How to find GCD, LCM on a set of numbers

Question

What would be the easiest way to calculate Greatest Common Divisor and Least Common Multiple on a set of numbers? What math functions can be used to find this information?

Answer 1

There is an Euclid's algorithm for GCD,

public int GCF(int a, int b) {
    if (b == 0) return a;
    else return (GCF (b, a % b));
}

By the way, a and b should be greater or equal 0, and LCM = |ab| / GCF(a, b)

Answer 2

int lcmcal(int i,int y)
{
    int n,x,s=1,t=1;
    for(n=1;;n++)
    {
        s=i*n;
        for(x=1;t<s;x++)
        {
            t=y*x;
        }
        if(s==t)
            break;
    }
    return(s);
}

Answer 3

int lcm = 1;
int y = 0;
boolean flag = false;
for(int i=2;i<=n;i++){
            if(lcm%i!=0){
                for(int j=i-1;j>1;j--){
                    if(i%j==0){
                        flag =true;
                        y = j;
                        break;
                    }
                }
                if(flag){
                    lcm = lcm*i/y;
                }
                else{
                    lcm = lcm*i;
                }
            }
            flag = false;
        }

here, first for loop is for getting every numbers starting from '2'. then if statement check whether the number(i) divides lcm if it does then it skip that no. and if it doesn't then next for loop is for finding a no. which can divides the number(i) if this happens we don't need that no. we only wants its extra factor. so here if the flag is true this means there already had some factors of no. 'i' in lcm. so we divide that factors and multiply the extra factor to lcm. If the number isn't divisible by any of its previous no. then when simply multiply it to the lcm.

Answer 4

There are no build in function for it. You can find the GCD of two numbers using Euclid's algorithm.

For a set of number

GCD(a_1,a_2,a_3,...,a_n) = GCD( GCD(a_1, a_2), a_3, a_4,..., a_n )

Apply it recursively.

Same for LCM:

LCM(a,b) = a * b / GCD(a,b)
LCM(a_1,a_2,a_3,...,a_n) = LCM( LCM(a_1, a_2), a_3, a_4,..., a_n )

Answer 5

If you can use Java 8 (and actually want to) you can use lambda expressions to solve this functionally:

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

public static int gcd(int... numbers) {
    return Arrays.stream(numbers).reduce(0, (x, y) -> gcd(x, y));
}

public static int lcm(int... numbers) {
    return Arrays.stream(numbers).reduce(1, (x, y) -> x * (y / gcd(x, y)));
}

I oriented myself on Jeffrey Hantin's answer, but

• calculated the gcd functionally
• used the varargs-Syntax for an easier API (I was not sure if the overload would work correctly, but it does on my machine)
• transformed the gcd of the numbers-Array into functional syntax, which is more compact and IMO easier to read (at least if you are used to functional programming)

This approach is probably slightly slower due to additional function calls, but that probably won't matter at all for the most use cases.

Answer 6

Basically to find gcd and lcm on a set of numbers you can use below formula,

LCM(a, b) X HCF(a, b) = a * b

Meanwhile in java you can use euclid's algorithm to find gcd and lcm, like this

public static int GCF(int a, int b)
{
    if (b == 0)
    {
       return a;
    }
    else
    {
       return (GCF(b, a % b));
    }
}

You can refer this resource to find examples on euclid's algorithm.