Write a program in java to find gcf of two numbers?

Answer 1

There are two basic steps to finding the greatest number. 1) convert the input from Strings to ints (or doubles or whatever number type you are expecting), 2) find the greatest number.

/*

For this example we're going to answer the question exactly,

with no regard to expandability (for dealing with more numbers).

*/

public static void main(String[] args) {

// step 1 - convert input from Strings to ints

int a = 0, b = 0, c = 0;

// we need to wrap a try-catch block around this to

// deal with any input that cannot be converted from

// a String to an int

try {

a = Integer.parseInt(args[0]);

b = Integer.parseInt(args[1]);

c = Integer.parseInt(args[2]);

}catch(NumberFormatException ex) {

// do whatever you need to do here to manage the

// exception

}

// step 2 - find the greatest number

int max = a;

if(b > max) {

max = b;

}

if(c > max) {

max = c;

}

// at this point the value in max is the greatest of the three

// numbers passed in through the command line

}

OR ELSE YOU SHOULD TRY THE ANOTHER STYLE FOR COMMAND LINE ARGUMENT , AS IT IS GIVEN BELOW.

class Max_command

{

public static void main(String args[])

{

for(String name : args)

{

System.out.println(name);

}

int i,max=0;

for(i=0;i<=args.length;i++)

{

int a=Integer.parseInt(args[i]);

if(a>max)

{

max=a;

}

if(i==args.length-1)

{

System.out.println("Max is --> "+max);

}

}

}

}

Answer 2

Since this sounds like a homework assignment, I'm not going give you source code, but I can give you some pseudo code to help you out.

[code]

//Find the smallest of the four numbers

min = a

IF b < min THEN min = b

IF c < min THEN min = c

IF d < min THEN min = d

FOR i = 1 TO min

//A number is only a common divisor if division does not produce a remainder

//when using it to divide any of the four numbers.

//The modulo operator, %, can be used to find the remainder.

IF a % i 0 THEN gcd = i

END FOR

[/code]

When writing the above code in Java, don't forget that you want to INCLUDE "min" in your for loop's iterations.

Answer 3

The following would be easy to program, but slow to run: try all the factors, starting at the smaller of the two numbers, and going down to one, until you find a common factor.

A little more tricky, but much faster for larger numbers, is the following property, which I will illustrate with an example. The greatest common factor of 14 and 10 is the same as the greatest common factor of 10 and 4, where 4 is calculated as 14 - 10 (or, faster, to avoid repeated subtraction, 14 % 10 - the remainder of the division). Repeat until you have a remainder of zero: (14, 10), (10, 4), (4, 2), (2, 0). The last non-zero number, in this case 2, is the greatest common factor.

The following would be easy to program, but slow to run: try all the factors, starting at the smaller of the two numbers, and going down to one, until you find a common factor.

A little more tricky, but much faster for larger numbers, is the following property, which I will illustrate with an example. The greatest common factor of 14 and 10 is the same as the greatest common factor of 10 and 4, where 4 is calculated as 14 - 10 (or, faster, to avoid repeated subtraction, 14 % 10 - the remainder of the division). Repeat until you have a remainder of zero: (14, 10), (10, 4), (4, 2), (2, 0). The last non-zero number, in this case 2, is the greatest common factor.

The following would be easy to program, but slow to run: try all the factors, starting at the smaller of the two numbers, and going down to one, until you find a common factor.

A little more tricky, but much faster for larger numbers, is the following property, which I will illustrate with an example. The greatest common factor of 14 and 10 is the same as the greatest common factor of 10 and 4, where 4 is calculated as 14 - 10 (or, faster, to avoid repeated subtraction, 14 % 10 - the remainder of the division). Repeat until you have a remainder of zero: (14, 10), (10, 4), (4, 2), (2, 0). The last non-zero number, in this case 2, is the greatest common factor.

The following would be easy to program, but slow to run: try all the factors, starting at the smaller of the two numbers, and going down to one, until you find a common factor.

A little more tricky, but much faster for larger numbers, is the following property, which I will illustrate with an example. The greatest common factor of 14 and 10 is the same as the greatest common factor of 10 and 4, where 4 is calculated as 14 - 10 (or, faster, to avoid repeated subtraction, 14 % 10 - the remainder of the division). Repeat until you have a remainder of zero: (14, 10), (10, 4), (4, 2), (2, 0). The last non-zero number, in this case 2, is the greatest common factor.

Answer 4

You can just use the GCD of any two of your numbers and find the GCD of it with your third number. Same for LCM.

public class Lcmgcd {

private static int gcd(int a, int b) { return (b == 0) ? a: gcd(b, a%b); }

private static int lcm(int a, int b) { return a * b / gcd(a, b); }

public static void main(String[] args) {

int[] n = {12, 16, 28};

System.out.println("GCD: " + gcd(n[2], gcd(n[0], n[1])) + "\tLCM: " + lcm(n[1],lcm(n[2],n[0])));

}

}

Answer 5

The following would be easy to program, but slow to run: try all the factors, starting at the smaller of the two numbers, and going down to one, until you find a common factor.

A little more tricky, but much faster for larger numbers, is the following property, which I will illustrate with an example. The greatest common factor of 14 and 10 is the same as the greatest common factor of 10 and 4, where 4 is calculated as 14 - 10 (or, faster, to avoid repeated subtraction, 14 % 10 - the remainder of the division). Repeat until you have a remainder of zero: (14, 10), (10, 4), (4, 2), (2, 0). The last non-zero number, in this case 2, is the greatest common factor.

Answer 6

public class GCD { public static int gcd(int a, int b) { return (b == 0 ? a : gcd(b, a % b)); } public static int gcd(int a, int b, int c) { return gcd(gcd(a, b), c); } }