D Numbers

D Number is a number N > 3 such that N divides k^(n-2)-k for all k with gcd(k, n) = 1, 1<k<n.

9, 15, 21, 33, 39, 51, 57, 63, 69, 87, 93…. 

Check if a number is a D-Number

Given a number N, the task is to check if N is an D Number or not. If N is an D Number then print “Yes” else print “No”.
Examples:

Input: N = 9 
Output: Yes 
Explanation:
9 is a D-number since it divides all the numbers 
2^7-2, 4^7-4, 5^7-5, 7^7-7 and 8^7-8, and 
2, 4, 5, 7, 8 are relatively prime to n.
Input: N = 16 
Output: No 

Approach: Since D Number is a number N > 3 such that N divides k^(n-2)-k for all k with gcd(k, n) = 1, 1<k<n. So in a loop of k from 2 to n-1, we will check if gcd of N and k is 1 or not.If it’s 1 then we will check if k^(n-2)-k is divisible by N or not.If not divisible we will return false.At last we will return true.
Below is the implementation of the above approach: 

C++

// C++ implementation
// for the above approach
#include<bits/stdc++.h>
using namespace std;

// Function to find the N-th
// icosikaipentagon number
int isDNum(int n)
{
    // number should be
    // greater than 3
    if (n < 4)
        return false;

    int numerator, hcf;

    // Check every k in range 2 to n-1
    for (int k = 2; k <= n; k++)
    {
        numerator = pow(k, n - 2) - k;
        hcf = __gcd(n, k);
    }

    // condition for D-Number
    if (hcf == 1 && (numerator % n) != 0)
        return false;

    return true;
}

// Driver Code
int main()
{
    int n = 15;
    int a = isDNum(n);
    if (a)
        cout << "Yes";
    else
        cout << "No";
}

// This code is contributed by Ritik Bansal

Java

// Java implementation for the
// above approach
import java.util.*;

class GFG{

// Function to find the N-th
// icosikaipentagon number
static boolean isDNum(int n)
{

    // Number should be
    // greater than 3
    if (n < 4)
        return false;

    int numerator = 0, hcf = 0;

    // Check every k in range 2 to n-1
    for(int k = 2; k <= n; k++)
    {
       numerator = (int)(Math.pow(k, n - 2) - k);
       hcf = __gcd(n, k);
    }

    // Condition for D-Number
    if (hcf == 1 && (numerator % n) != 0)
        return false;

    return true;
}

static int __gcd(int a, int b)
{
    return b == 0 ? a : __gcd(b, a % b);
}

// Driver Code
public static void main(String[] args)
{
    int n = 15;
    boolean a = isDNum(n);

    if (a)
        System.out.print("Yes");
    else
        System.out.print("No");
}
}

// This code is contributed by Amit Katiyar

Python3

# Python3 implementation
# for the above approach

import math

# Function to find the N-th
# icosikaipentagon number
def isDNum(n):
    # number should be
        # greater than 3
    if n < 4:
        return False

    # Check every k in range 2 to n-1
    for k in range(2, n):
        numerator = pow(k, n - 2) - k
        hcf = math.gcd(n, k)

        # condition for D-Number
        if(hcf ==1 and (numerator % n) != 0):
            return False
    return True

# Driver code
n = 15
if isDNum(n):
    print("Yes")
else:
    print("No")

C#

// C# implementation for the
// above approach
using System;
class GFG{

// Function to find the N-th
// icosikaipentagon number
static bool isDNum(int n)
{

    // Number should be
    // greater than 3
    if (n < 4)
        return false;

    int numerator = 0, hcf = 0;

    // Check every k in range 2 to n-1
    for(int k = 2; k <= n; k++)
    {
        numerator = (int)(Math.Pow(k, n - 2) - k);
        hcf = __gcd(n, k);
    }

    // Condition for D-Number
    if (hcf == 1 && (numerator % n) != 0)
        return false;

    return true;
}

static int __gcd(int a, int b)
{
    return b == 0 ? a : __gcd(b, a % b);
}

// Driver Code
public static void Main(String[] args)
{
    int n = 15;
    bool a = isDNum(n);

    if (a)
        Console.Write("Yes");
    else
        Console.Write("No");
}
}

// This code contributed by Princi Singh

Javascript

<script>
// Javascript implementation for the
// above approach


    // Function to find the N-th
    // icosikaipentagon number
    function isDNum( n) {

        // Number should be
        // greater than 3
        if (n < 4)
            return false;

        let numerator = 0, hcf = 0;

        // Check every k in range 2 to n-1
        for ( k = 2; k <= n; k++) {
            numerator = parseInt( (Math.pow(k, n - 2) - k));
            hcf = __gcd(n, k);
        }

        // Condition for D-Number
        if (hcf == 1 && (numerator % n) != 0)
            return false;

        return true;
    }

    function __gcd( a, b) {
        return b == 0 ? a : __gcd(b, a % b);
    }

    // Driver Code

        let n = 15;
        let a = isDNum(n);

        if (a)
            document.write("Yes");
        else
            document.write("No");


// This code contributed by Rajput-Ji

</script>

Output:

Yes

Time Complexity: O(1)
Reference: OEIS