Determine whether the given integer N is a Peculiar Number or not

Given an integer N, our task is the determine if the integer N is Peculiar Number. If it is then print “yes” otherwise output “no”.

The peculiar number is the number which is three times the sum of digits of the number.

Examples:

Input: N = 27
Output: Yes
Explanation:
Digit sum for 27 is 9 and 3 * 9 = 27 which is equal to N. Hence the output is yes.

Input: N = 36
Output: No
Explanation:
Digit sum for 36 is 9 and 3 * 9 = 27 which is not equal to N. Hence the output is no.



Approach:

To solve the problem mentioned above we have to first find the sum of the digits of a number N. Then check if the sum of digits of the number multiplied by 3 is actually the number N itself. If it is then print Yes otherwise the output is no.

Below is the implementation of the above approach:

C++

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// C++ implementation to check if the
// number is peculiar
  
#include <bits/stdc++.h>
using namespace std;
  
// Function to find sum of digits
// of a number
int sumDig(int n)
{
    int s = 0;
  
    while (n != 0) {
        s = s + (n % 10);
  
        n = n / 10;
    }
  
    return s;
}
  
// Function to check if the 
// number is peculiar
bool Pec(int n)
{
    // Store a duplicate of n
    int dup = n;
  
    int dig = sumDig(n);
  
    if (dig * 3 == dup)
        return true;
  
    else
        return false;
}
  
// Driver code
int main()
{
    int n = 36;
  
    if (Pec(n) == true)
        cout << "Yes" << endl;
  
    else
        cout << "No" << endl;
  
    return 0;
}

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Java

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// Java implementation to check if the
// number is peculiar
import java.io.*;
  
class GFG{
  
// Function to find sum of digits
// of a number
static int sumDig(int n)
{
    int s = 0;
  
    while (n != 0)
    {
        s = s + (n % 10);
        n = n / 10;
    }
    return s;
}
  
// Function to check if number is peculiar
static boolean Pec(int n)
{
      
    // Store a duplicate of n
    int dup = n;
    int dig = sumDig(n);
  
    if (dig * 3 == dup)
        return true;
    else
        return false;
}
  
// Driver code
public static void main (String[] args) 
{
    int n = 36;
  
    if (Pec(n) == true)
        System.out.println("Yes");
    else
        System.out.println("No");
}
}
  
// This code is contributed by shubhamsingh10

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Python3

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# Python3 implementation to check if the
# number is peculiar
  
# Function to get sum of digits
# of a number 
def sumDig(n):
      
    s = 0
    while (n != 0): 
        s = s + int(n % 10
        n = int(n / 10
      
    return s
      
# Function to check if the  
# number is peculiar     
def Pec(n):
      
    dup = n
    dig = sumDig(n)
  
    if(dig * 3 == dup):
        return "Yes"
    else :
        return "No"
          
# Driver code 
n = 36
  
if Pec(n) == True:
    print("Yes")
else:
    print("No")
  
# This code is contributed by grand_master

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C#

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// C# implementation to check if the
// number is peculiar
using System;
  
class GFG{
  
// Function to find sum of digits
// of a number
static int sumDig(int n)
{
    int s = 0;
  
    while (n != 0)
    {
        s = s + (n % 10);
        n = n / 10;
    }
    return s;
}
  
// Function to check if the number is peculiar
static bool Pec(int n)
{
      
    // Store a duplicate of n
    int dup = n;
    int dig = sumDig(n);
  
    if (dig * 3 == dup)
        return true;
    else
        return false;
}
  
// Driver code
public static void Main() 
{
    int n = 36;
  
    if (Pec(n) == true)
        Console.Write("Yes");
    else
        Console.Write("No");
}
}
  
// This code is contributed by Akanksha_Rai

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Output:

No

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