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 r 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; } |
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 = 1second = 2c = 1n = 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. ?> |
YES
Time Complexity: O(N)
Auxiliary Space: O(N)
