Given a number N, the task is to check whether the number is divisible by 43 or not.
Examples:
Input: N = 2795
Output: yes
Explanation:
43 * 65 = 2795
Input: N = 11094
Output: yes
Explanation:
43 * 258 = 11094
Approach: The divisibility test of 43 is:
- Extract the last digit.
- Add 13 * last digit from the remaining number obtained after removing the last digit.
- Repeat the above steps until a two-digit number, or zero, is obtained.
- If the two-digit number is divisible by 43, or it is 0, then the original number is also divisible by 43.
For example:
If N = 11739 Step 1: N = 11739 Last digit = 9 Remaining number = 1173 Adding 13 times last digit Resultant number = 1173 + 13*9 = 1290 Step 2: N = 1290 Since 129 is divisible by 43 as 43 * 3 = 129 Therefore N = 11739 is also divisible by 43
Below is the implementation of the above approach:
C++
// C++ program to check whether a number// is divisible by 43 or not#include<bits/stdc++.h>#include<stdlib.h>using namespace std;// Function to check if the number is divisible by 43 or not bool isDivisible(int n) { int d; // While there are at least two digits while (n / 100) { // Extracting the last d = n % 10; // Truncating the number n /= 10; // adding thirteen times the last // digit to the remaining number n = abs(n+(d * 13)); } // Finally return if the two-digit // number is divisible by 43 or not return (n % 43 == 0) ;}// Driver Code int main() { int N = 2795; if (isDivisible(N)) cout<<"Yes"<<endl ; else cout<<"No"<<endl ; return 0; } // This code is contributed by ANKITKUMAR34 |
Java
// Java program to check whether a number// is divisible by 43 or notclass GFG{// Function to check if the number is divisible by 43 or not static boolean isDivisible(int n) { int d; // While there are at least two digits while ((n / 100) > 0) { // Extracting the last d = n % 10; // Truncating the number n /= 10; // adding thirteen times the last // digit to the remaining number n = Math.abs(n+(d * 13)); } // Finally return if the two-digit // number is divisible by 43 or not return (n % 43 == 0) ;} // Driver Code public static void main(String[] args) { int N = 2795; if (isDivisible(N)) System.out.print("Yes"); else System.out.print("No"); } } // This code is contributed by PrinciRaj1992 |
Python 3
# Python program to check whether a number# is divisible by 43 or not# Function to check if the number is # divisible by 43 or not def isDivisible(n) : # While there are at least two digits while n // 100 : # Extracting the last d = n % 10 # Truncating the number n //= 10 # Adding thirteen times the last # digit to the remaining number n = abs(n+(d * 13)) # Finally return if the two-digit # number is divisible by 43 or not return (n % 43 == 0) # Driver Code if __name__ == "__main__" : N = 2795 if (isDivisible(N)): print("Yes") else : print("No") |
C#
// C# program to check whether a number// is divisible by 43 or notusing System; class GFG { // Function to check if the number is divisible by 43 or not static bool isDivisible(int n) { int d; // While there are at least two digits while (n / 100 > 0) { // Extracting the last d = n % 10; // Truncating the number n /= 10; // adding thirteen times the last // digit to the remaining number n = Math.Abs(n + (d * 13)); } // Finally return if the two-digit // number is divisible by 43 or not return (n % 43 == 0) ;}// Driver Code public static void Main() { int N = 2795; if (isDivisible(N)) Console.WriteLine("Yes"); else Console.WriteLine("No"); } }// This code is contributed by AbhiThakur |
Javascript
<script>//javascript program to check whether a number// is divisible by 43 or not// Function to check if the number is divisible by 43 or notfunction isDivisible(n){ let d; // While there are at least two digits while(parseInt(n/100) > 0) { // Extracting the last d = n % 10; // Truncating the number n = parseInt(n / 10) // adding thirteen times the last // digit to the remaining number n = Math.abs(n+(d * 13)); } // Finally return if the two-digit // number is divisible by 43 or not return (n % 43 == 0) ;}// Driver Code let N = 2795; if (isDivisible(N)) document.write("Yes"); else document.write("No");// This code is contributed by vaibhavrabadiya117.</script> |
Output:
Yes
Time Complexity: O(log10N)
Auxiliary Space: O(1)
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