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 ith step by (ans * 10 + digit[i]) % 41 using associative property.
  4. Check for the final value of ans if it divisible by 41 r not.

Below is the implementation of the above approach.

C++

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// 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;
}

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Java














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// 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 



















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Python3














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# 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 



















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C#














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// 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 



















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PHP














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<?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. 


?> 



















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Output:


YES





Time Complexity: O(N)

Auxiliary Space: O(N)


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