Program to check if N is a Pentadecagonal Number
Given a number N, the task is to check if N is a Pentadecagon Number or not. If the number N is an Pentadecagon Number then print “Yes” else print “No”.
Pentadecagon Number is a 15-sided polygon..The first few Pentadecagon numbers are 1, 15, 42, 82, 135, 201, …
Examples:
Input: N = 15
Output: Yes
Explanation:
Second Pentadecagon number is 15.
Input: N = 30
Output: No
Approach:
- The Kth term of the Pentadecagon number is given as
- As we have to check that the given number can be expressed as a Pentadecagon Number or not. This can be checked as:
=>
=>
- If the value of K calculated using the above formula is an integer, then N is a Pentadecagon Number.
- Else N is not a Pentadecagon Number.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
bool isPentadecagon( int N)
{
float n
= (11 + sqrt (104 * N + 121))
/ 26;
return (n - ( int )n) == 0;
}
int main()
{
int N = 15;
if (isPentadecagon(N)) {
cout << "Yes" ;
}
else {
cout << "No" ;
}
return 0;
}
|
Java
import java.io.*;
import java.util.*;
class GFG {
public static boolean isPentadecagon( int N)
{
double n = ( 11 + Math.sqrt( 104 * N +
121 )) / 26 ;
return (n - ( int )n) == 0 ;
}
public static void main(String[] args)
{
int N = 15 ;
if (isPentadecagon(N))
{
System.out.println( "Yes" );
}
else
{
System.out.println( "No" );
}
}
}
|
Python3
from math import sqrt
def isPentadecagon(N):
n = ( 11 + sqrt( 104 * N + 121 )) / 26 ;
return (n - int (n) = = 0 );
if __name__ = = "__main__" :
N = 15 ;
if (isPentadecagon(N)):
print ( "Yes" );
else :
print ( "No" );
|
C#
using System;
class GFG {
public static bool isPentadecagon( int N)
{
double n = (11 + Math.Sqrt(104 * N +
121)) / 26;
return (n - ( int )n) == 0;
}
public static void Main(String[] args)
{
int N = 15;
if (isPentadecagon(N))
{
Console.WriteLine( "Yes" );
}
else
{
Console.WriteLine( "No" );
}
}
}
|
Javascript
<script>
function isPentadecagon(N)
{
var n = (11 + Math.sqrt(104 * N + 121))
/ 26;
return (n - parseInt(n)) == 0;
}
var N = 15;
if (isPentadecagon(N)) {
document.write( "Yes" );
}
else {
document.write( "No" );
}
</script>
|
Time Complexity: O(logN) since inbuilt sqrt function has been used
Auxiliary Space: O(1)
Last Updated :
19 Sep, 2022
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