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Least Common Denominator (LCD)
  • Difficulty Level : Easy
  • Last Updated : 01 Apr, 2021

The lowest Common Denominator or Least Common Denominator is the Least Common Multiple of the denominators of a set of fractions.
 

Common denominator : when the denominators of two or more fractions are the same. 
Least Common denominator is the smallest of all common denominators. 
Why do we need LCD
It simplifies addition, subtraction and comparing fraction.
Common Denominator can be simply evaluated by multiplying the denominators. In this case, 3 * 6 = 18
 

But that may not always be least common denominator, as in this case LCD = 6 and not 18. LCD is actually LCM of denominators.
Examples : 
 



LCD for fractions 5/12 and 7/15 is 60.
We can write both fractions as 25/60 and
28/60 so that they can be added and 
subtracted easily.

LCD for fractions 1/3 and 4/7 is 21.

 

Example Problem : Given two fractions, find their sum using least common dominator.
Examples
 

Input :  1/6  +  7/15    
Output : 19/30
         Explanation : LCM of 6 and 15 is 30. 
         So, 5/30  +  14/30 = 19/30 
Input :  1/3  +  1/6
Output : 3/6
         Explanation : LCM of 3 and 6 is 6. 
         So, 2/6  +  1/6 = 3/6

Note* These answers can be further simplified by Anomalous cancellation. 
 

 

C++




// C++ Program to determine
// LCD of two fractions and
// Perform addition on fractions
#include <iostream>
using namespace std;
 
// function to calculate gcd
// or hcf of two numbers.
int gcd(int a, int b)
{
    if (a == 0)
        return b;
    return gcd(b % a, a);
}
 
// function to calculate
// lcm of two numbers.
int lcm(int a, int b)
{
    return (a * b) / gcd(a, b);
}
 
void printSum(int num1, int den1,
              int num2, int den2)
{
    // least common multiple
    // of denominators LCD
    // of 6 and 15 is 30.
    int lcd = lcm(den1, den2);
 
    // Computing the numerators for LCD:
    // Writing 1/6 as 5/30 and 7/15 as
    // 14/30
    num1 *= (lcd / den1);
    num2 *= (lcd / den2);
 
    // Our sum is going to be res_num/lcd
    int res_num = num1 + num2;
    cout << res_num << "/" << lcd;
}
 
// Driver Code
int main()
{
    // First fraction is 1/6
    int num1 = 1, den1 = 6;
 
    // Second fraction is 7/15
    int num2 = 7, den2 = 15;
 
    printSum(num1, den1, num2, den2);
    return 0;
}

Java




// Java Program to determine LCD of two
// fractions and Perform addition on
// fractions
public class GFG {
     
    // function to calculate gcd or
    // hcf of two numbers.
    static int gcd(int a, int b)
    {
        if (a == 0)
            return b;
             
        return gcd(b % a, a);
    }
     
    // function to calculate lcm of
    // two numbers.
    static int lcm(int a, int b)
    {
        return (a * b) / gcd(a, b);
    }
     
    static void printSum(int num1, int den1,
                         int num2, int den2)
    {
         
        // least common multiple of
        // denominators LCD of 6 and 15
        // is 30.
        int lcd = lcm(den1, den2);
     
        // Computing the numerators for LCD:
        // Writing 1/6 as 5/30 and 7/15 as
        // 14/30
        num1 *= (lcd / den1);
        num2 *= (lcd / den2);
     
        // Our sum is going to be res_num/lcd
        int res_num = num1 + num2;
         
        System.out.print( res_num + "/" + lcd);
    }
         
    // Driver code
    public static void main(String args[])
    {
         
        // First fraction is 1/6
        int num1 = 1, den1 = 6;
 
        // Second fraction is 7/15
        int num2 = 7, den2 = 15;
 
        printSum(num1, den1, num2, den2);
    }
}
 
// This code is contributed by Sam007.

Python3




# python Program to determine
# LCD of two fractions and
# Perform addition on fractions
 
# function to calculate gcd
# or hcf of two numbers.
def gcd(a, b):
     
    if (a == 0):
        return b
    return gcd(b % a, a)
 
 
# function to calculate
# lcm of two numbers.
def lcm(a, b):
     
    return (a * b) / gcd(a, b)
 
 
def printSum(num1, den1,
                  num2, den2):
                 
    # least common multiple
    # of denominators LCD
    # of 6 and 15 is 30.
    lcd = lcm(den1, den2);
     
    # Computing the numerators
    # for LCD: Writing 1/6 as
    # 5/30 and 7/15 as 14/30
    num1 *= (lcd / den1)
    num2 *= (lcd / den2)
 
    # Our sum is going to be
    # res_num/lcd
    res_num = num1 + num2;
    print( int(res_num) , "/" ,
                       int(lcd))
 
# Driver Code
# First fraction is 1/6
num1 = 1
den1 = 6
 
# Second fraction is 7/15
num2 = 7
den2 = 15
printSum(num1, den1, num2, den2);
 
# This code is contributed
# by Sam007

C#




// C# Program to determine LCD of two
// fractions and Perform addition on
// fractions
using System;
 
class GFG {
 
    // function to calculate gcd or
    // hcf of two numbers.
    static int gcd(int a, int b)
    {
        if (a == 0)
            return b;
             
        return gcd(b % a, a);
    }
     
    // function to calculate lcm of
    // two numbers.
    static int lcm(int a, int b)
    {
        return (a * b) / gcd(a, b);
    }
     
    static void printSum(int num1, int den1,
                         int num2, int den2)
    {
         
        // least common multiple of
        // denominators LCD of 6 and 15
        // is 30.
        int lcd = lcm(den1, den2);
     
        // Computing the numerators for LCD:
        // Writing 1/6 as 5/30 and 7/15 as
        // 14/30
        num1 *= (lcd / den1);
        num2 *= (lcd / den2);
     
        // Our sum is going to be res_num/lcd
        int res_num = num1 + num2;
         
        Console.Write( res_num + "/" + lcd);
    }
         
    // Driver code
    public static void Main ()
    {
         
        // First fraction is 1/6
        int num1 = 1, den1 = 6;
 
        // Second fraction is 7/15
        int num2 = 7, den2 = 15;
 
        printSum(num1, den1, num2, den2);
    }
}
 
// This code is contributed by Sam007.

PHP




<?php
// PHP Program to determine
// LCD of two fractions and
// Perform addition on fractions
 
// function to calculate gcd
// or hcf of two numbers.
function gcd($a,$b)
{
    if ($a == 0)
        return $b;
    return gcd($b % $a, $a);
}
 
// function to calculate
// lcm of two numbers.
function lcm($a,$b)
{
    return ($a * $b) / gcd($a, $b);
}
 
function printSum($num1, $den1,
                  $num2, $den2)
{
    // least common multiple
    // of denominators
    // LCD of 6 and 15 is 30.
    $lcd = lcm($den1, $den2);
 
    // Computing the numerators for LCD:
    // Writing 1/6 as 5/30 and 7/15 as
    // 14/30
    $num1 *= ($lcd / $den1);
    $num2 *= ($lcd / $den2);
 
    // Our sum is going to be res_num/lcd
    $res_num = $num1 + $num2;
    echo $res_num . "/" . $lcd;
}
    // Driver Code
    // First fraction is 1/6
    $num1 = 1;
    $den1 = 6;
 
    // Second fraction is 7/15
    $num2 = 7;
    $den2 = 15;
 
    printSum($num1, $den1, $num2, $den2);
 
// This code is contributed by Sam007.
?>

Javascript




<script>
// javascript Program to determine LCD of two
// fractions and Perform addition on
// fractions
 
    // function to calculate gcd or
    // hcf of two numbers.
    function gcd(a , b)
    {
        if (a == 0)
            return b;
 
        return gcd(b % a, a);
    }
 
    // function to calculate lcm of
    // two numbers.
    function lcm(a , b)
    {
        return (a * b) / gcd(a, b);
    }
 
    function printSum(num1 , den1 , num2 , den2)
    {
 
        // least common multiple of
        // denominators LCD of 6 and 15
        // is 30.
        var lcd = lcm(den1, den2);
 
        // Computing the numerators for LCD:
        // Writing 1/6 as 5/30 and 7/15 as
        // 14/30
        num1 *= (lcd / den1);
        num2 *= (lcd / den2);
 
        // Our sum is going to be res_num/lcd
        var res_num = num1 + num2;
 
        document.write(res_num + "/" + lcd);
    }
 
    // Driver code
     
        // First fraction is 1/6
        var num1 = 1, den1 = 6;
 
        // Second fraction is 7/15
        var num2 = 7, den2 = 15;
 
        printSum(num1, den1, num2, den2);
 
// This code is contributed by todaysgaurav
</script>

Output : 
 

19/30

This article is contributed by Shubham Rana. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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