通过7检查除数

给定一个数字，检查它是否可以被7整除。不允许使用模运算符，也不允许使用浮点运算。
一种简单的方法是重复减法。以下是另一种有趣的方法。
可以通过递归方法检查7除数。当且仅当a – 2b可被7整除时，形式10a + b的数字可被7整除。换句话说，从剩余数字形成的数字中减去最后一位的两倍。继续执行此操作，直到数量很少。
例如：数字371：37 –(2×1)= 37 – 2 = 35； 3 –(2×5)= 3 – 10 = -7;因此，由于-7可被7整除，因此371可被7整除。
以下是上述方法的实现

C++

// A Program to check whether a number is divisible by 7
#include 
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 
 
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

Python

# 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.

PHP

=0 )
        echo("Divisible");
    else
        echo("Not Divisible");
     
// This code is contributed by vt_m.
?>

Javascript

 
// 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
 

输出：

Divisible

这是如何运作的?设“ b”为数字“ n”的最后一位，设“ a”为除以“ b”时得到的数字。
数字的表示形式也可以乘以除数相对质数的任何数字，而不会改变其除数。观察到7除以21后，我们可以执行以下操作：

10.a + b 

乘以2后，变为

20.a + 2.b 

然后

21.a - a + 2.b 

消除21的倍数

-a + 2b

并乘以-1得到

a - 2b