Skip to content
Related Articles

Related Articles

To check whether a large number is divisible by 7

View Discussion
Improve Article
Save Article
  • Difficulty Level : Medium
  • Last Updated : 24 Sep, 2022
View Discussion
Improve Article
Save Article

You are given an n-digit large number, you have to check whether it is divisible by 7. 
A (r+1)-digit integer n whose digital form is (ar ar-1 ar-2….a2 a1 a0) is divisible by 7 if and only if the alternate series of numbers (a2 a1 a0) – (a5 a4 a3) + (a8 a7 a6) – … is divisible by 7. 

The triplets of digits within parenthesis represent a 3-digit number in digital form.

The given number n can be written as a sum of powers of 1000 as follows. 
n= (a2 a1 a0) + (a5 a4 a3)*1000 + (a8 a7 a6)*(1000*1000) +…. 
As 1000 = (-1)(mod 7), 1000 as per congruence relation. 
For a positive integer n, two numbers a and b are said to be congruent modulo n, if their difference 
(a – b) is an integer multiple of n (that is, if there is an integer k such that a – b = kn). This congruence relation is typically considered when a and b are integers, and is denoted 
{\displaystyle a\equiv b{\pmod {n}}.}
Hence we can write: 
n = { (a2a1a0) + (a5a4a3)* (-1) + (a8a7a6)* (-1)*(-1)+…..}(mod 7), 
Thus n is divisible by 7 if and if only if the series is divisible by 7. 
 

Examples:  

Input : 8955795758
Output : Divisible by 7
       Explanation:
       We express the number in terms of triplets 
       of digits as follows.
                (008)(955)(795)(758)
       Now, 758- 795 + 955 - 8 = 910, which is 
       divisible by 7

Input : 100000000000
Output : Not Divisible by 7
       Explanation:
       We express the number in terms of triplets 
       of digits as follows.
                (100)(000)(000)(000)
       Now, 000- 000 + 000 - 100 = -100, which is 
       not divisible by 7

Note that the number of digits in n may not be multiple of 3. In that case, we pass zero(s) on the left side of the remaining digits(s) after taking out all the triplets (from the right side of n) to form the last triplet. 
A simple and efficient method is to take input in form of a string (make its length in form of 3*m by adding 0 to left of the number if required) and then you have to add the digits in blocks of three from the right to left until it becomes a 3 digit number to form an alternate series and check whether the series is divisible by 7 or not. 

Here the program implementation to check the divisibility of 7 is done. 

Recommended Practice

C++




// C++ code to check divisibility of a
// given large number by 7
#include<bits/stdc++.h>
using namespace std;
 
int isdivisible7(string num)
{
    int n = num.length(), gSum=0;
    if (n == 0)
        return 1;
 
    // Append required 0s at the beginning.
    if (n % 3 == 1) {
        num="00" + num;
        n += 2;
    }
    else if (n % 3 == 2) {
        num= "0" + num;
        n++;
    }
 
    // add digits in group of three in gSum
    int i, GSum = 0, p = 1;
    for (i = n - 1; i >= 0; i--) {
 
        // group saves 3-digit group
        int group = 0;
        group += num[i--] - '0';
        group += (num[i--] - '0') * 10;
        group += (num[i] - '0') * 100;
 
        gSum = gSum + group * p;
 
        // generate alternate series of plus
        // and minus
        p *= (-1);
    }
 
    return (gSum % 7 == 0);
}
 
// Driver code
int main()
{
    // Driver method
    string num= "8955795758";
    if (isdivisible7(num))
        cout << "Divisible by 7";
    else
        cout << "Not Divisible by 7";
    return 0;
}

C




// C code to check divisibility of a
// given large number by 7
#include <stdio.h>
#include <string.h>
int isdivisible7(char num[])
{
    int n = strlen(num), gSum=0;
    char final[n+3];
    if (n == 0 && num[0] == '\n')
        return 1;
 
    // Append required 0s at the beginning.
    if (n % 3 == 1) {
        final[0]='0';
        final[1]='0';
        strcat(final,num);
        n += 2;
    }
    else if (n % 3 == 2) {
        final[0]='0';
        strcat(final,num);
        n++;
    }
 
    // add digits in group of three in gSum
    int i, GSum = 0, p = 1;
    for (i = n - 1; i >= 0; i--) {
 
        // group saves 3-digit group
        int group = 0;
        group += final[i--] - '0';
        group += (final[i--] - '0') * 10;
        group += (final[i] - '0') * 100;
 
        gSum = gSum + group * p;
 
        // generate alternate series of plus
        // and minus
        p *= (-1);
    }
 
    return (gSum % 7 == 0);
}
 
// Driver code
int main()
{
    // Driver method
    char num[] = "8955795758";
    if (isdivisible7(num))
        printf("Divisible by 7");
    else
        printf("Not Divisible by 7");
    return 0;
}

Java




// Java code to check divisibility of a given large number by 7
 
class Test {
    // Method to check divisibility
    static boolean isDivisible7(String num)
    {
        int n = num.length();
        if (n == 0 && num.charAt(0) == '0')
            return true;
 
        // Append required 0s at the beginning.
        if (n % 3 == 1)
            num = "00" + num;
        if (n % 3 == 2)
            num = "0" + num;
        n = num.length();
 
        // add digits in group of three in gSum
        int gSum = 0, p = 1;
        for (int i = n - 1; i >= 0; i--) {
 
            // group saves 3-digit group
            int group = 0;
            group += num.charAt(i--) - '0';
            group += (num.charAt(i--) - '0') * 10;
            group += (num.charAt(i) - '0') * 100;
            gSum = gSum + group * p;
            // generate alternate series of plus and minus
            p = p * -1;
        }
 
        // calculate result till 3 digit sum
        return (gSum % 7 == 0);
    }
 
    // Driver method
    public static void main(String args[])
    {
        String num = "8955795758";
 
        System.out.println(isDivisible7(num) ? "Divisible by 7" : "Not Divisible  by 7");
    }
}

Python3




# Python 3 code to check divisibility
# of a given large number by 7
 
def isdivisible7(num):
    n = len(num)
    if (n == 0 and num[0] == '\n'):
        return 1
 
    # Append required 0s at the beginning.
    if (n % 3 == 1) :
        num = "00" + str(num)
        n += 2
     
    elif (n % 3 == 2) :
        num = "0" + str(num)
        n += 1
 
    # add digits in group of three in gSum
    GSum = 0
    p = 1
    i = n-1
    while i>=0 :
 
        # group saves 3-digit group
        group = 0
        group += ord(num[i]) - ord('0')
        i -= 1
        group += (ord(num[i]) - ord('0')) * 10
        i -= 1
        group += (ord(num[i]) - ord('0')) * 100
 
        GSum = GSum + group * p
 
        # generate alternate series of
        # plus and minus
        p *= (-1)
        i -= 1
 
    return (GSum % 7 == 0)
 
# Driver code
if __name__ == "__main__":
     
    num = "8955795758"
    if (isdivisible7(num)):
        print("Divisible by 7")
    else :
        print("Not Divisible by 7")
 
# This code is contributed by ChitraNayal

C#




// C# code to check divisibility of a
// given large number by 7
using System;
 
class GFG {
 
    // Method to check divisibility
    static bool isDivisible7(String num)
    {
        int n = num.Length;
        if (n == 0 && num[0] == '0')
            return true;
 
        // Append required 0s at the beginning.
        if (n % 3 == 1)
            num = "00" + num;
 
        if (n % 3 == 2)
            num = "0" + num;
 
        n = num.Length;
 
        // add digits in group of three in gSum
        int gSum = 0, p = 1;
        for (int i = n - 1; i >= 0; i--) {
 
            // group saves 3-digit group
            int group = 0;
            group += num[i--] - '0';
            group += (num[i--] - '0') * 10;
            group += (num[i] - '0') * 100;
            gSum = gSum + group * p;
 
            // generate alternate series
            // of plus and minus
            p = p * -1;
        }
 
        // calculate result till 3 digit sum
        return (gSum % 7 == 0);
    }
 
    // Driver code
    static public void Main()
    {
        String num = "8955795758";
 
        // Function calling
        Console.WriteLine(isDivisible7(num) ? "Divisible by 7" : "Not Divisible by 7");
    }
}
 
// This code is contributed by Ajit.

PHP




<?php
// PHP code to check divisibility of
// a given large number by 7
 
// Function to check divisibility
function isDivisible7($num)
{
    $n = strlen($num) ;
    if ($n == 0 && $num[0] == '0')
        return true;
 
    // Append required 0s at the beginning.
    if ($n % 3 == 1)
        $num = "00" . $num;
    if ($n % 3 == 2)
        $num = "0". $num;
    $n = strlen($num);
 
    // add digits in group of three in gSum
    $gSum = 0 ;
    $p = 1;
    for ($i = $n - 1; $i >= 0; $i--)
    {
 
        // group saves 3-digit group
        $group = 0;
        $group += $num[$i--] - '0';
        $group += ($num[$i--] - '0') * 10;
        $group += ($num[$i] - '0') * 100;
        $gSum = $gSum + $group * $p;
         
        // generate alternate series
        // of plus and minus
        $p = $p * -1;
    }
 
    // calculate result till 3 digit sum
    return ($gSum % 7 == 0);
}
 
// Driver Code
$num = "8955795758";
 
echo (isDivisible7($num) ?
        "Divisible by 7" :
        "Not Divisible by 7");
 
// This code is contributed by Ryuga
?>

Javascript




<script>
 
// Javascript code to check divisibility of
// a given large number by 7
 
// Function to check divisibility
function isDivisible7(num)
{
    let n = num.length;
     
    if (n == 0 && num[0] == '0')
        return true;
 
    // Append required 0s at the beginning.
    if (n % 3 == 1)
        num = "00" + num;
    if (n % 3 == 2)
        num = "0" + num;
         
    n = num.length;
 
    // Add digits in group of three in gSum
    gSum = 0 ;
    let p = 1;
     
    for(let i = n - 1; i >= 0; i--)
    {
         
        // Group saves 3-digit group
        group = 0;
        group += num[i--] - '0';
        group += (num[i--] - '0') * 10;
        group += (num[i] - '0') * 100;
        gSum = gSum + group * p;
         
        // Generate alternate series
        // of plus and minus
        p = p * -1;
    }
 
    // Calculate result till 3 digit sum
    return (gSum % 7 == 0);
}
 
// Driver Code
let num = "8955795758";
 
document.write(isDivisible7(num) ?
               "Divisible by 7" :
               "Not Divisible by 7");
 
// This code is contributed by _saurabh_jaiswal
     
</script>

Output

Divisible by 7

Time Complexity: O(n), where n is the number of digits in num.
Auxiliary Space: O(1)

Method 2: Checking given number is divisible by 7 or not by using modulo division operator “%”.  

C++




#include <iostream>
using namespace std;
int main()
{
    //input
    long long int n=100000000000;
     
     
    // finding given number is divisible by 7 or not
    if (n%7==0)
    {
        cout << "Yes";
    }
    else
    {
        cout << "No";
    }
   
    return 0;
}

Java




// Java code
// To check whether the given number is divisible by 7 or not
 
import java.io.*;
import java.util.*;
  
class GFG
{
   
  public static void main(String[] args)
  {
    //input
    long n=100000000000L;
    // finding given number is divisible by 7 or not
     
    if ((n)%7==0)
    {
        System.out.println("Yes");
    }
    else
    {
        System.out.println("No");
    }
     
  }
}
//this code is contributed by aditya942003patil

Python3




# Python code
# To check whether the given number is divisible by 7 or not
 
#input
n=100000000000
# the above input can also be given as n=input() -> taking input from user
# finding given number is divisible by 7 or not
if int(n)%7==0:
  print("Yes")
else:
  print("No")
   
  # this code is contributed by gangarajula laxmi

C#




// c# code to check whether the given
// number is diivisible by 7 or not
using System;
  
public class GFG {
      
    public static void Main()
    {
        //input
        long n=100000000000;
         
        // the above input can also be given as n=input() -> taking input from user
        // finding given number is divisible by 7 or not
        if (n%7==0)
        {
            Console.Write("Yes");
        }
        else
        {
            Console.Write("No");
        }
    }
}
// This code is contributed by aditya942003patil

Javascript




<script>
        // JavaScript code for the above approach
        // To check whether the given number is divisible by 7 or not
 
        //input
        var n = 100000000000
        // the above input can also be given as n=input() -> taking input from user
        // finding given number is divisible by 7 or not
        if (n % 7 == 0)
            document.write("Yes")
        else
            document.write("No")
 
    // This code is contributed by Potta Lokesh
    </script>

PHP




<?php
   //input
    $n=100000000000;
     
    // the above input can also be given as n=input() -> taking input from user
    // finding given number is divisible by 7 or not
    if ($n%7==0)
    {
        echo "Yes";
    }
    else
    {
        echo "No";
    
 
// This code is contributed by laxmigangarajula03
?>

Output

No

Time Complexity: O(1)
Auxiliary Space: O(1)

This article is contributed by Sruti Rai. 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.
 


My Personal Notes arrow_drop_up
Recommended Articles
Page :

Start Your Coding Journey Now!