检查数字是否可被41整除

给定一个数字，任务是快速检查数字是否可被41整除。

例子：

Input : x  = 123
Output : Yes

Input : 104413920565933
Output : YES

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4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被41整除，则给定数字可被41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4 *(前一个数字的最后一位)
• 视需要重复上述三个步骤。

插图：

Illustration 1:
30873-->3087-4*3=3075-->307-4*5=287-->28-4*7=0
As the remainder is zero, 30873 is divisible by 41

Illustration 2:
104413920565933 --> 10441392056593 - 4*3= 10441392056581
10441392056581 --> 1044139205658 - 4*1 = 1044139205654
1044139205654 --> 104413920565 - 4*4 = 104413920549
104413920549 --> 10441392054 - 4*9 = 10441392018
10441392018 --> 1044139201 - 4*8 = 1044139169
1044139169 --> 104413916 - 4*9 = 104413880
104413880 --> 10441388 - 4*0 = 10441380
10441388 --> 1044138 - 4*8 = 1044106
1044106 --> 104410 - 4*6 = 104386
104386 --> 10438 - 4*6 = 10414
10414 --> 1041 - 4*4 = 1025
1025 --> 102 - 4*5 =82
Now, 82%41 = 0 --> 82 is divisible by 41 and hence, 104413920565933 is divisible by 41

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 41. Then 
 

$\overline{a b c}\equiv$

0 (mod 41) 
100a+10b+c

$\equiv$

0 (mod 41) 
10(10a+b)+c

$\equiv$

0 (mod 41) 
10

$\overline{a b}$

+c

$\equiv$

0 (mod 41)
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 41. 
It can be observed that the smallest n which satisfies this property is -4 as -40

$\equiv$

1 mod 41. 
Now we can multiply the original equation 10

$\overline{a b}$

+c

$\equiv$

0 (mod 41) 
by -4 and simplify it: 
-40

$\overline{a b}$

-4c

$\equiv$

0 (mod 41) 
 

$\overline{a b}$

-4c

$\equiv$

0 (mod 41) 
We have found out that if 

$\overline{a b c}\equiv$

0 (mod 41) then, 
 

$\overline{a b}$

-4c

$\equiv$

0 (mod 41). 
In other words, to check if a 3-digit number is divisible by 41, 
we can just remove the last digit, multiply it by 4, 
and then subtract it from the rest of the two digits. 
 

下面是上述方法的实现：

C++

// CPP program to validate above logic
#include 
using namespace std;
 
// Function to check if the number
// is divisible by 41 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 four times
        // the last digit from the
        // remaining number
        n -= d * 4;
    }
 
    // return true if number is divisible by 41
    return (n % 41 == 0);
}
 
int main()
{
    long long int n = 104413920565933;
    if (isDivisible(n))
        cout << "Yes" << endl;
    else
        cout << "No" << endl;
    return 0;
}

Java

// Java program to validate above logic
 
class GFG {
 
// Function to check if the number
// is divisible by 41 or not
    static boolean isDivisible(long n) {
        while (n / 100 != 0) {
// Extracting the last digit
            int d = (int) (n % 10);
 
// Truncating the number
            n /= 10;
 
// Subtracting the four times
// the last digit from the
// remaining number
            n -= d * 4;
        }
 
// return true if number
// is divisible by 41
        return (n % 41 == 0);
    }
 
    public static void main(String[] args) {
        long n = 104413920565933L;
        if (isDivisible(n)) {
            System.out.println("Yes");
        } else {
            System.out.println("No");
        }
 
    }
}
// This code is contributed by RAJPUT-JI

Python3

# Python3 Program to validate above logic
 
# Function to check if the number
# is divisible by 41 or not
def isDivisible(n) :
 
    while n // 100 :
 
        # Extracting the last digit
        d = n % 10
 
        # Truncating the number
        n //= 10
 
        # Subtracting the four times
        # the last digit from the 
        # remaining number
        n -= d * 4
 
    # return true if number is divisible by 41
    return n % 41 == 0
     
 
     
# Driver Code
if __name__ == "__main__" :
 
    n = 104413920565933
     
    if isDivisible(n) :
        print("Yes")
 
    else :
        print("No")
         
# This code is contributed by ANKITRAI1

C#

// C# program to validate above logic
using System;
 
class GFG
{
// Function to check if the number
// is divisible by 41 or not
static bool isDivisible(long n)
{
    while (n / 100 != 0)
    {
        // Extracting the last digit
        int d = (int)(n % 10);
 
        // Truncating the number
        n /= 10;
 
        // Subtracting the four times
        // the last digit from the
        // remaining number
        n -= d * 4;
    }
 
    // return true if number
    // is divisible by 41
    return (n % 41 == 0);
}
 
// Driver Code
static public void Main ()
{
    long n = 104413920565933;
    if (isDivisible(n))
        Console.Write("Yes");
    else
        Console.Write("No");
}
}
 
// This code is contributed by Raj

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％41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。