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

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 i^th 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 <bits/stdc++.h>
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 <stdbool.h>
#include <stdio.h>

// 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

<?php
// PHP program to check a
// large number divisible
// by 41 or not

// Check if a number is
// divisible by 41 or not
function DivisibleBy41($first, $second, $c, $n)
{
    // array to store
    // all the digits
    $digit[$n] = range(1, $n);

    // base values
    $digit[0] = $first;
    $digit[1] = $second;

    // calculate remaining digits
    for ($i = 2; $i < $n; $i++)
        $digit[$i] = ($digit[$i - 1] * $c +
                      $digit[$i - 2]) % 10;

    // calculate answer
    $ans = $digit[0];
    for ($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
$first = 1;
$second = 2;
$c = 1;
$n = 3;

if (DivisibleBy41($first, $second, $c, $n))
    echo "YES";
else
    echo "NO";

// This code is contributed by Mahadev.
?>

Javascript

<script>

// Javascript program to check
// a large number divisible
// by 41 or not

// Check if a number is
// divisible by 41 or not
function DivisibleBy41(first, second,  c, n)
{
    // array to store
    // all the digits
    let digit = new Array(n).fill(0);

    // base values
    digit[0] = first;
    digit[1] = second;

    // calculate remaining
    // digits
    for (let i = 2; i < n; i++)
        digit[i] = (digit[i - 1] * c +
                    digit[i - 2]) % 10;

    // calculate answer
    let ans = digit[0];
    for (let i = 1; i < n; i++)
        ans = (ans * 10 +
               digit[i]) % 41;

    // check for
    // divisibility
    if (ans % 41 == 0)
        return true;
    else
        return false;
}

// driver program

    let first = 1, second = 2, c = 1, n = 3;

    if (DivisibleBy41(first, second, c, n))
        document.write("YES");
    else
        document.write("NO");

 // This code is contributed by susmitakundugoaldanga.
</script>

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 <iostream>
#include <vector>
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<Integer> 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.