GCD of digits of a given number

Last Updated : 17 Feb, 2023

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# 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 :

Time complexity: O(logn)

Auxiliary Space: O(1)

Maximize GCD of all possible pairs from 1 to N

Dibyendu Roy Chaudhuri

Improve

Article Tags :

GCD of digits of a given number

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