Given a number N, the task is to check if for every value of i (0 <= i <= len), the first i digits of a number is divisible by (len – i + 1) or not, where len is the number of digits in N. If found to be true, then print “Yes”. Otherwise, print “No”.
Examples:
Input: N = 52248.
Output: Yes
Explanation:
- 5 is divisible by 5
- 52 is divisible by 4
- 522 is divisible by 3
- 5224 is divisible by 2
- 52248 is divisible by 1
Input: N = 59268
Output : No
Approach: The idea is to traverse all the prefixes of the given number and for each prefix, check if it is satisfies the condition or not.
Follow the steps below to solve the problem:
- Initialize a variable, say i as 1, to maintain the value of (len – i + 1).
- Iterate while n is greater than 0.
- If N is not divisible by i, then return false.
- Otherwise, divide N by 10 and increase the value of i by 1.
- Finally, return true.
Below is the implementation of the above approach.
C++
// C++ program for the above approach#include <iostream>using namespace std;// Function to check if all prefixes// of a number is divisible by// remaining count of digits or notbool prefixDivisble(int n){ int i = 1; while (n > 0) { // Traverse and check divisibility // for each updated number if (n % i != 0) return false; // Update the original number n = n / 10; i++; } return true;}// Driver Codeint main(){ // Given Input int n = 52248; // Function Call if (prefixDivisble(n)) cout << "Yes" << endl; else cout << "No" << endl; return 0;} |
Java
// Java program for the above approachimport java.io.*;class GFG{ // Function to check if all prefixes// of a number is divisible by// remaining count of digits or notpublic static boolean prefixDivisble(int n){ int i = 1; while (n > 0) { // Traverse and check divisibility // for each updated number if (n % i != 0) return false; // Update the original number n = n / 10; i++; } return true;}// Driver Codepublic static void main (String[] args) { // Given Input int n = 52248; // Function Call if (prefixDivisble(n)) System.out.println("Yes"); else System.out.println("No");}}// This code is contributed by lokeshpotta20. |
Python3
# Python3 program for the above approach# Function to check if all prefixes of # a number is divisible by remaining count # of digits or notdef prefixDivisble(n): i = 1 while n > 0: # Traverse and check divisibility # for each updated number if n % i != 0: return False # Update the original number n = n // 10 i += 1 return True# Driver Code# Given Inputn = 52248# Function Callif (prefixDivisble(n)): print("Yes")else: print("No")# This code is contributed by abhivick07 |
C#
// C# program for the above approachusing System;using System.Collections.Generic;class GFG{ // Function to check if all prefixes// of a number is divisible by// remaining count of digits or notstatic bool prefixDivisble(int n){ int i = 1; while (n > 0) { // Traverse and check divisibility // for each updated number if (n % i != 0) return false; // Update the original number n = n / 10; i++; } return true;}// Driver Codepublic static void Main(){ // Given Input int n = 52248; // Function Call if (prefixDivisble(n)) Console.Write("Yes"); else Console.Write("No");}}// This code is contributed by ipg2016107 |
Javascript
<script>// Javascript program for the above approach// Function to check if all prefixes// of a number is divisible by// remaining count of digits or notfunction prefixDivisble(n) { let i = 1; while (n > 0) { // Traverse and check divisibility // for each updated number if (n % i != 0) return false; // Update the original number n = parseInt(n / 10); i++; } return true;}// Driver Code// Given Inputlet n = 52248;// Function Callif (prefixDivisble(n) == true) document.write("Yes");else document.write("No");// This code is contributed by lokeshpotta20.</script> |
Output
Yes
Time Complexity : O(len) where len is the number of digits in N.
Auxiliary Space : O(1)
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