Program to find G.C.D and L.C.M of two numbers

[thiruvasagamani]

Find G.C.D and L.C.M of two numbers

Code Select Expand

#include <stdio.h>
#include <conio.h>

int gcd( int m, int n)
{
while( m!= n) // execute loop until m == n
{
if( m > n)
m= m - n; // large - small , store the results in large variable
else
n= n - m;
}
return ( m); // m or n is GCD
}

int lcm( int m, int n)
{
int gcd( int, int);
return( m * n / gcd (m , n)); // product of 2 numbers / gcd is lcm
}

main()
{
int m, n , lcm( int, int), gcd( int ,int);
printf(" enter anyu 2 values \n");
scanf("%d%d", &m,&n);
printf(" gcd is %d\n", gcd( m, n));
printf(" lcm is %d\n", lcm(m , n));
getch();
}
o/p:
If u enter m = 12 , n= 15 output gcd is 3 and output lcm is 60

NOTE 1:
gcd(12,15) = 3
gcd(12, 18) = 6
gcd(4 , 7) = 1

NOTE 2:

LCM of 12, 15 = 12 * 15 / gcd(12, 15);

[Sam Opoka]

Find G.C.D and L.C.M of two numbers

Code Select Expand

#include <stdio.h>
#include <conio.h>

int gcd( int m, int n)
{
while( m!= n) // execute loop until m == n
{
if( m > n)
m= m - n; // large - small , store the results in large variable
else
n= n - m;
}
return ( m); // m or n is GCD
}

int lcm( int m, int n)
{
int gcd( int, int);
return( m * n / gcd (m , n)); // product of 2 numbers / gcd is lcm
}

main()
{
int m, n , lcm( int, int), gcd( int ,int);
printf(" enter anyu 2 values \n");
scanf("%d%d", &m,&n);
printf(" gcd is %d\n", gcd( m, n));
printf(" lcm is %d\n", lcm(m , n));
getch();
}
o/p:
If u enter m = 12 , n= 15 output gcd is 3 and output lcm is 60

NOTE 1:
gcd(12,15) = 3
gcd(12, 18) = 6
gcd(4 , 7) = 1

NOTE 2:

LCM of 12, 15 = 12 * 15 / gcd(12, 15);

[Elvin-Aze]

#include<iostream>
#include<cstdlib>
long a,b,m,n,c;
using namespace std;
int main()
{cin>>m>>n;
a=m;
b=n;
while(n!=0)
{
c=m%n;
m=n;
n=c;}
   
    if(m!=1)
    cout<<a/m<<" "<<b/m<<endl;
    else
    cout<<a<<" "<<b<<endl;
   
    system("pause");
    return 0;
    }
use it will udnerstand best and quickest way

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Run the program and wait, and in order to increase your chances, install the program on all computers available to you, at work, with your friends, with your relatives, you can also ask your classmates to use the program, so your chances will increase tenfold!
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