Given an integer N, the task is to check if N is a Zuckerman Number.
Zuckerman Number is a number which is divisible by the product of its digits
Examples:
Input: N = 115
Output: Yes
Explanation:
1*1*5 = 5 and 115 % 5 = 0
Input: N = 28
Output: No
Approach: The idea is to find the product of digits of N and check if N is divisible by its product of digits or not. If yes then the number N is a Zuckerman Number.
Below is the implementation of the above approach:
C++
// C++ implementation to check if N// is a Zuckerman number#include <bits/stdc++.h>using namespace std;// Function to get product of digitsint getProduct(int n){ int product = 1; while (n != 0) { product = product * (n % 10); n = n / 10; } return product;}// Function to check if N is an// Zuckerman numberbool isZuckerman(int n){ return n % getProduct(n) == 0;}// Driver codeint main(){ int n = 115; if (isZuckerman(n)) cout << "Yes"; else cout << "No"; return 0;} |
Java
// Java implementation to check if N// is a Zuckerman numberclass GFG{ // Function to get product of digitsstatic int getProduct(int n){ int product = 1; while (n != 0) { product = product * (n % 10); n = n / 10; } return product;}// Function to check if N is an// Zuckerman numberstatic boolean isZuckerman(int n){ return n % getProduct(n) == 0;}// Driver code public static void main(String[] args) { int n = 115; if (isZuckerman(n)) { System.out.println("Yes"); } else { System.out.println("No"); }} } // This code is contributed by shubham |
Python 3
# Python3 implementation to check if N# is a Zuckerman number# Function to get product of digitsdef getProduct(n): product = 1 while (n > 0): product = product * (n % 10) n = n // 10 return product# Function to check if N is an# Zuckerman numberdef isZuckerman(n): return n % getProduct(n) == 0# Driver codeN = 115if (isZuckerman(N)): print("Yes")else: print("No") # This code is contributed by Vishal Maurya |
C#
// C# implementation to check if N// is a Zuckerman numberusing System;class GFG{ // Function to get product of digitsstatic int getProduct(int n){ int product = 1; while (n != 0) { product = product * (n % 10); n = n / 10; } return product;} // Function to check if N is an// Zuckerman numberstatic bool isZuckerman(int n){ return n % getProduct(n) == 0;} // Driver code public static void Main(String[] args) { int n = 115; if (isZuckerman(n)) { Console.WriteLine("Yes"); } else { Console.WriteLine("No"); }} } // This code is contributed by Rohit_ranjan |
Javascript
<script>// Javascript implementation to check if N// is a Zuckerman number // Function to get product of digits function getProduct( n) { let product = 1; while (n != 0) { product = product * (n % 10); n = parseInt(n / 10); } return product; } // Function to check if N is an // Zuckerman number function isZuckerman( n) { return n % getProduct(n) == 0; } // Driver code let n = 115; if (isZuckerman(n)) { document.write("Yes"); } else { document.write("No"); }// This code is contributed by todaysgaurav</script> |
Output
Yes
Time Complexity: O(log10n)
References: OEIS
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