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# Program to check if N is a Icositetragonal number

• Last Updated : 23 Jun, 2021

Given an integer N, the task is to check if it is a icositetragonal number or not.

icositetragonal number
s a class of figurate number. It has a 24-sided polygon called Icositetragon. The N-th Icositetragonal number count’s the number of dots and all others dots are surrounding with a common sharing corner and make a pattern
The first few icositetragonal numbers are 1, 24, 69, 136, 225, 336, …

Examples:

Input: N = 24
Output: Yes
Explanation:
Second icositetragonal number is 24.
Input: N = 30
Output: No

Approach:

1. The Kth term of the icositetragonal number is given as 2. As we have to check that the given number can be expressed as a icositetragonal number or not. This can be checked as follows –

=> => 1.
2. Finally, check the value of computed using this formulae is an integer, which means that N is a icositetragonal number.

Below is the implementation of the above approach:

## C++

 // C++ implementation to check that// a number is icositetragonal number or not #include  using namespace std; // Function to check that the// number is a icositetragonal numberbool isicositetragonal(int N){    float n        = (10 + sqrt(44 * N + 100))          / 22;     // Condition to check if the    // number is a icositetragonal number    return (n - (int)n) == 0;} // Driver Codeint main(){    int i = 24;     // Function call    if (isicositetragonal(i)) {        cout << "Yes";    }    else {        cout << "No";    }    return 0;}

## Java

 // Java implementation to check that// a number is icositetragonal number or notclass GFG{ // Function to check that the// number is a icositetragonal numberstatic boolean isicositetragonal(int N){    float n = (float)((10 + Math.sqrt(44 * N +                                      100)) / 22);     // Condition to check if the    // number is a icositetragonal number    return (n - (int)n) == 0;} // Driver Codepublic static void main(String[] args){    int i = 24;     // Function call    if (isicositetragonal(i))    {        System.out.print("Yes");    }    else    {        System.out.print("No");    }}} // This code is contributed by 29AjayKumar

## Python3

 # Python3 implementation to check that# a number is icositetragonal number# or notimport math # Function to check that the number# is a icositetragonal numberdef isicositetragonal(N):     n = (10 + math.sqrt(44 * N + 100)) / 22     # Condition to check if the number    # is a icositetragonal number    return (n - int(n)) == 0 # Driver Codei = 24 # Function callif (isicositetragonal(i)):    print("Yes")else:    print("No") # This code is contributed by divyamohan123

## C#

 // C# implementation to check that// a number is icositetragonal number or notusing System;class GFG{ // Function to check that the// number is a icositetragonal numberstatic bool isicositetragonal(int N){    float n = (float)((10 + Math.Sqrt(44 * N +                                      100)) / 22);     // Condition to check if the    // number is a icositetragonal number    return (n - (int)n) == 0;} // Driver Codepublic static void Main(){    int i = 24;     // Function call    if (isicositetragonal(i))    {        Console.Write("Yes");    }    else    {        Console.Write("No");    }}} // This code is contributed by Akanksha_Rai

## Javascript

 
Output:
Yes

Time Complexity: O(1)

Auxiliary Space: O(1)

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