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:
- Initialize the first digit to ans.
- Iterate for all n-1 digits.
- Compute ans at every ith step by (ans * 10 + digit[i]) % 41 using associative property.
- 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> |
YES
Time Complexity: O(N)
Auxiliary Space: O(N)
Approach: Alternating Digit Sum approach
- Calculate the first two digits of the large number.
- 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.
- 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.
- 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 } } |
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.
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