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

• Difficulty Level : Medium
• Last Updated : 23 Jun, 2022

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++

 `// C++ implementation to check if the``// number is peculiar` `#include ``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;``}`

## Java

 `// 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`

## Python3

 `# 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`

## C#

 `// 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`

## Javascript

 ``

Output:

`No`

Time Complexity: O(log10n), time used to find the sum of digits of a number
Auxiliary Space: O(1), as no extra space is required

My Personal Notes arrow_drop_up