void main()
{
int n1,n2,n3,i,gcd,s;
printf("enter number n1,n2,n3:-");
scanf("%d %d %d",&n1,&n2,&n3);
if(n1 s=n1; else if(n2 s=n2; else s=n3; for(i=1;i<=s;i++) { if(n1%i==0&&n2%i==0&&n3%i==0) gcd=i; } printf("gcd=%d",gcd); }
void main()
{
int gcd(int x,int y);
int a,b,c;
printf("enter two number");
scanf("%d %d",&a,&b);
c=gcd(a,b);
printf("%d",c);
}
int gcd(int x,int y)
{
int temp1,temp2,temp;
temp1=(x>=0)?x:-x;
temp2=(y>=0)?y:-y;
if(temp1 { temp=temp1; temp1=temp2; temp2=temp; } while(temp2>0) { temp=temp1%temp2; temp1=temp2; temp2=temp; } return (temp1); }
LCM AND GCD OF TWO NUMBER.
#include
void main()
{
int n1,n2,i,lcm,gcd,min=0,f;
printf("enter number :-");
scanf("%d %d",&n1,&n2);
if(n1 min=n1; else min=n2; for(i=1;i<=min;i++) { if(n1%i==0&&n2%i==0) gcd=i; } f=n1*n2; lcm=f/gcd; printf("gcd=%d",gcd); printf("\nlcm=%d",lcm); }
if the gcd and lcm are given and one of the numbers are also given,multiply the gcd and lcm and divide them by the given number
You need at least two numbers to find either of those.
If you have two numbers m and n and their gcd (or gcf), g then their LCM = m*n/g so LCM = 72*252/36 = 2*252 = 504.
If you have the gcd or the LCM of two numbers, call them a and b, you can use the relationship that gcd(a,b) = (a multiplied by b) divided by LCM (a,b) where LCM or gcd (a,b) means the LCM or a and b. This means the gcd multiplied by the LCM is the same as the product of two numbers. Let's assume you have neither. There are several ways to do this. One way to approach both problems at once is to factor each number into primes. You can use these prime factorizations to find both the LCM and gcd To compute the Greatest common divisor, list the common prime factors and raise each to the least multiplicities that occurs among the several whole numbers. To compute the least common multiple, list all prime factors and raise each to the greatest multiplicities that occurs among the several whole numbers.
write a shell program for finding out gcd of three given numbers? write a shell program for finding out gcd of three given numbers? write a shell program for finding out gcd of three given numbers? check bellow link http://bashscript.blogspot.com/2009/08/gcd-of-more-than-two-numbers.html
800
The following function will return the GCD or LCM of two arguments (x and y) depending on the value of the fct argument (GCD or LCM). enum FUNC {GCD, LCM}; int gcd_or_lcm(FUNC fct, int x, int y) { int result = 0; switch (fct) { case (GCD): result = gcd (x, y); break; case (LCM): result = lcm (x, y); break; } return result; }
The GCF is 39 The LCM is 2457.
I think you mean either the GCD or the LCM? Not sure which since they are relatively prime, the LCM will the the product of the three numbers and the GCD is 1
The LCM of any two numbers can be found with the following formula:LCM(a,b) = (ab) / GCD (a,b).The GCD of two numbers is best found with the Euclidean algorithm which is as follows:GCD(a,b) =a --if b = 0or GCD(b, a mod b) otherwiseIn the example given we have GCD(42,7) = GCD(7, 0) = 7Then LCM(42,7) = (7*42)/7 = 42Note: mod is the operation of dividing one number by another and taking the remainder. e.g. 7 mod 4 = 3, 12 mod 6 = 0.
If we multiply the gcd and the LCM, we get the numbers.Call the numbers a and b. So 16(LCM)=ab3584=ab let's all the LCM, x 16x=a(3584/a)using the information above.x= 1/16(3584)or x=224 So the LCM is 224 we can just say the (gcd)LCM=ab=3584, so just divide 3584 by 16.
The LCM of 7 & 11 is 77.