Check if the given number is divisible by 71 or not
Given a number N, the task is to check whether the number is divisible by 71 or not.
Examples:
Input: N = 25411681
Output: yes
Explanation:
71 * 357911 = 25411681Input: N = 5041
Output: yes
Explanation:
71 * 71 = 5041
Approach: The divisibility test of 71 is:
- Extract the last digit.
- Subtract 7 * 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 71, or it is 0, then the original number is also divisible by 71.
For example:
If N = 5041 Step 1: N = 5041 Last digit = 1 Remaining number = 504 Subtracting 7 times last digit Resultant number = 504 - 7*1 = 497 Step 2: N = 497 Last digit = 7 Remaining number = 49 Subtracting 7 times last digit Resultant number = 49 - 7*7 = 0 Step 3: N = 0 Since N is a two-digit number, and 0 is divisible by 71 Therefore N = 5041 is also divisible by 71
Below is the implementation of the above approach:
C++
// C++ program to check whether a number // is divisible by 71 or not #include<bits/stdc++.h> #include<stdlib.h> using namespace std; // Function to check if the number is divisible by 71 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; // Subtracting seven times the last // digit to the remaining number n = abs (n - (d * 7)); } // Finally return if the two-digit // number is divisible by 71 or not return (n % 71 == 0) ; } // Driver Code int main() { int N = 5041; 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 71 or not import java.util.*; class GFG{ // Function to check if the number is divisible by 71 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 ; // Subtracting seven times the last // digit to the remaining number n = Math.abs(n - (d * 7 )); } // Finally return if the two-digit // number is divisible by 71 or not return (n % 71 == 0 ) ; } // Driver Code public static void main(String args[]){ int N = 5041 ; if (isDivisible(N)) System.out.println( "Yes" ) ; else System.out.println( "No" ); } } // This code is contributed by AbhiThakur |
Python 3
# Python program to check whether a number # is divisible by 71 or not # Function to check if the number is # divisible by 71 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 # Subtracting seven times the last # digit to the remaining number n = abs (n - (d * 7 )) # Finally return if the two-digit # number is divisible by 71 or not return (n % 71 = = 0 ) # Driver Code if __name__ = = "__main__" : N = 5041 if (isDivisible(N)) : print ( "Yes" ) else : print ( "No" ) |
C#
// C# program to check whether a number // is divisible by 71 or not using System; class GFG { // Function to check if the number is divisible by 71 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; // Subtracting fourteen times the last // digit to the remaining number n = Math.Abs(n - (d * 7)); } // Finally return if the two-digit // number is divisible by 71 or not return (n % 71 == 0); } // Driver Code public static void Main() { int N = 5041; if (isDivisible(N)) Console.WriteLine( "Yes" ); else Console.WriteLine( "No" ); } } // This code is contributed by mohit kumar 29. |
Javascript
<script> // Javascript program to check whether a number // is divisible by 71 or not // Function to check if the number is divisible by 71 or not function isDivisible(n) { let d; // While there are at least two digits while (Math.floor(n / 100) <=0) { // Extracting the last d = n % 10; // Truncating the number n = Math.floor(n/10); // Subtracting seven times the last // digit to the remaining number n = Math.abs(n - (d * 7)); } // Finally return if the two-digit // number is divisible by 71 or not return (n % 71 == 0) ; } // Driver Code let N = 5041; if (isDivisible(N)) document.write( "Yes" ) ; else document.write( "No" ); // This code is contributed by patel2127 </script> |
Output:
Yes
Time Complexity: O(log10N)
Auxiliary Space: O(1)
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