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 <bits/stdc++.h>using namespace std;// Function to find sum of digits// of a numberint 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 peculiarbool 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 codeint 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 peculiarimport java.io.*;class GFG{// Function to find sum of digits// of a numberstatic 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 peculiarstatic 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 codepublic 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 = 36if Pec(n) == True: print("Yes")else: print("No")# This code is contributed by grand_master |
C#
// C# implementation to check if the// number is peculiarusing System;class GFG{// Function to find sum of digits// of a numberstatic 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 peculiarstatic 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 codepublic 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
<script>// Javascript implementation to check if the// number is peculiar// Function to find sum of digits// of a numberfunction sumDig(n){ var s = 0; while (n != 0) { s = s + (n % 10); n = parseInt(n / 10); } return s;}// Function to check if the // number is peculiarfunction Pec(n){ // Store a duplicate of n var dup = n; var dig = sumDig(n); if (dig * 3 == dup) return true; else return false;}// Driver codevar n = 36;if (Pec(n) == true) document.write( "Yes");else document.write( "No" );// This code is contributed by noob2000.</script> |
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
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