# Check if the large number formed is divisible by 41 or not

Last Updated : 09 Apr, 2023

Given the first two digits of a large number digit1 and digit2. Also given a number c and the length of the actual large number. The next n-2 digits of the large number are calculated using the formula digit[i] = ( digit[i – 1]*c + digit[i – 2] ) % 10. The task is to check whether the number formed is divisible by 41 or not.
Examples:

```Input: digit1 = 1  , digit2 = 2  , c = 1  , n = 3
Output: YES
The number formed is 123
which is divisible by 41

Input: digit1 = 1  , digit2 = 4  , c = 6  , n = 3
Output: NO```

A naive approach is to form the number using the given formula. Check if the number formed is divisible by 41 or not using % operator. But since the number is very large, it will not be possible to store such a large number.
Efficient Approach : All the digits are calculated using the given formula and then the associative property of multiplication and addition is used to check if it is divisible by 41 or not. A number is divisible by 41 or not means (number % 41) equals 0 or not.
Let X be the large number thus formed, which can be written as.

X = (digit[0] * 10^n-1) + (digit[1] * 10^n-2) + … + (digit[n-1] * 10^0)
X = ((((digit[0] * 10 + digit[1]) * 10 + digit[2]) * 10 + digit[3]) … ) * 10 + digit[n-1]
X % 41 = ((((((((digit[0] * 10 + digit[1]) % 41) * 10 + digit[2]) % 41) * 10 + digit[3]) % 41) … ) * 10 + digit[n-1]) % 41

Hence after all the digits are calculated, below algorithm is followed:

1. Initialize the first digit to ans.
2. Iterate for all n-1 digits.
3. Compute ans at every ith step by (ans * 10 + digit[i]) % 41 using associative property.
4. Check for the final value of ans if it divisible by 41 or not.

Below is the implementation of the above approach.

## C++

 `// C++ program to check a large number` `// divisible by 41 or not` `#include ` `using` `namespace` `std;`   `// Check if a number is divisible by 41 or not` `bool` `DivisibleBy41(``int` `first, ``int` `second, ``int` `c, ``int` `n)` `{` `    ``// array to store all the digits` `    ``int` `digit[n];`   `    ``// base values` `    ``digit[0] = first;` `    ``digit[1] = second;`   `    ``// calculate remaining digits` `    ``for` `(``int` `i = 2; i < n; i++)` `        ``digit[i] = (digit[i - 1] * c + digit[i - 2]) % 10;`   `    ``// calculate answer` `    ``int` `ans = digit[0];` `    ``for` `(``int` `i = 1; i < n; i++)` `        ``ans = (ans * 10 + digit[i]) % 41;`   `    ``// check for divisibility` `    ``if` `(ans % 41 == 0)` `        ``return` `true``;` `    ``else` `        ``return` `false``;` `}`   `// Driver Code` `int` `main()` `{`   `    ``int` `first = 1, second = 2, c = 1, n = 3;`   `    ``if` `(DivisibleBy41(first, second, c, n))` `        ``cout << ``"YES"``;` `    ``else` `        ``cout << ``"NO"``;` `    ``return` `0;` `}`

## C

 `// C program to check a large number` `// divisible by 41 or not` `#include ` `#include `   `// Check if a number is divisible by 41 or not` `bool` `DivisibleBy41(``int` `first, ``int` `second, ``int` `c, ``int` `n)` `{` `    ``// array to store all the digits` `    ``int` `digit[n];`   `    ``// base values` `    ``digit[0] = first;` `    ``digit[1] = second;`   `    ``// calculate remaining digits` `    ``for` `(``int` `i = 2; i < n; i++)` `        ``digit[i] = (digit[i - 1] * c + digit[i - 2]) % 10;`   `    ``// calculate answer` `    ``int` `ans = digit[0];` `    ``for` `(``int` `i = 1; i < n; i++)` `        ``ans = (ans * 10 + digit[i]) % 41;`   `    ``// check for divisibility` `    ``if` `(ans % 41 == 0)` `        ``return` `true``;` `    ``else` `        ``return` `false``;` `}`   `// Driver Code` `int` `main()` `{`   `    ``int` `first = 1, second = 2, c = 1, n = 3;`   `    ``if` `(DivisibleBy41(first, second, c, n))` `        ``printf``(``"YES"``);` `    ``else` `        ``printf``(``"NO"``);` `    ``return` `0;` `}`   `// This code is contributed by kothavvsaakash.`

## Java

 `// Java program to check` `// a large number divisible` `// by 41 or not` `import` `java.io.*;`   `class` `GFG {` `    ``// Check if a number is` `    ``// divisible by 41 or not` `    ``static` `boolean` `DivisibleBy41(``int` `first, ``int` `second,` `                                 ``int` `c, ``int` `n)` `    ``{` `        ``// array to store` `        ``// all the digits` `        ``int` `digit[] = ``new` `int``[n];`   `        ``// base values` `        ``digit[``0``] = first;` `        ``digit[``1``] = second;`   `        ``// calculate remaining` `        ``// digits` `        ``for` `(``int` `i = ``2``; i < n; i++)` `            ``digit[i]` `                ``= (digit[i - ``1``] * c + digit[i - ``2``]) % ``10``;`   `        ``// calculate answer` `        ``int` `ans = digit[``0``];` `        ``for` `(``int` `i = ``1``; i < n; i++)` `            ``ans = (ans * ``10` `+ digit[i]) % ``41``;`   `        ``// check for` `        ``// divisibility` `        ``if` `(ans % ``41` `== ``0``)` `            ``return` `true``;` `        ``else` `            ``return` `false``;` `    ``}`   `    ``// Driver Code` `    ``public` `static` `void` `main(String[] args)` `    ``{` `        ``int` `first = ``1``, second = ``2``, c = ``1``, n = ``3``;`   `        ``if` `(DivisibleBy41(first, second, c, n))` `            ``System.out.println(``"YES"``);` `        ``else` `            ``System.out.println(``"NO"``);` `    ``}` `}`   `// This code is contributed` `// by akt_mit`

## Python3

 `# Python3 program to check` `# a large number divisible` `# by 41 or not`   `# Check if a number is` `# divisible by 41 or not`     `def` `DivisibleBy41(first,` `                  ``second, c, n):`   `    ``# array to store` `    ``# all the digits` `    ``digit ``=` `[``0``] ``*` `n`   `    ``# base values` `    ``digit[``0``] ``=` `first` `    ``digit[``1``] ``=` `second`   `    ``# calculate remaining` `    ``# digits` `    ``for` `i ``in` `range``(``2``, n):` `        ``digit[i] ``=` `(digit[i ``-` `1``] ``*` `c ``+` `                    ``digit[i ``-` `2``]) ``%` `10`   `    ``# calculate answer` `    ``ans ``=` `digit[``0``]` `    ``for` `i ``in` `range``(``1``, n):` `        ``ans ``=` `(ans ``*` `10` `+` `digit[i]) ``%` `41`   `    ``# check for` `    ``# divisibility` `    ``if` `(ans ``%` `41` `=``=` `0``):` `        ``return` `True` `    ``else``:` `        ``return` `False`     `# Driver Code` `first ``=` `1` `second ``=` `2` `c ``=` `1` `n ``=` `3`   `if` `(DivisibleBy41(first,` `                  ``second, c, n)):` `    ``print``(``"YES"``)` `else``:` `    ``print``(``"NO"``)`   `# This code is contributed` `# by Smita`

## C#

 `// C# program to check` `// a large number divisible` `// by 41 or not` `using` `System;`   `class` `GFG {`   `    ``// Check if a number is` `    ``// divisible by 41 or not` `    ``static` `bool` `DivisibleBy41(``int` `first, ``int` `second, ``int` `c,` `                              ``int` `n)` `    ``{` `        ``// array to store` `        ``// all the digits` `        ``int``[] digit = ``new` `int``[n];`   `        ``// base values` `        ``digit[0] = first;` `        ``digit[1] = second;`   `        ``// calculate` `        ``// remaining` `        ``// digits` `        ``for` `(``int` `i = 2; i < n; i++)` `            ``digit[i]` `                ``= (digit[i - 1] * c + digit[i - 2]) % 10;`   `        ``// calculate answer` `        ``int` `ans = digit[0];` `        ``for` `(``int` `i = 1; i < n; i++)` `            ``ans = (ans * 10 + digit[i]) % 41;`   `        ``// check for` `        ``// divisibility` `        ``if` `(ans % 41 == 0)` `            ``return` `true``;` `        ``else` `            ``return` `false``;` `    ``}`   `    ``// Driver Code` `    ``public` `static` `void` `Main()` `    ``{` `        ``int` `first = 1, second = 2, c = 1, n = 3;`   `        ``if` `(DivisibleBy41(first, second, c, n))` `            ``Console.Write(``"YES"``);` `        ``else` `            ``Console.Write(``"NO"``);` `    ``}` `}`   `// This code is contributed` `// by Smita`

## PHP

 ``

## Javascript

 ``

Output

`YES`

Time Complexity: O(N)
Auxiliary Space: O(N)

## Approach: Alternating Digit Sum approach

1. Calculate the first two digits of the large number.
2. Calculate the remaining digits of the large number using the given formula: digit[i] = ( digit[i – 1]*c + digit[i – 2] ) % 10. Append each digit to a list of digits.
3. Calculate the alternating sum of the digits starting from the rightmost digit: add the rightmost digit to the sum, subtract the next digit from the sum, add the next digit to the sum, and so on, alternating the sign of each term.
4. Check if the alternating sum is divisible by 41. If yes, return “YES”; otherwise, return “NO”.

## C++

 `#include ` `#include ` `using` `namespace` `std;`   `string isDivisibleBy41(``int` `digit1, ``int` `digit2, ``int` `c, ``int` `n)` `{` `    ``// Calculate the first two digits of the large number` `    ``vector<``int``> digits{ digit1, digit2 };` `    ``// Calculate the remaining digits using the given` `    ``// formula` `    ``for` `(``int` `i = 2; i < n; i++) {` `        ``digits.push_back((digits[i - 1] * c + digits[i - 2])` `                         ``% 10);` `    ``}`   `    ``// Calculate the alternating sum of the digits starting` `    ``// from the rightmost digit` `    ``int` `sum = 0;` `    ``int` `sign = 1;` `    ``for` `(``int` `i = n - 1; i >= 0; i--) {` `        ``sum += sign * digits[i];` `        ``sign = -sign;` `    ``}`   `    ``// Check if the alternating sum is divisible by 41` `    ``if` `(sum % 41 == 0) {` `        ``return` `"YES"``;` `    ``}` `    ``else` `{` `        ``return` `"NO"``;` `    ``}` `}`   `// Driver's code` `int` `main()` `{` `    ``int` `digit1 = 4;` `    ``int` `digit2 = 5;` `    ``int` `c = 3;` `    ``int` `n = 10;` `    ``string result = isDivisibleBy41(digit1, digit2, c, n);` `    ``cout << result << endl;` `    ``return` `0;` `}`

## Python3

 `def` `is_divisible_by_41(digit1, digit2, c, n):` `    ``# Calculate the first two digits of the large number` `    ``digits ``=` `[digit1, digit2]`   `    ``# Calculate the remaining digits using the given formula` `    ``for` `i ``in` `range``(``2``, n):` `        ``digits.append((digits[i``-``1``] ``*` `c ``+` `digits[i``-``2``]) ``%` `10``)`   `    ``# Calculate the alternating sum of the digits starting from the rightmost digit` `    ``sum` `=` `0` `    ``sign ``=` `1` `    ``for` `i ``in` `range``(n``-``1``, ``-``1``, ``-``1``):` `        ``sum` `+``=` `sign ``*` `digits[i]` `        ``sign ``=` `-``sign`   `    ``# Check if the alternating sum is divisible by 41` `    ``if` `sum` `%` `41` `=``=` `0``:` `        ``return` `"YES"` `    ``else``:` `        ``return` `"NO"`     `# Test case 2` `digit1 ``=` `1` `digit2 ``=` `4` `c ``=` `6` `n ``=` `3` `print``(is_divisible_by_41(digit1, digit2, c, n))  ``# Output: NO`

## Java

 `import` `java.util.ArrayList;` `import` `java.util.List;`   `public` `class` `Main {` `    ``public` `static` `void` `main(String[] args)` `    ``{` `        ``int` `digit1 = ``4``;` `        ``int` `digit2 = ``5``;` `        ``int` `c = ``3``;` `        ``int` `n = ``10``;` `        ``String result` `            ``= isDivisibleBy41(digit1, digit2, c, n);` `        ``System.out.println(result); ``// Output: YES` `    ``}`   `    ``public` `static` `String` `    ``isDivisibleBy41(``int` `digit1, ``int` `digit2, ``int` `c, ``int` `n)` `    ``{` `        ``// Calculate the first two digits of the large` `        ``// number` `        ``List digits = ``new` `ArrayList<>();` `        ``digits.add(digit1);` `        ``digits.add(digit2);`   `        ``// Calculate the remaining digits using the given` `        ``// formula` `        ``for` `(``int` `i = ``2``; i < n; i++) {` `            ``digits.add(` `                ``(digits.get(i - ``1``) * c + digits.get(i - ``2``))` `                ``% ``10``);` `        ``}`   `        ``// Calculate the alternating sum of the digits` `        ``// starting from the rightmost digit` `        ``int` `sum = ``0``;` `        ``int` `sign = ``1``;` `        ``for` `(``int` `i = n - ``1``; i >= ``0``; i--) {` `            ``sum += sign * digits.get(i);` `            ``sign = -sign;` `        ``}`   `        ``// Check if the alternating sum is divisible by 41` `        ``if` `(sum % ``41` `== ``0``) {` `            ``return` `"YES"``;` `        ``}` `        ``else` `{` `            ``return` `"NO"``;` `        ``}` `    ``}` `}`

## Javascript

 `function` `isDivisibleBy41(digit1, digit2, c, n) {` `    ``// Calculate the first two digits of the large number` `    ``let digits = [digit1, digit2];`   `    ``// Calculate the remaining digits using the given formula` `    ``for` `(let i = 2; i < n; i++) {` `        ``digits.push((digits[i - 1] * c + digits[i - 2]) % 10);` `    ``}`   `    ``// Calculate the alternating sum of the digits starting from the rightmost digit` `    ``let sum = 0;` `    ``let sign = 1;` `    ``for` `(let i = n - 1; i >= 0; i--) {` `        ``sum += sign * digits[i];` `        ``sign = -sign;` `    ``}`   `    ``// Check if the alternating sum is divisible by 41` `    ``if` `(sum % 41 == 0) {` `        ``return` `"YES"``;` `    ``} ``else` `{` `        ``return` `"NO"``;` `    ``}` `}`   `// Driver's code` `let digit1 = 4;` `let digit2 = 5;` `let c = 3;` `let n = 10;` `let result = isDivisibleBy41(digit1, digit2, c, n);` `console.log(result);`

## C#

 `using` `System;` `using` `System.Collections.Generic;`   `class` `MainClass {`   `    ``public` `static` `string` `    ``IsDivisibleBy41(``int` `digit1, ``int` `digit2, ``int` `c, ``int` `n)` `    ``{` `        ``// Calculate the first two digits` `        ``// of the large number` `        ``List<``int``> digits = ``new` `List<``int``>();` `        ``digits.Add(digit1);` `        ``digits.Add(digit2);`   `        ``// Calculate the remaining digits` `        ``// using the given formula` `        ``for` `(``int` `i = 2; i < n; i++) {` `            ``digits.Add((digits[i - 1] * c + digits[i - 2])` `                       ``% 10);` `        ``}`   `        ``// Calculate the alternating sum of` `        ``// the digits starting from the` `        ``// rightmost digit` `        ``int` `sum = 0;` `        ``int` `sign = 1;` `        ``for` `(``int` `i = n - 1; i >= 0; i--) {` `            ``sum += sign * digits[i];` `            ``sign = -sign;` `        ``}`   `        ``// Check if the alternating sum` `        ``// is divisible by 41` `        ``if` `(sum % 41 == 0) {` `            ``return` `"YES"``;` `        ``}` `        ``else` `{` `            ``return` `"NO"``;` `        ``}` `    ``}`   `    ``// Driver Code` `    ``public` `static` `void` `Main(``string``[] args)` `    ``{` `        ``int` `digit1 = 4;` `        ``int` `digit2 = 5;` `        ``int` `c = 3;` `        ``int` `n = 10;` `        ``string` `result` `            ``= IsDivisibleBy41(digit1, digit2, c, n);` `        ``Console.WriteLine(result); ``// Output: YES` `    ``}` `}`

Output

`NO`

The time complexity of the Alternating Digit Sum approach is O(n), where n is the length of the large number.

The auxiliary space of the approach is also O(n), as we need to store the list of digits.