检查数字是否可被47整除

给定数字N ，任务是检查数字是否可被47整除。

例子：

Input: N = 1645
Output: yes
Explanation:
47 * 35 = 1645

Input: N = 4606
Output: yes
Explanation:
47 * 98 = 4606

方法： 47的可除性检验为：

1. 提取最后一位数字。
2. 从除去最后一位后获得的剩余号码中减去14 *最后一位。
3. 重复上述步骤，直到获得两位数字或零。
4. 如果两位数可被47整除或为0，则原始数字也可被47整除。

例如：

If N = 59173

Step 1:
  N = 59173
  Last digit = 3
  Remaining number = 5917
  Subtracting 14 times last digit
  Resultant number = 5917 - 14*3 = 5875

Step 2:
  N = 5875
  Last digit = 5
  Remaining number = 587
  Subtracting 14 times last digit
  Resultant number = 587 - 14*5 = 517

Step 3:
  N = 517
  Last digit = 7
  Remaining number = 51
  Subtracting 14 times last digit
  Resultant number = 51 - 14*7 = -47

Step 4:
  N = -47
  Since N is a two-digit number,
  and -47 is divisible by 47

Therefore N = 59173 is also divisible by 47

下面是上述方法的实现：

C++

// C++ program to check whether a number
// is divisible by 47 or not
#include
#include
  
using namespace std;
  
// Function to check if the number is  divisible by 47 or not 
bool isDivisible(int n)  
{
    int d;
    // While there are at least two digits 
    while (n / 100) 
    {
   
        // Extracting the last 
        d = n % 10;
   
        // Truncating the number 
        n /= 10;
   
        // Subtracting fourteen times the last 
        // digit to the remaining number 
        n = abs(n-(d * 14));
    }
    // Finally return if the two-digit
    // number is divisible by 47 or not
    return (n % 47 == 0) ;
}
  
// Driver Code 
int main() {
    int N = 59173;
   
    if (isDivisible(N)) 
        cout<<"Yes"<

---

Java

// Java program to check whether a number
// is divisible by 47 or not
import java.util.*;
  
class GFG{
   
// Function to check if the number is  divisible by 47 or not 
static boolean isDivisible(int n)  
{
    int d;
      
    // While there are at least two digits 
    while ((n / 100) > 0) 
    {
    
        // Extracting the last 
        d = n % 10;
    
        // Truncating the number 
        n /= 10;
    
        // Subtracting fourteen times the last 
        // digit to the remaining number 
        n = Math.abs(n - (d * 14));
    }
  
    // Finally return if the two-digit
    // number is divisible by 47 or not
    return (n % 47 == 0) ;
}
   
// Driver Code 
public static void main(String[] args) {
    int N = 59173;
    
    if (isDivisible(N)) 
        System.out.print("Yes") ;
    else
        System.out.print("No");
      
 }     
}    
  
// This code is contributed by PrinciRaj1992

Python 3

# Python program to check if a number
# is divisible by 47 or not
  
# Function to check if the number is 
# divisible by 47 or not 
def isDivisible(n) : 
   
    # While there are at least two digits 
    while n // 100 : 
   
        # Extracting the last 
        d = n % 10
   
        # Truncating the number 
        n //= 10
   
        # Subtracting fourteen times the last 
        # digit to the remaining number 
        n = abs(n- (d * 14))
   
    # Finally return if the two-digit
    # number is divisible by 43 or not
    return (n % 47 == 0) 
   
# Driver Code 
if __name__ == "__main__" : 
       
    n = 59173
  
    if (isDivisible(n)) : 
        print("Yes") 
    else : 
        print("No")

C#

// C# program to check whether a number
// is divisible by 47 or not
using System; 
          
class GFG 
{ 
      
// Function to check if the number is divisible by 47 or not 
static bool isDivisible(int n) 
{
    int d;
      
    // While there are at least two digits 
    while (n / 100 > 0) 
    {
      
        // Extracting the last 
        d = n % 10;
      
        // Truncating the number 
        n /= 10;
      
        // Subtracting fourteen times the last 
        // digit to the remaining number 
        n = Math.Abs(n - (d * 14));
    }
      
    // Finally return if the two-digit
    // number is divisible by 47 or not
    return (n % 47 == 0);
}
      
// Driver Code 
public static void Main() 
{ 
    int N = 59173;
      
    if (isDivisible(N)) 
        Console.WriteLine("Yes");
    else
        Console.WriteLine("No");
} 
} 
  
// This code is contributed by mohit kumar 29.

输出：

Yes