Check divisibility by 7

Given a number N, the task is to check if it is divisible by 7 or not.
Note: You are not allowed to use the modulo operator, floating point arithmetic is also not allowed. 

Naive approach: A simple method is repeated subtraction. Following is another interesting method.
Divisibility by 7 can be checked by a recursive method. A number of the form 10a + b is divisible by 7 if and only if a – 2b is divisible by 7. In other words, subtract twice the last digit from the number formed by the remaining digits. Continue to do this until a small number. 

Example: the number 371: 37 – (2×1) = 37 – 2 = 35; 3 – (2 × 5) = 3 – 10 = -7; thus, since -7 is divisible by 7, 371 is divisible by 7. 
Following is the implementation of the above method 

C++

// A Program to check whether a number is divisible by 7
#include <bits/stdc++.h>
using namespace std;

int isDivisibleBy7( int num )
{
    // If number is negative, make it positive
    if( num < 0 )
        return isDivisibleBy7( -num );

    // Base cases
    if( num == 0 || num == 7 )
        return 1;
    if( num < 10 )
        return 0;

    // Recur for ( num / 10 - 2 * num % 10 )
    return isDivisibleBy7( num / 10 - 2 *
            ( num - num / 10 * 10 ) );
}

// Driver code
int main()
{
    int num = 616;
    if( isDivisibleBy7(num ) )
        cout << "Divisible" ;
    else
        cout << "Not Divisible" ;
    return 0;
}

// This code is contributed by rathbhupendra

C

// A Program to check whether a number is divisible by 7
#include <stdio.h>

int isDivisibleBy7( int num )
{
    // If number is negative, make it positive
    if( num < 0 )
        return isDivisibleBy7( -num );

    // Base cases
    if( num == 0 || num == 7 )
        return 1;
    if( num < 10 )
        return 0;

    // Recur for ( num / 10 - 2 * num % 10 )
    return isDivisibleBy7( num / 10 - 2 * ( num - num / 10 * 10 ) );
}

// Driver program to test above function
int main()
{
    int num = 616;
    if( isDivisibleBy7(num ) )
        printf( "Divisible" );
    else
        printf( "Not Divisible" );
    return 0;
}

Java

// Java program to check whether a number is divisible by 7
import java.io.*;

class GFG
{
    // Function to check whether a number is divisible by 7
    static boolean isDivisibleBy7(int num)
    {
        // If number is negative, make it positive
        if( num < 0 )
            return isDivisibleBy7( -num );

        // Base cases
        if( num == 0 || num == 7 )
            return true;
        if( num < 10 )
            return false;

        // Recur for ( num / 10 - 2 * num % 10 )
        return isDivisibleBy7( num / 10 - 2 * ( num - num / 10 * 10 ) );
    }

    // Driver program
    public static void main (String[] args)
    {
        int num = 616;
        if(isDivisibleBy7(num))
            System.out.println("Divisible");
        else
            System.out.println("Not Divisible");
    }
}

// Contributed by Pramod Kumar

Python3

# Python program to check whether a number is divisible by 7

# Function to check whether a number is divisible by 7
def isDivisibleBy7(num) :

    # If number is negative, make it positive
    if num < 0 :
        return isDivisibleBy7( -num )

    # Base cases
    if( num == 0 or num == 7 ) :
        return True

    if( num < 10 ) :
        return False

    # Recur for ( num / 10 - 2 * num % 10 )
    return isDivisibleBy7( num // 10 - 2 * ( num - num // 10 * 10 ) )

# Driver program
num = 616
if(isDivisibleBy7(num)) :
    print ("Divisible")
else :
    print ("Not Divisible")

# This code is contributed by Nikita Tiwari

C#

// C# program to check whether a
// number is divisible by 7
using System;

class GFG {

    // Function to check whether a
    // number is divisible by 7
    static bool isDivisibleBy7(int num)
    {

        // If number is negative,
        // make it positive
        if( num < 0 )
            return isDivisibleBy7(-num);

        // Base cases
        if( num == 0 || num == 7 )
            return true;
        if( num < 10 )
            return false;

        // Recur for ( num / 10 - 2 * num % 10 )
        return isDivisibleBy7(num / 10 - 2 *
                             ( num - num / 10 * 10 ));
    }

    // Driver Code
    public static void Main ()
    {
        int num = 616;
        if(isDivisibleBy7(num))
            Console.Write("Divisible");
        else
            Console.Write("Not Divisible");
    }
}

// This code is contributed by Nitin Mittal.

Javascript

<script>

// js Program to check whether
// a number is divisible by 7

// Function to check whether a
// number is divisible by 7
function isDivisibleBy7( num )
{

    // If number is negative,
    // make it positive
    if( num < 0 )
        return isDivisibleBy7( -num );

    // Base cases
    if( num == 0 || num == 7 )
        return 1;
    if( num < 10 )
        return 0;

    // Recur for ( num / 10 - 2 * num % 10 )
    return isDivisibleBy7(num / 10 - 2 *
                         (num - num / 10 * 10 ) );
}

    // Driver Code
    let num = 616;
    if( isDivisibleBy7(num )>=0 )
        document.write("Divisible");
    else
        document.write("Not Divisible");

// This code is contributed by sravan kumar

</script>

PHP

<?php
// PHP Program to check whether
// a number is divisible by 7

// Function to check whether a
// number is divisible by 7
function isDivisibleBy7( $num )
{

    // If number is negative,
    // make it positive
    if( $num < 0 )
        return isDivisibleBy7( -$num );

    // Base cases
    if( $num == 0 || $num == 7 )
        return 1;
    if( $num < 10 )
        return 0;

    // Recur for ( num / 10 - 2 * num % 10 )
    return isDivisibleBy7($num / 10 - 2 *
                         ($num - $num / 10 * 10 ) );
}

    // Driver Code
    $num = 616;
    if( isDivisibleBy7($num )>=0 )
        echo("Divisible");
    else
        echo("Not Divisible");

// This code is contributed by vt_m.
?>

Output

Divisible

Time Complexity: O(log n)
Auxiliary Space: O(log n)

How does this work? Let ‘b’ be the last digit of a number ‘n’ and let ‘a’ be the number we get when we split off ‘b’. 
The representation of the number may also be multiplied by any number relatively prime to the divisor without changing its divisibility. After observing that 7 divides 21, we can perform the following: 

 10.a + b 

after multiplying by 2, this becomes  

 20.a + 2.b 

and then 

 21.a - a + 2.b 

Eliminating the multiple of 21 gives 

 -a + 2b

and multiplying by -1 gives 

 a - 2b

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.