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 --> 83 is divisible by 41 and hence, 104413920565933 is divisible by 41

Mathematical Proof :

Let be any number such that =100a+10b+c .

Now assume that is divisible by 41. Then

0 (mod 41)

100a+10b+c 0 (mod 41)

10(10a+b)+c 0 (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 of 1.

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 if 0 (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; ` `} ` |

*chevron_right*

*filter_none*

## 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 ` |

*chevron_right*

*filter_none*

## Python 3

`# Python 3 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 ` |

*chevron_right*

*filter_none*

## 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 ` |

*chevron_right*

*filter_none*

## 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 ` `?> ` |

*chevron_right*

*filter_none*

**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.

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the **DSA Self Paced Course** at a student-friendly price and become industry ready.

## Recommended Posts:

- Find permutation of n which is divisible by 3 but not divisible by 6
- Check if a large number is divisible by 11 or not
- Check if a large number is divisible by 3 or not
- Check if a large number is divisible by 4 or not
- Check if a large number is divisible by 8 or not
- Check if a large number is divisible by 6 or not
- Check if a large number is divisible by 9 or not
- Check if a large number is divisible by 5 or not
- Check a large number is divisible by 16 or not
- Check if a large number is divisible by 25 or not
- Check if LCM of array elements is divisible by a prime number or not
- Check if a large number is divisible by 13 or not
- Check if the large number formed is divisible by 41 or not
- Check if a number is divisible by 23 or not
- Check if any large number is divisible by 19 or not
- Check if any large number is divisible by 17 or not
- Check if a large number is divisible by 2, 3 and 5 or not
- Check if a large number is divisible by 75 or not
- Check if the number formed by the last digits of N numbers is divisible by 10 or not
- Check if the number is divisible 43 or not

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.