检查是否有大数可被17整除

给定一个数字，任务是快速检查该数字是否可被17整除。
例子：

Input : x = 34
Output : Yes

Input : x = 47
Output : No

解决该问题的方法是提取最后一位数字，并从剩余数字中减去最后一位数字的5倍，然后重复此过程，直到获得两位数字为止。如果获得的两位数可以被17整除，则给定的数字可以被17整除。

方法：

• 每次提取数字/截断数字的最后一位
• 从截断的数字中减去5 *(前一个数字的最后一位)
• 视需要重复上述三个步骤。

插图：

3978-->397-5*8=357-->35-5*7=0. 
So 3978 is divisible by 17.

Mathematical Proof :
Let $\overline{a b c}$ be any number such that $\overline{a b c}$=100a+10b+c .
Now assume that $\overline{a b c}$ is divisible by 17. Then
$\overline{a b c}\equiv$ 0 (mod 17)
100a+10b+c$\equiv$ 0 (mod 17)
10(10a+b)+c$\equiv$ 0 (mod 17)
10$\overline{a b}$+c$\equiv$ 0 (mod 17)

Now that we have separated the last digit from the number, we have to find a way to use it.
Make the coefficient of $\overline{a b}$ 1.
In other words, we have to find an integer such that n such that 10n$\equiv$1 mod 17.
It can be observed that the smallest n which satisfies this property is -5 as -50$\equiv$1 mod 17.
Now we can multiply the original equation 10$\overline{a b}$+c$\equiv$ 0 (mod 17)
by -5 and simplify it:
-50$\overline{a b}$-5c$\equiv$ 0 (mod 17)
$\overline{a b}$-5c$\equiv$ 0 (mod 17)
We have found out that if $\overline{a b c}\equiv$ 0 (mod 17) then,
$\overline{a b}$-5c$\equiv$ 0 (mod 17).
In other words, to check if a 3-digit number is divisible by 17,
we can just remove the last digit, multiply it by 5,
and then subtract it from the rest of the two digits.

程序 ：

C++

// CPP Program to validate the above logic
#include 
  
using namespace std;
  
// Function to check if the
// number is divisible by 17 or not
bool isDivisible(long long int n)
{
  
    while (n / 100) 
    {
        // Extracting the last digit
        int d = n % 10;
  
        // Truncating the number
        n /= 10;
  
        // Subtracting the five times the
        // last digit from the remaining number
        n -= d * 5;
    }
  
    // Return n is divisible by 17
    return (n % 17 == 0);
}
  
// Driver code
int main()
{
    long long int n = 19877658;
    if (isDivisible(n))
        cout << "Yes" << endl;
    else
        cout << "No" << endl;
    return 0;
}

Java

// Java Program to validate the above logic
  
import java.io.*;
  
class GFG {
  
  
// Function to check if the
// number is divisible by 17 or not
 static boolean isDivisible(long n)
{
  
    while (n / 100>0) 
    {
        // Extracting the last digit
        long d = n % 10;
  
        // Truncating the number
        n /= 10;
  
        // Subtracting the five times the
        // last digit from the remaining number
        n -= d * 5;
    }
  
    // Return n is divisible by 17
    return (n % 17 == 0);
}
  
// Driver code
  
    public static void main (String[] args) {
    long n = 19877658;
    if (isDivisible(n))
        System.out.println( "Yes");
    else
        System.out.println( "No");
    }
}
// This code is contributed by inder_verma.

Python 3

# Python 3 Program to validate
# the above logic 
  
# Function to check if the 
# number is divisible by 17 or not 
def isDivisible(n) :
  
    while(n // 100) :
  
        # Extracting the last digit 
        d = n % 10
  
        # Truncating the number 
        n //= 10
  
        # Subtracting the five times  
        # the last digit from the 
        # remaining number
        n -= d * 5
  
    # Return n is divisible by 17 
    return (n % 17 == 0)
  
# Driver Code
if __name__ == "__main__" :
  
    n = 19877658
      
    if isDivisible(n) :
        print("Yes")
    else :
        print("No")
  
# This code is contributed
# by ANKITRAI1

C#

// C# Program to validate the above logic
   
using System;
   
class GFG {
   
   
// Function to check if the
// number is divisible by 17 or not
 static bool isDivisible(long n)
{
   
    while (n / 100>0) 
    {
        // Extracting the last digit
        long d = n % 10;
   
        // Truncating the number
        n /= 10;
   
        // Subtracting the five times the
        // last digit from the remaining number
        n -= d * 5;
    }
   
    // Return n is divisible by 17
    return (n % 17 == 0);
}
   
// Driver code
   
    public static void Main () {
    long n = 19877658;
    if (isDivisible(n))
        Console.Write( "Yes");
    else
        Console.Write( "No");
    }
}

输出：

Yes

请注意，上面的程序可能没有多大意义，因为可以简单地执行n％23来检查可除性。该程序的目的是验证该概念。另外，如果输入数字很大并指定为字符串，那么这可能是一种有效的方法。