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Check whether a given number is an ugly number or not

  • Difficulty Level : Basic
  • Last Updated : 03 Mar, 2021

Given an integer N, the task is to find out whether the given number is an Ugly number or not . 

Ugly numbers are numbers whose only prime factors are 2, 3 or 5.

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Examples: 

Input: N = 14 
Output: No 
Explanation: 
14 is not ugly since it includes another prime factor 7.



Input: N = 6 
Output: Yes 
Explanation: 
6 is a ugly since it includes 2 and 3. 
 

Approach: The idea is to use recursion to solve this problem and check if a number is divisible by 2,  3 or 5. If yes then divide the number by that and recursively check that a number is an ugly number or not. If at any time, there is no such divisor, then return false, else true.

Below is the implementation of the above approach:
 

C++




// C++ implementation to check
// if a number is an ugly
// number or not
 
#include <stdio.h>
#include <stdlib.h>
 
// Function to check if a number
// is an ugly number or not
int isUgly(int n)
{
    // Base Cases
    if (n == 1)
        return 1;
    if (n <= 0)
        return 0;
 
    // Condition to check if the
    // number is divided by 2, 3, or 5
    if (n % 2 == 0) {
        return (isUgly(n / 2));
    }
    if (n % 3 == 0) {
        return (isUgly(n / 3));
    }
    if (n % 5 == 0) {
        return (isUgly(n / 5));
    }
 
    // Otherwise return false
    return 0;
}
// Driver Code
int main()
{
    int no = isUgly(14);
    if (no == 1)
        printf("Yes");
    else
        printf("No");
    return 0;
}

Java




// Java implementation to
// check if a number is ugly number
class GFG {
 
    // Function to check the ugly
    // number
    static int isUgly(int n)
    {
        // Base Cases
        if (n == 1)
            return 1;
        if (n <= 0)
            return 0;
 
        // Condition to check if
        // a number is divide by
        // 2, 3, or 5
        if (n % 2 == 0) {
            return (isUgly(n / 2));
        }
        if (n % 3 == 0) {
            return (isUgly(n / 3));
        }
        if (n % 5 == 0) {
            return (isUgly(n / 5));
        }
        return 0;
    }
 
    // Driver Code
    public static void main(String args[])
    {
        int no = isUgly(14);
        if (no == 1)
            System.out.println("Yes");
        else
            System.out.println("No");
    }
}

Python3




# Python3 implementation to check
# if a number is an ugly number
# or not
 
# Function to check if a number
# is an ugly number or not
def isUgly(n):
 
    # Base Cases
    if (n == 1):
        return 1
    if (n <= 0):
        return 0
 
    # Condition to check if the
    # number is divided by 2, 3, or 5
    if (n % 2 == 0):
        return (isUgly(n // 2))
         
    if (n % 3 == 0):
        return (isUgly(n // 3))
     
    if (n % 5 == 0):
        return (isUgly(n // 5))
 
    # Otherwise return false
    return 0
 
# Driver Code
if __name__ == "__main__":
 
    no = isUgly(14)
     
    if (no == 1):
        print("Yes")
    else:
        print("No")
 
# This code is contributed by chitranayal

C#




// C# implementation to check
// if a number is ugly number
using System;
 
class GFG{
 
// Function to check the ugly
// number
static int isUgly(int n)
{
     
    // Base Cases
    if (n == 1)
        return 1;
    if (n <= 0)
        return 0;
 
    // Condition to check if
    // a number is divide by
    // 2, 3, or 5
    if (n % 2 == 0)
    {
        return (isUgly(n / 2));
    }
    if (n % 3 == 0)
    {
        return (isUgly(n / 3));
    }
    if (n % 5 == 0)
    {
        return (isUgly(n / 5));
    }
    return 0;
}
 
// Driver Code
public static void Main(String []args)
{
    int no = isUgly(14);
     
    if (no == 1)
        Console.WriteLine("Yes");
    else
        Console.WriteLine("No");
}
}
 
// This code is contributed by amal kumar choubey

Javascript




<script>
 
// Javascript implementation to check
// if a number is an ugly
// number or not
 
// Function to check if a number
// is an ugly number or not
function isUgly(n)
{
    // Base Cases
    if (n == 1)
        return 1;
    if (n <= 0)
        return 0;
 
    // Condition to check if the
    // number is divided by 2, 3, or 5
    if (n % 2 == 0) {
        return (isUgly(n / 2));
    }
    if (n % 3 == 0) {
        return (isUgly(n / 3));
    }
    if (n % 5 == 0) {
        return (isUgly(n / 5));
    }
 
    // Otherwise return false
    return 0;
}
// Driver Code
  
    let no = isUgly(14);
    if (no == 1)
        document.write("Yes");
    else
        document.write("No");
     
 
// This code is contributed by Mayank Tyagi
 
</script>
Output: 
No

 




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