Given a number N, the task is to check whether the number is divisible by 31 or not.
Examples:
Input: N = 1922
Output: Yes
Explanation:
31 * 62 = 1922
Input: N = 2722400
Output: No
Approach: The divisibility test of 31 is:
- Extract the last digit.
- Subtract 3 * 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 31, or it is 0, then the original number is also divisible by 31.
For example:
If N = 49507 Step 1: N = 49507 Last digit = 7 Remaining number = 4950 Subtracting 3 times last digit Resultant number = 4950 - 3*7 = 4929 Step 2: N = 4929 Last digit = 9 Remaining number = 492 Subtracting 3 times last digit Resultant number = 492 - 3*9 = 465 Step 3: N = 465 Last digit = 5 Remaining number = 46 Subtracting 3 times last digit Resultant number = 46 - 3*5 = 31 Step 4: N = 31 Since N is a two-digit number, and 31 is divisible by 31 Therefore N = 49507 is also divisible by 31
Below is the implementation of the above approach:
C++
// C++ program to check whether a number// is divisible by 31 or not#include<bits/stdc++.h>#include<stdlib.h>using namespace std;// Function to check if the number is divisible by 31 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 three times the last // digit to the remaining number n = abs(n-(d * 3)); } // Finally return if the two-digit // number is divisible by 31 or not return (n % 31 == 0) ;}// Driver Code int main() { int N = 1922; 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 31 or notimport java.util.*;class GFG{ // Function to check if the number is divisible by 31 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 three times the last // digit to the remaining number n = Math.abs(n - (d * 3)); } // Finally return if the two-digit // number is divisible by 31 or not return (n % 31 == 0) ;} // Driver Code public static void main(String[] args) { int N = 1922; 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 31 or not# Function to check if the number is # divisible by 31 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 three times the last # digit to the remaining number n = abs(n-(d * 3)) # Finally return if the two-digit # number is divisible by 31 or not return (n % 31 == 0) # Driver Code if __name__ == "__main__" : n = 1922 if (isDivisible(n)) : print("Yes") else : print("No") |
C#
// C# program to check whether a number// is divisible by 31 or notusing System;class GFG{ // Function to check if the number is divisible by 31 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 three times the last // digit to the remaining number n = Math.Abs(n - (d * 3)); } // Finally return if the two-digit // number is divisible by 31 or not return (n % 31 == 0) ;} // Driver Code public static void Main(String[] args) { int N = 1922; if (isDivisible(N)) Console.Write("Yes"); else Console.Write("No");} }// This code is contributed by Rajput-Ji |
Javascript
<script>// Javascript program to check whether a number// is divisible by 31 or not // Function to check if the number is divisible by 31 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 three times the last // digit to the remaining number n = Math.abs(n - (d * 3)); } // Finally return if the two-digit // number is divisible by 31 or not return (n % 31 == 0) ;}// Driver Code let N = 1922; if (isDivisible(N) != 0) document.write("Yes") ; else document.write("No");// This code is contributed by sanjoy_62.</script> |
Output:
Yes
Time Complexity: O(log10N)
Auxiliary Space: O(1)
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