Given a number, the task is to quickly check if the number is divisible by 41 or not.

**Examples:**

Input :x = 123Output :YesInput :104413920565933Output :YES

A solution to the problem is to extract the last digit and subtract 4 times of last digit from remaining number and repeat this process until a two digit number is obtained. If the obtained two digit number is divisible by 41, then the given number is divisible by 41.**Approach:**

- Extract the last digit of the number/truncated number every time
- Subtract 4*(last digit of the previous number) from the truncated number
- Repeat the above three steps as long as necessary.

**Illustrations:**

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 :

Letbe any number such that

=100a+10b+c .

Now assume thatis divisible by 41. Then

0 (mod 41)

100a+10b+c0 (mod 41)

10(10a+b)+c0 (mod 41)

10+c

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 of1.

In other words, we have to find an integer such that n such that 10n1 mod 41.

It can be observed that the smallest n which satisfies this property is -4 as -401 mod 41.

Now we can multiply the original equation 10+c

0 (mod 41)

by -4 and simplify it:

-40-4c

0 (mod 41)

-4c

0 (mod 41)

We have found out that if0 (mod 41) then,

-4c

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.

Below is the implementation of above approach:

## C++

`// CPP program to validate above logic` `#include <bits/stdc++.h>` `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` |

## PHP

`<?php` `// PHP program to validate above logic` `// Function to check if the number` `// is divisible by 41 or not` `function` `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` `$n` `= 104413920565933;` `if` `(isDivisible(` `$n` `))` ` ` `echo` `"Yes"` `.` `"\n"` `;` `else` ` ` `echo` `"No"` `.` `"\n"` `;` `// This code is contributed` `// by ChitraNayal` `?>` |

**Output:**

Yes

Note that the above program may not make a lot of sense as could simply do n % 41 to check for divisibility. The idea of this program is to validate the concept. Also, this might be an efficient approach if input number is large and given as string.

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