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GCD of digits of a given number

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  • Difficulty Level : Easy
  • Last Updated : 10 Jun, 2022
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Given a number n, find GCD of its digits.

Examples : 

Input  : 345
Output : 1
GCD of 3, 4 and 5 is 1.

Input  : 2448
Output : 2
GCD of 2, 4, 4 and 8 is 2

We traverse the digits of number one by one using below loop  

digit = n mod 10;
n  = n / 10; 

While traversing digits, we keep track of current GCD and keep updating GCD by finding GCD of current digit with current GCD. 

C++




// CPP program to find GCD of digits of a number
#include<iostream>
#include<algorithm>
using namespace std;
 
int digitGCD(int n)
{
    int gcd = 0;
    while (n > 0)
    {       
        gcd = __gcd(n%10, gcd);
 
        // If at point GCD becomes 1,
        // return it
        if (gcd == 1)
           return 1;
 
        n = n/10;
    }
    return gcd;
}
 
// driver code
int main()
{
    long n = 2448;
    cout << digitGCD(n);
    return 0;
}

Java




// Java program to find GCD of digits of a number
 
class GFG
{
    // Recursive function to return gcd of a and b
    static int __gcd(int a, int b)
    {
        return b == 0 ? a : __gcd(b, a % b);
         
    }
    static int digitGCD(int n)
    {
        int gcd = 0;
        while (n > 0)
        {    
            gcd = __gcd(n % 10, gcd);
     
            // If at point GCD becomes 1,
            // return it
            if (gcd == 1)
            return 1;
     
            n = n / 10;
        }
        return gcd;
    }
     
    // Driver code
    public static void main (String[] args)
    {
        int n = 2448;
        System.out.print(digitGCD(n));
    }
}
 
// This code is contributed by Anant Agarwal.

Python3




# Python program to find
# GCD of digits of a number
 
# Recursive function to return gcd of a and b
def __gcd(a,b):
    return a if(b==0) else __gcd(b, a % b)
     
def digitGCD(n):
 
    gcd = 0
    while (n > 0):
     
        gcd = __gcd(n % 10, gcd)
  
        # If at point GCD becomes 1,
        # return it
        if (gcd == 1):
           return 1
  
        n = n // 10
     
    return gcd
 
#Driver code
n = 2448
print(digitGCD(n))
 
# This code is contributed
# by Anant Agarwal.

C#




// C# program to find GCD of
// digits of a number
using System;
 
class GFG {
     
    // Recursive function to return
    // gcd of a and b
    static int __gcd(int a, int b)
    {
        return b == 0 ? a : __gcd(b, a % b);
         
    }
     
    static int digitGCD(int n)
    {
        int gcd = 0;
        while (n > 0)
        {    
            gcd = __gcd(n % 10, gcd);
     
            // If at point GCD becomes 1,
            // return it
            if (gcd == 1)
            return 1;
     
            n = n / 10;
        }
        return gcd;
    }
     
    // Driver code
    public static void Main ()
    {
        int n = 2448;
        Console.Write(digitGCD(n));
    }
}
 
// This code is contributed by Nitin Mittal.

PHP




<?php
// PHP program to find GCD
// of digits of a number
 
// Recursive function to
// return gcd of a and b
function __gcd($a,$b)
{
    return $b == 0 ? $a :
      __gcd($b, $a % $b);
         
}
     
function digitGCD($n)
{
    $gcd = 0;
    while ($n > 0)
    {    
        $gcd = __gcd($n % 10, $gcd);
 
        // If at point GCD
        // becomes 1, return it
        if ($gcd == 1)
        return 1;
 
        $n = $n / 10;
    }
    return $gcd;
}
 
// Driver code
$n = 2448;
echo digitGCD($n);
 
// This code is contributed by Sam007
?>

Javascript




<script>
// javascript program to find GCD of digits of a number   
// Recursive function to return gcd of a and b
    function __gcd(a, b)
    {
        return b == 0 ? a : __gcd(b, a % b);
 
    }
 
    function digitGCD(n)
    {
        var gcd = 0;
        while (n > 0)
        {
            gcd = __gcd(n % 10, gcd);
 
            // If at point GCD becomes 1,
            // return it
            if (gcd == 1)
                return 1;
 
            n = parseInt(n / 10);
        }
        return gcd;
    }
 
    // Driver code
    var n = 2448;
    document.write(digitGCD(n));
 
// This code is contributed by aashish1995
</script>

Output : 

2

Time complexity: O(logn)

Auxiliary Space: O(1)

This article is contributed by Dibyendu Roy Chaudhuri. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
 


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