**HCF:**

- HCF Stands For Higest Common Factor.
- HCF is also Known as GCD (Greatest Common Divisor).
- HCF of Two Number is defined as the Largest Number that Divides Both Numbers Completely with Remainder Zero.

**LCM:**

- LCM Stands For
- LCM of two Numbers Defined as the Smallest Number that is Multiple of both Numbers.

**Statement of C Program: Find The LCM and HCF of Two Numbers**:

#include<stdio.h>

#include<conio.h>

void main()

{

**int**num1 , num2 , lcm , gcd , remainder , numerator , denominator ;

clrscr();

printf( "Enter two numbers\n");

scanf(" %d%d " , &num1 , &num2 );

if (num1 > num2)

{

numerator=num1;

denominator=num2;

}

else

{

numerator = num2 ;

denominator = num1 ;

}

remainder = num1

**%**num2;**while**( remainder != 0)

{

numerator = denominator;

denominator = remainder;

remainder = numerator

**%**denominator;}

gcd = denominator;

lcm = (num1 * num2 )

**/**gcd;printf("GCD of %d and %d =%d\n" , num1, num2, gcd);

printf(" LCM of %d and %d= %d\n" , num1, num2, lcm);

getch();

}

Output:

Enter two numbers

5

15

GCD of 5 and 15 = 5

LCM of 5 and 15 = 15

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ReplyDeletedissertation topics coming up

hey dear there is a bug in the program :

ReplyDeletejust type num1 a smaller value then num 2 ...

the program will not generate desired output.. here is the simple modification :

int main()

{

int num1 , num2 , lcm , gcd , remainder , numerator , denominator;

printf( "Enter two numbers\n");

scanf("%d%d" , &num1 , &num2 );

if (num1 > num2)

{

numerator=num1;

denominator=num2;

remainder = num1 % num2;//now at correct place :)

}

else

{

numerator=num2;

denominator=num1;

remainder = num2 % num1;// this was missing :)

}

while ( remainder != 0)

{

numerator = denominator;

denominator = remainder;

remainder = numerator % denominator;

}

gcd = denominator;

lcm = (num1 * num2 ) / gcd;

printf("GCD of %d and %d =%d\n" , num1, num2, gcd);

printf(" LCM of %d and %d= %d\n" , num1, num2, lcm);

getch();

return 0;

}

http://www.compprog.com/2013/07/find-highest-common-factor.html This will describe your question better

ReplyDelete