c.
  • Find the maximum among the three.
  • Run a reverse loop from maximum number to 1.
  • Check for every number if all three number is divisible by that particular number or not.
  • If it is so then break the loop and print that number.
#include <iostream>
using namespace std;

int gcd (int a, int b, int c) 
{
  
  int maxi = 0;
  
  maxi = max(a, max(b, c));
  
  for(int i = maxi; i>1; i--)
  {
      if(( a%i == 0 ) and ( b%i == 0 ) and ( c%i == 0 ))
      {
          return i;
          break;
      }
  }
  
  return 1;
}

int main () 
{
  int a, b, c;
  
  cout<<"Enter the numbers :";
  cin >> a >> b >> c;
  
  
  
  cout << "GCD of " << a << ", " << b << " and " << c << " is " << gcd (a, b, c);
  
  return 0;
}
Output :

Enter the numbers : 20 45 30

GCD of 20, 45 and 30 is 5